Chapter3 Nucleus 110422

“The Emergent Universe - how did we get from there to here?” Chapter 3: Protons and Neutrons – Nuclei

Chapter 3 – Nuclei...............................................................................................................53.1 Perspective...........................................................................................................8

In the last chapter….................................................................................................8In this chapter….......................................................................................................9In the next chapter…................................................................................................9

3.2 Introducing nuclides- the nuclear force..............................................................10some new terminology – nucleon, nuclide, nucleus, element.................................10

3.2.1 The nuclear force without pions..................................................................103.2.2 the nuclear potential energy graph.............................................................113.2.3 Mass loss and binding energy....................................................................12

mass-energy bank accounts...................................................................................14a shocking result?...................................................................................................14

3.2.4 Pions and the nuclear force - a technical section.......................................143.2.5 A technical note on data sources and calculations.....................................15

sources...................................................................................................................15Calculations............................................................................................................17the atomic mass unit (amu)....................................................................................17mass-energy accounting........................................................................................18

3.2.6 the simplest nuclide – the deuteron............................................................18the proton-neutron balance....................................................................................19why don’t diprotons and dineutrons exist? - a technical section.............................21

3.2.7 Conclusion..................................................................................................223.3 Building nucleon clusters – a simple nuclear model..........................................23

varying the proton:neutron ratio..............................................................................23varying the cluster size...........................................................................................25Summary................................................................................................................27

3.4 Real nuclei..........................................................................................................273.4.1 what does a nucleus look like?...................................................................27

the nucleus has a fuzzy boundary..........................................................................28some real nuclei.....................................................................................................28how big is a nucleon?............................................................................................30A football as heavy as Everest...............................................................................31

3.4.2 the nuclide plot............................................................................................313.4.3 the nuclear valley........................................................................................343.4.4 The nuclear binding energy and mass curves............................................37

carbon-12 – doing the accounting for mass and binding energy............................39energy “invested” in matter.....................................................................................39mass per nucleon and binding energy per nucleon – a technical section...............40

3.5 Cluster-12 - an individual cluster seeks stability.................................................413.5.1 the cluster-12 family in nuclide-space.........................................................413.5.2 the cluster-12 family in the nuclear valley...................................................44

3.6 An inner structure to the nuclear cluster.............................................................453.6.1 nucleon pairs - clusters-100 and -101.........................................................45

cluster-101..............................................................................................................45cluster-100..............................................................................................................46clusters 100 and 101 in the nuclear valley..............................................................47Patterns of stability.................................................................................................48

3.6.2 Magic numbers...........................................................................................49magic numbers 50 and 82......................................................................................50

3.6.3 Super-heavy clusters and the "Island of Stability".......................................513.7 Fusion and fission – changing cluster size..........................................................51

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3.7.1 fusion - the energetics of cluster growth.....................................................52small clusters are isolated by their mutual repulsion..............................................53

3.7.2 fission - the energetics of cluster splitting...................................................53the spectrum of nuclear fission reactions...............................................................53the pattern of decay of the known nuclides............................................................54the three decay modes of nuclide (77p,90n)..........................................................56competition between nucleon configurations..........................................................57

3.7.3 Spontaneous fission...................................................................................58fission is energetically viable for clusters >~100 nucleons.....................................58spontaneous fission of very heavy nuclei...............................................................59the nucleus as a liquid drop....................................................................................60why there are stable nuclides >100 nucleons.........................................................61

3.7.4 alpha-decay................................................................................................62why is alpha-decay so common?............................................................................62alpha-decay is viable for clusters >145 nucleons...................................................62why there are stable nuclides >145 nucleons.........................................................64alpha particles in a potential well............................................................................64inside the nuclear well............................................................................................66outside the nucleus.................................................................................................66tumbling and tunnelling...........................................................................................67a “cloud” of possibilities?........................................................................................69

3.7.5 Why there is an arc of stable nuclides........................................................69motive, means and opportunity..............................................................................69a surprising conclusion?.........................................................................................70

3.8 Nuclear reactions...............................................................................................71nuclear Lego...........................................................................................................71nuclear and chemical reactions..............................................................................72

3.9 Life in the nuclear valley....................................................................................723.9.1 a balance of conflicting factors creates the nuclear valley..........................723.9.2 The pathways towards stability...................................................................72

the limited options for nuclide decay.......................................................................72average mass/nucleon is absolute, but stability is relative.....................................73decay pathways......................................................................................................74nuclide “ancestors” and their “descendents”...........................................................75

3.9.3 a rain of nucleons.......................................................................................75draining the nuclear valley......................................................................................76

3.10 The emergent nuclide........................................................................................77nucleosynthesis – making nuclei............................................................................78

3.11 Nucleosynthesis 1 - The first quarter of an hour................................................783.11.1 A time-line – linking temperature, time and energy......................................793.11.2 Before the first threshold – temperature, T > 1015 K (energy, E > 100 GeV, time, t < 10-10 seconds)...............................................................................................803.11.3 Threshold for creating W/Z particles (mass ~80GeV), T ~ 1015 K (E ~ 100 GeV, t ~ 10-10 seconds)..............................................................................................803.11.4 Threshold for creating protons and neutrons (mass ~940 MeV), T ~ 1013 K (E ~ 1000 MeV, t ~ 10-6 seconds)...............................................................................813.11.5 Threshold for creating electrons (mass ~ 0.5 MeV), T ~ 6 x 109 K (E ~ 0.5 MeV, t ~ 1 second).....................................................................................................82

the end of neutron creation.....................................................................................83one tick of the clock................................................................................................83



3.11.6 Protons and neutrons start to combine, T ~ 109 K (E ~ 100 keV, t ~ 200 seconds) – the first nuclei..........................................................................................84

the deuteron...........................................................................................................84The first clusters of protons and neutrons..............................................................86the creation of helium-4..........................................................................................87

3.11.7 The end of nucleus formation – T ~ 3 x 108K (E ~ 30 keV, t ~ 13 minutes). 893.11.8 The threshold for ionisation - T ~ 3,000 K (E ~ 0.3 eV, t ~ 300,000 years) - the first neutral atoms.................................................................................................90

Space becomes transparent...................................................................................91The first atoms........................................................................................................92

3.11.9 review..........................................................................................................92From radiation-dominated to matter-dominated.....................................................92

3.12 Nucleosynthesis 2 - the next 14 billion years.....................................................943.12.1 The cosmic microwave background.............................................................94

photons are stretched by expanding space............................................................94ripples in the microwave background.....................................................................96

3.12.2 Collapsing gas clouds..................................................................................96gravity takes over the show....................................................................................96the first stages of collapse......................................................................................97pumping up a bike tyre...........................................................................................97

3.12.3 Stars and the elements..............................................................................100The basic nucleosynthesis reactions....................................................................100the creation of the elements.................................................................................101

3.12.4 Describing nuclear reactions......................................................................1023.12.5 Hydrogen-burning - the proton-proton chain..............................................104

a very slow nuclear reaction.................................................................................106the proton-proton chain........................................................................................107

3.12.6 The carbon (CNO) cycle............................................................................1123.12.7 Helium burning - the " triple alpha" process...............................................114

a trio of cosmic coincidences................................................................................1173.12.8 Filling in the gaps - capturing protons and neutrons..................................118

The p-process – making proton-rich nuclides.......................................................118Neutron capture processes..................................................................................119

3.12.9 The life cycles of stars...............................................................................122stars of different masses......................................................................................122Our sun - the life cycle of a star of 1 solar mass...................................................124matter under pressure..........................................................................................125

3.12.10 The alpha-process - burning all the way to iron....................................128carbon burning......................................................................................................128oxygen burning.....................................................................................................131silicon burning.......................................................................................................131an onion-like structure..........................................................................................134

3.12.11 running out of nuclear fuel......................................................................135the “death” of a star..............................................................................................136a technical note - why not nickel-62?...................................................................136

3.12.12 Supernova..............................................................................................138The Core's Tale....................................................................................................139The outer layers’ tale............................................................................................141Supernova SN1987A............................................................................................145

3.13 Review.............................................................................................................1483.13.1 Seeding inter-stellar space - the "cosmic stock-pot"..................................148



3.13.2 Reviewing nucleosynthesis processes......................................................1493.13.3 Flooding the nuclear valley........................................................................150

the array of nuclide species..................................................................................1513.13.4 Nuclides and the stellar eco-system..........................................................151

3.14 The emergent atom..........................................................................................1513.14.1 From nuclides to atoms.............................................................................151

parallels and contrasts..........................................................................................152Element abundances and ancestries....................................................................158

3.14 The emergent atom..........................................................................................1603.15 The next chapter..............................................................................................163References..................................................................................................................164



Chapter 3 – NucleiQuotes

… who knows what mysteries are hidden within the nucleus of an atom, which,

although a million million times smaller than the smallest living thing, is yet a universe

in itself?

Louis de Broglie, “Matter and Light – the New Physics”, W.W. Norton, 1939, p. 10

Neutron - walking into a bar: "A pint of beer please. How much will that be?"

Barman: "For you, no charge."

Atoms, instead of being placed in the universe fully formed, had come in kit form. …

The atoms forged long ago in the fireball of the big bang, in countless stars across the

length and breadth of the galaxy, became incorporated into human beings. They

became, in short, the atoms of curiosity. Now, why should the universe be constructed

in such a way that atoms acquire the ability to be curious about themselves?

Chown, Furnace, p. 216

In all the great cities of the world we have detached ourselves from night. If you are a

city-dweller who doesn't believe this, travel at least a hundred miles into the

countryside, mount the highest hill and stare at the sky. It is not the same sky at all. In a

city, the stars overhead glitter like lights on a distant roof-top, and the sky begins

beyond the horizon. On a clear night in the mountains, you become part of the sky. The

stars reach out and touch you, and suddenly you feel the embrace of a galaxy.

Krauss, Atom, p.96

… it is in the highest degree unlikely that this earth and sky is the only one to have

been created and that all those particles of matter outside are accomplishing nothing.

This follows from the fact that our world has been made by nature through the

spontaneous and casual collision and the multifarious, accidental, random and

purposeless congregation and coalescence of atoms whose suddenly formed

combinations could serve on each occasion as the starting-point of substantial fabrics –

earth and sea and sky and the races of living creatures.

Lucretius, “On the Nature of the Universe”, trans. Ronald Latham. Penguin, 1968….p. 91

Look around. Our familiar world is built from the debris of stars. The rocks beneath our

feet, the steel and glass in our skyscrapers, the air we breathe - all are made from


McNeil, 23/04/11,

The account here is not so much composed, as discovered. Writing this has been more like exploring an unknown landscape, criss-crossing the terrain to find and map all the major features, camping out in places, to explore them more fully, and viewing the terrain from the vantage point of a new idea or bit of understanding. This is not a full account of events; that is covered very well by other people, more qualified than me. This is an attempt to find a narrative that takes us “from there to here”. I use lots of diagrams; if I can draw a diagram of something then I feel I understand it. A good diagram can be more effective than a page or two of text in communicating to the reader, for many readers will be visual thinkers. This is the first full chapter to be completed. I’ve been through it again and again, and each time it gets a little bit clearer and better organised. There are omissions, errors, things not understood or explained well. But I now have a continuous narrative that connects quarks to atoms – from a deep level that’s very hard to understand, to a world we can relate to. I’ve taken this as far as I can for now. Version 1.0 is going on-line; it’s time it was “out there”. I hope for some constructive feedback, to correct and improve it - 22 April 2011.


atoms in very hot places: the interiors of stars and the outrushing shock waves of a

supernova. Blown away into space, these atoms later condensed into the sun and

planets of out solar system.

Laurence Marschall, p. 197.

It is a grand scheme, no doubt about it. The destructive power of supernovae is,

paradoxically, a major agent for creation and change in the universe. Supernovae

produce and distribute the elements, develop the solar system, and shape the

evolution of life on one of its planets. Supernovae are at the root of our existence.

Laurence Marschall, p.215.

The stuff of which we are made, was “cooked” once, in a star, and spit out.

Richard Feynman, p. 61.

How I'm rushing through this! How much each sentence in this brief story contains.

"The stars are made of the same atoms as the earth." I usually pick one small topic like

this to give a lecture on. Poets say science takes away from the beauty of the stars-

mere globs of gas atoms. Nothing is "mere." I too can see the stars on a desert night,

and feel them. But do I see less or more? The vastness of the heavens stretches my

imagination - stuck on this carousel my little eye can catch one-million-year-old light. A

vast pattern-of which I am a part - perhaps my stuff was belched from some forgotten

star, as one is belching there. Or see them with the greater eye of Palomar, rushing all

apart from some common starting point when they were perhaps all together. What is

the pattern, or the meaning, or the why? It does not do harm to the mystery to know a

little about it. For far more marvellous is the truth than any artists of the past imagined!

Why do the poets of the present not speak of it? What men are poets who can speak of

Jupiter if he were like a man, but if he is an immense spinning sphere of methane and

ammonia must be silent?

Richard Feynman, footnote to p. 59.

I celebrate myself, and sing myself,

And what I assume you shall assume,

For every atom belonging to me as good as belongs to you.

Walt Whitman, Song of Myself, Leaves of Grass (1881-82)



SUBTLE. Why, what have you observ'd, sir, in our art,

Seems so impossible?

SURLY. But your whole work, no more.

That you should hatch gold in a furnace, sir,

As they do eggs in Egypt!

SUBTLE. Sir, do you Believe that eggs are hatch'd so?

SURLY. If I should?

SUBTLE. Why, I think that the greater miracle.

No egg but differs from a chicken more

Than metals in themselves.

SURLY. That cannot be.

The egg's ordain'd by nature to that end,

And is a chicken in potentia.

SUBTLE. The same we say of lead and other metals,

Which would be gold, if they had time.

MAMMON. And that Our art doth further.

SUBTLE. Ay, for 'twere absurd To think that nature in the earth bred gold

Perfect in the instant: something went before.

There must be remote matter.

SUBTLE, the Alchemist, PERTINAX SURLY, a Gamester, SIR EPICURE MAMMON, a

Knight, in , “The Alchemist”, Ben Jonson, 1610.

..the universe is set up in such a way that the production of carbon, oxygen and

nitrogen… …is an inevitable consequence of the life cycles of stars, and it is inevitable

that planets like the Earth will form around stars like the sun and be laced with complex

organic molecules, originally from interstellar clouds, by the arrival of comets. We are

made of stardust because we are a natural consequence of the existence of stars…

John Gribbin, p.186



I cannot believe that a sense requires that we be ignorant and that wonder fades in the

face of knowledge. If that is so, then it is a poor sense of wonder.

Robert Gilmore, page x

nucleons represent a level in the organisation of matter having exceptional stability

unique in the universe...the proton and the neutron...[are]...a hundred thousand times

smaller than the smallest atom...Yet as small as that might be, there is a world hidden

inside each one.

Timothy Paul Smith, p.5

Quantum mechanics tells us that if a process is not strictly forbidden, then it must

occur. Pairs of every conceivable article and antiparticle are constantly being created

and destroyed at every location across the universe.

Kaufmann and Freedman, p.733

3.1 PerspectiveIn the last chapter…

… we saw the materialisation of energy and the emergence of discrete particle-waves,

which exist in three dimensions of space and one of time. We encountered the quantum

vacuum, matter and anti-matter, and the uncertainties inherent in a material world of

particle-waves.

We met the Standard Model of fermions and bosons. We saw how fermions always

remain separate, becoming the matter of the physical universe, while bosons freely

mingle and merge, and mediate the interactions, or “forces”, between the fermions. Of the

fermions, the leptons stay single, while quarks are bound by gluons into groups - either in

duos (mesons) or trios (baryons). Of all the combinations of quarks, only the lightest, the

the proton (uud), is stable. The ever so slightly heavier neutron (udd) is nearly stable, and

an isolated neutron has an average lifetime of about 15 minutes before it decays to a

proton.

We met (1) the weak interaction carried by the W / Z0 exchange particles, that changes

the flavour of quarks, and bridges the divide between quarks and leptons and matter and

anti-matter, and (2) the colour interaction, that binds quarks into clusters. We saw how the

colour interaction “leaks” out of protons and neutrons, to give the powerful but very short

range nuclear force, mediated by virtual pions.


McNeil, 04/23/11,

The half life is the time it takes for half a sample of items to decay. Decay is statistical, like flipping a coin, so the larger the number of events the closer they will be to the average. The more times you flip a coin, the closer to 50% will be the proportion of heads. If you start with 1000 neutrons there will, on average, be 500 left after 614 seconds, a little over 10 minutes. After another 614 seconds the number will have halved again to 250, and so on. The neutron's lifetime, the average time it will survive in isolation, is given by… lifetime=half life / 0.693 = 886 s. See the web-site of the Brookhaven National Laboratory, at… http://www.nndc.bnl.gov/chart/help/glossary.jsp#halflife

..., 24/04/11,

References: As far as possible, every statement is backed up with a reference to at least one established authority, and most things have been checked with two or more. I’ve included a lot of quotes because the writers have expressed things better than I could. I make occasional references to HyperPhysics, Rod Nave’s excellent and comprehensive web-site, covering just about every topic in physics. Any aspect of this chapter can be followed up there. Similarly, I’ve made a few specific references to articles in Wikipedia, but its coverage seems to be truly universal; there seems to be an entry for everything, written by someone, somewhere, with knowledge and interest. Wikipedia is an excellent starting point for finding things out, with good links to other academic and authoritative sources, both in print and on the internet. I use Wikipedia, not so much as a reference, rather as a source of further information for the reader. Wikipedia is a reliable resource. A fairly recent academic comparison found that Wikipedia (4 errors/article) was nearly as reliable as Britannica (3 errors/article) in its coverage of scientific topics. http://news.cnet.com/Study-Wikipedia-as-accurate-as-Britannica/2100-1038_3-5997332.html (accessed 9 April 2011).


We saw the emergence of the proton and neutron, the quark trios bound by the colour

interaction. We also saw how the extension of this, the nuclear force, can bind protons

and neutrons into clusters – we saw the first hint of the emergent nucleus.

In this chapter…

… we will see how the nuclear force binds protons and neutrons into hundreds of stable

nuclei. We will see the delicate balance of energies that sets a limit on the maximum

stable nucleus size, and decides the precise ratio of protons and neutrons for stability. We

will see why there is only a finite number of stable nuclei, and why the mid-sized nuclei

are most stable. We will see all the ways that a combination of protons and neutrons

reduces its size and adjusts its ratio to arrive at stability.

We will see our material universe of radiation (photons) and matter (protons, neutrons and

electrons) emerge from the cosmic fireball, and within a few minutes assemble the first

few nuclei. We will see how over the next few billion years successive generations of stars

go on to create all the known nuclei.

This chapter is wholly about quarks, and how the protons and neutrons they create,

themselves combine into big nuclear clusters. Finally, we will see how these nuclei

acquire electrons, to create the atoms , and we see the emergence of the ~100 or so

chemical elements of our atomic world.

In the next chapter…

…we'll see how, on their stable nuclear foundations, the electrons bind atoms into a rich

and subtle diversity of different chemical substances. We see solids and liquids emerge

as states of matter in addition to gases.



3.2 Introducing nuclides- the nuclear forceWe now leave behind the world of quarks and gluons, and now regard protons and

neutrons as unitary particles that attract each other by the nuclear force, mediated by the

exchange of virtual pions. We will now look at how protons and neutrons gather in

clusters, bound by this short range nuclear force.

some new terminology – nucleon, nuclide, nucleus, element

We need, at this point, to introduce some important terminology, and also explain a

necessary inconsistency. I will call a cluster of nucleons a “nucleus”, and may also give it

a chemical name, such as helium-4 or lead-208. Historically, the nucleus was identified as

the centre of every atom, defining its identity as a chemical element. But we have only just

encountered the nucleons’ ability to cluster; there are as yet no atoms, and the concept of

a chemical element is meaningless. The name “nucleus” implicitly denotes something that

is at the heart of a bigger structure, and we are looking at nuclear clusters in their own

terms – we don’t even know yet what sort of clusters they can form.

But “nucleus” is the accepted term, and other terms are derived from it: so, “nuclei” is the

plural, protons and neutrons are “nucleons”, equal citizens of the cluster community, and

a “nuclide” is a cluster of a specific number of protons and neutrons, for example, (6p,6n).

This is a member of the cluster-12 family of nuclides, all of which contain 12 nucleons;

other members are (5p,7n) and (7p,5n).

Our atomic world is made of a hundred or so chemical elements - carbon, hydrogen,

oxygen, iron and so on. What defines each element is the number of protons in the nuclei

of its atoms - for example helium has 2, carbon has 6, and oxygen has 8 - while the

number of neutrons can vary. So I will refer to some nuclear clusters by their chemical

element name, for example, helium-4 (2p,2n). This is the nucleus of helium (symbol He),

which comprises 2 protons and 2 neutrons. I will refer to the more common elements by

their chemical symbols, thus He for helium. This is rather cumbersome, we will see the

foundations of our atomic world appear, and also be able to follow what the nucleons are

doing.

3.2.1 The nuclear force without pions

Let's now consider a proton and a neutron, close enough for each to be within the other’s

pion "cloud", so they experience the attraction of the nuclear force – figure 3.1.


..., 23/04/11,

I’ll introduce isotopes at the end of this chapter.

..., 23/04/11,

A rough figure. We have practical uses for elements up to number 98 (Emsley), so I’ve gone for the round number.

..., 23/04/11,

In the sense that the nuclear force between two nucleons is similar, regardless of their identity – Smith, p.35.

..., 23/04/11,

The standard scientific terms were devised in the context of “going inwards”, finding smaller and more fundamental things inside bigger things. Thus we have found atoms to contain nuclei, these to contain nucleons, and these to contain quarks. The term “nucleus” implicitly denotes something that is at the heart of a bigger structure. This book is not a scientific description or a historical sequence, but an account of the emergence of the hierarchy of structures in the physical universe. Thus we have found that quarks are bound into protons and neutrons, and that these are bound into “clusters”. We are going to find out how these clusters behave in their own terms; none of them is yet the “nucleus” of any bigger structure. Throughout this, we must be aware that a nuclear cluster is not a static thing, like a bag of oranges, but a endlessly bustling community of nucleons, binding themselves together by their continual exchange of pions. So, I'll use "cluster" where the f'll use "clusters"rs" wxed centre of a bigger structure.ther by their continual exchange of pions. ��ormation or dynamics of grouping is involved, and I'll use "nucleus" as a generic name for a non-specific aggregate of nucleons, that is established and fixed. A reader might reasonably think that this is rather fussy. But a “cluster” is a simple emergent structure of protons and neutrons, free of connotations. The term “nucleus” becomes fully appropriate at the end of this chapter, when we let nuclides acquire electrons, and create the atoms of the chemical elements. There is an analogy between protons and neutrons in a nucleus, bound by quark-gluon interactions, and atoms in a molecule, bound by electromagnetic interactions. So “nuclei may just be the molecules of particle physics” – Williams, p.179. The set of nuclides with the same total of nucleons is an “isobar”, but I think the term “cluster family” will be more amenable to the general reader, without losing clarity or rigour. A nuclide exists in its own right, with its own internal relationships. It sustains a bigger structure when it acquires electrons, and becomes an atom, expressing a chemical character. In a similar, but much more elaborate way, a human nuclide couple become the centre of a bigger structure when they have children and sustain a family, with its new and richer interactions in the social world.

..., 04/23/11,

Going through this section yet again. It’s like trying to hold a handful of cooked spaghetti – so many loose ends keep on sticking out – 11 April 2011.


Figure 3.1: A proton and neutron, close enough to be within range of their pion “clouds”, and so experiencing the attractive nuclear force. This is a deuteron – the simplest nuclear cluster.In the space between the nucleons, we can perhaps see in our imagination "the pions

fluttering back and forth" creating an "invisible, evanescent web" between the proton and

neutron, thus binding them together. We have our first nuclear cluster, the simplest

nuclide, known as a deuteron.

Smith sums the nuclear force like this: “it is repulsive at about half a fermi [fm], it is

attractive at a fermi, and it essentially vanishes at about 3-4 fermis. That means that two

nucleons will be oblivious to each other if they are separated by more than 4 fermis, and

they cannot get closer to each other than half a fermi. The whole regime of nuclear

physics is essentially defined by this distance scale.”

So we will go with a simple and pragmatic description of the nuclear force as (1) mediated

by pions, and (2) being very strong, but (3) with a very short range of attraction of ~3-4 fm.

We will see that we can understand nucleon clusters very well just with ideas (2) and (3).

3.2.2 the nuclear potential energy graph

We're now in a position to read and understand the nuclear potential energy graph, where

we bring a neutron and a proton together (figure 3.2).


n0 p+

nucleons - a fermions that resists merging

clouds of pions - bosons that readily merge

the "evanescent web" of pions "fluttering back and forth" binding the proton and neutron together

..., 23/04/11,

I'm using the graph available at… http://universe-review.ca/F14-nucleus.htm (reference kindly supplied by Don Lincoln). This is for a proton and a neutron (a deuterium nucleus), where there is no electrical repulsion. Coordinates of a number of points on the graph were obtained using image editing software, and the curve then plotted using EXCEL. This graph is consistent with others: the nucleon-nucleon potential given by Bertulani, p.83, Turnbull, p.164, and the inter-nucleon force in Smith, p.36. The graph shows the net force, so there are dots, zero length arrows, at 0.8 and 2.5 fm. The arrows below the graph are an attempt to show the ranges of the forces, with the fade effects showing how their strength varies with distance. The nucleons packed within a nucluclear cluster behave as if they each have a “hard repulsive core at separations less than about 0.5 fm” - Williams, p.155. The nuclear potential of the deuteron given by Barrow and Tipler has a similar shape, but a minimum energy of ~-50 MeV – p.319.

..., 23/04/11,

Smith, p. 35.

..., 23/04/11,

These quotes from Timothy Smith, p.150, and from Frank Close, "The Particle Odyssey", p.73. See back to section 3.4.7. The situation is complicated: at very close range nucleons appear to exchange three pions or more, at intermediate distances they exchange two pions, at large distances the nuclear force is "transmitted" by single pion exchange - Bertulani, p.77 and 84.


Figure 3.2: The forces acting on two nucleons close to each other. The arrows on the graph give an idea of the relative sizes and directions of the net force. The shaded block arrows below the graph give some idea of the ranges and varying strength of the forces – darker shading means strongerThe graph shows the potential energy (PE) of the system of two nucleons as zero when

their centres are 2.5 fm apart (a), because they are outside the range of the nuclear force .

At closer distances (b), the nucleons attract each other, and the system's PE becomes

negative - we have to supply energy to pull them apart. The slope of the graph is steepest

- the attractive force is strongest - at a bit more than 1 fm separation (c). But now the

nucleons themselves are getting so close that we see the fermion repulsion force begin to

act. At about 0.9 fm separation (d) the attractive and repulsive forces are balanced - the

net force is zero, and the system's PE is a minimum. If we try to push the nucleons closer

than this (e), then the repulsion force very quickly dominates, and the PE steeply

increases.

3.2.3 Mass loss and binding energy

The nuclear PE curve shows how the proton-neutron pair bound into a deuteron have a

negative PE, sitting at the bottom of an energy “well”, since the system of two nucleons

has less energy when they are together, than when they are apart. We can visualise the


-100

-50

0

50

100

150

200

250

300

0.0 0.5 1.0 1.5 2.0 2.5

separation (fm)

Pot

entia

l Ene

rgy

(MeV

)

attraction

repulsion

(a) the attraction is negligible beyond

about 2 fm

(c) the attraction is greatest at a bit more than 1 fm

(d) the forces of repulsion and attraction are balanced

(e) the two nucleons

resist merging and repel each

other

(b) a weak attraction

..., 23/04/11,

There may be a problem concerning the size of a nucleon. The graph suggests that the nucleons in the deuteron have an effective radius of about 0.4-0.5 fm, so when they butt up against each other, their centres are about 0.9 fm apart. Ths doesn’t fit with the radius value of ~1 fm, given earlier – Bertulani, p.98. I can’t explain this at present – April 2011.

..., 23/04/11,

Technically, we say that the size and direction of the net force is given by the slope of the PE line. Where the slope is zero, at 2.5 and 0.9 fm, the net force is zero. Where the slope is positive, upwards and to the right, between 0.8 and 2.2 fm, the net force is attraction. Closer than about 0.9 fm, the slope is negative and repulsion is dominant.

..., 23/04/11,

Unlike the magnetic field, the nuclear force does fall to zero beyond a certain distance – Smith, fig. 3.1 and p.47.


separate nucleons “falling” together into this potential energy “well”, as they are attracted

by the nuclear force – figure 3.3.

Figure 3.3: The proton and neutron “fall” into energy well created by the attractive nuclear force.The total mass-energy of the system of two nucleons comprises the masses of the

particles and any energy bound up in their interaction. When the nucleons are separate,

there is no interaction, and all the system’s energy is in its mass. In combining, the

nucleons lose the binding energy, so the system’s energy is the total mass of the separate

nucleons minus the binding energy. In a sense, the binding energy is "paid for" by the

nucleons' mass-energy. The equivalence of mass and energy means that a system of

nucleons has less mass when bound in a nucleus than when separate; the deuteron

weighs less than the separate proton and neutron. The "'missing mass' is emitted in the

form of gamma radiation when the proton and neutron join to form the deuteron."

The energy that is needed to separate the bound nucleons is called, not surprisingly, the

nuclear binding energy. "The binding energy of a nucleus, which is conceptually the

energy needed to separate all the nucleons in the nucleus, is easily calculated if we

remember that it should be equal to the mass loss when the nucleus is formed ." Or, as a

word equation...


the binding energy is lost from the system

force is needed to separate the

bound nucleons

Separate nucleons…beyond the reach of the short-range nuclear forces…there is no binding energy…all mass-

energy of the system is wholly in the masses.

Nucleons "fall" into the binding energy "well", and are bound by the nuclear force … the total mass-

energy of the system is in the masses and the binding energy … the binding energy is released, leaving the mass of the nucleus less than the sum

of its separate parts.

np

bindingenergy

PE

of t

he s

yste

m

0

(a) (b)

np

binding

energy

a deuteron – the simplest nuclide

..., 23/04/11,

Bertulani, p.101.

..., 23/04/11,

Bertulani, p.36.

..., 23/04/11,

The same is true for a chemical reaction. In a combustion reaction, say, the mass of the products is less than the mass of the reactants. The difference is due to the heat energy released by the reaction. The thing here is that the energies are tiny, by comparison with nuclear events, and the mass loss is effectively negligible. A useful review of binding energy is at… http://en.wikipedia.org/wiki/Binding_energy (accessed 15 April 2011). All bound systems have lost mass, but nuclear particles are distinctive in that their mass loss is a significant fraction of the mass of the whole system - Bertulani, p.101. Binding energy is “a general property imposed by the theory of relativity to all bound systems” – Bertulani, p.37.

..., 23/04/11,

This will have been covered in chapter 0.


binding energy = mass-energy of separate nucleons - mass-energy of bound nuclear

cluster

mass-energy bank accounts

We can liken this to someone with two bank accounts - say, a current and a deposit

account. Money can be transferred between the accounts, but the total sum always

remains the same. In the same way, when nucleons are separate all their mass-energy is

in the "mass" account, and when they are bound by the strong nuclear force into a

nucleus, some of the "mass" account is transferred to the "binding energy" account. But

the total must always remain the same.

a shocking result?

This has been known for about a century, and the explanation in terms of the equivalence

of mass and energy is universally accepted and understood. Nonetheless, it may well be

deeply unsettling, even shocking, for it undermines our everyday experience of matter as

inviolable. How can the mass decrease? How can “stuff” disappear? It takes an effort to

look at our everyday world and see that “matter-stuff” has at some time in the past been

made from “energy-stuff”. Yet our continued existence depends on the radiation from the

sun, the result of vast amounts of matter transforming every second to light energy. We

have met an enormously important principle of nuclear interactions.

3.2.4 Pions and the nuclear force - a technical section

This is a technical bit, trying to get the nature of the nuclear force clear, that the reader

can skip without losing the thread.

The basic nucleon-nucleon (NN) interaction can be modelled in terms of the exchange of

pions. At large distances, more than about 2 fm, it is the exchange of single pions (OPE –

one pion exchange): “The force field between two protons, or two neutrons, can only be

produced by the exchange of neutral pions. Between a proton and a neutron the

exchange can be done by means of charged pions.” The medium range interaction

(between 1 and 2 fm, and in the strongly attractive part of the potential/distance graph in

figure xXx) “can be described by the exchange of two pions”. The short-range interaction

“is due to the exchange of three pions or more”, where the pions resonate and combine to

make a higher energy meson, that is responsible for a strong repulsion. Finally, at very

close range, the nucleons are assumed each to have some sort of “core”, that makes


..., 23/04/11,

Bertulani, p.78,91.

..., 23/04/11,

Bertulani, p. 86.

..., 23/04/11,

Bertulani, p.81. However, Bertulani states that there are difficulties with this approach. For example, it can call for a pair of pions that comprise a meson that does not exist as a free particle, and so this “departs from the idea that real, physical particles mediate the NN interaction” –Bertulani, p.95. Williams writes of the exchange of heavier exchange particles – p.179.

..., 23/04/11,

Bertulani, p.81.

..., 23/04/11,

Bertulani, p.77, 80, 91, and also Williams, p.179.

..., 23/04/11,

Bertulani discusses this in some detail – ch.3. Smith describes pion exchange processes, and other quark exchanges, as models of the nuclear force – ch.3.

..., 23/04/11,

We’ll look at the physics of this in a later section in this chapter.

..., 23/04/11,

I've certainly found this so, and I’ve had to revise my “everyday” perception of the fixed mass of things.

..., 23/04/11,

Ignoring, of course, interest payments – let’s keep this simple!


them repel very strongly, so the NN potential rises towards infinity for distances less than

about 0.4 fm.

Bertulani reminds us that the nuclear force binding nucleons is a residual of the colour

force operating inside every nucleon - “we can demystify the [single pion interaction] in the

sense that the exchange of real particles (pions) is, in fact, not its essential element … on

a deeper level it is an effect of the color force between color-polarized composite

particles”. Though both nucleons are colour-neutral, each is a composite of three colour-

charged quarks, which can individually interact when they get close enough. So when

nucleons get very close, “they start to touch and overlap”, and can exchange quarks and

gluons. At small distances the nucleon-nucleon interaction becomes very complicated.

The interaction between two nucleons is analogous to the van der Waals force between

two molecules that are electrically neutral overall, but which contain a distribution of

electrical charge within them. The force between two molecules can be described in terms

of the exchange of photons of radiation. “The pions take the place of the photons in the

case of nuclear forces…In this way, the nuclei are bound by a type of van der Waals

force.”

Pions are described as the exchange particles that mediate, or “carry”, the nuclear force.

Nucleon interactions can effectively be described in terms of pion exchange, though

“There are small details about the nuclear force that are not completely accounted for by

the pion. However, with the addition of other, less abundant particles … which have also

been observed, a complete description of the nuclear force does exist.”

3.2.5 A technical note on data sources and calculations

sources

Nuclear data has been obtained from the Atomic Mass Data Centre (AMDC), at…

http://www.nndc.bnl.gov/amdc/

From here you can get to the 2003 Atomic Mass Evaluation (AME2003) and the AMDC

mass.mas03 database, at http://www.nndc.bnl.gov/amdc/web/masseval.html, which is based

on…

G. Audi, A.H. Wapstra and C. Thibault, “The AME2003 atomic mass evaluation (II).

Tables, graphs and references”, Nuclear Physics A, volume 729 (2003), pages 337-676

(available as the file “Ame2003b.pdf” in the AMDC files, at…


..., 23/04/11,

Smith, p.49.

..., 23/04/11,

Heyde; Bertulani, section 1.3 and ch.3; Smith, ch.3; Close, “Onion” p.53 and “Odyssey”, p.90.

..., 23/04/11,

Bertulani, p.84 and figure 3.4.

..., 23/04/11,

We will meet the van der Waals force when we look at bulk matter.

..., 23/04/11,

Bertulani, p.83.

..., 23/04/11,

Bertulani, p.95, and Smith, fig.3.8 and p.47.

..., 23/04/11,

Bertulani, p.83.

http://www.nndc.bnl.gov/amdc/web/masseval.html

http://www.nndc.bnl.gov/amdc/


http://www.nndc.bnl.gov/amdc/masstables/Ame2003/filel.html

A prior paper describes how the data was evaluated and prepared…

A.H. Wapstra, G. Audi and C. Thibault, “The AME2003 atomic mass evaluation (I),

Evaluation of input data, adjustment procedures", Nuclear Physics A, volume 729 (2003),

pages 129-336 (file “Ame2003a.pdf” in the AMDC files).

AMDC also provide the NUBASE2003 nuclide table, which gives mass excess and decay

characteristics, and is based on…

G. Audi, O. Bersillon, J. Blachot and A.H. Wapstra, “The NUBASE evaluation of nuclear

and decay properties”, Nuclear Physics A, volume 729 (2003), pages 3-128, which can be

downloaded from…http://www.nndc.bnl.gov/amdc/web/nubase_en.html

The NUBASE2003 table can be downloaded as an ASCII file from…s

http://www.nndc.bnl.gov/amdc/nubase/nubtab03.asc

AMDC provide the Windows program NUCLEUS to display the contents of the NUBASE

table…

http://www.nndc.bnl.gov/amdc/web/nubdisp_en.html

Nuclide data and graphs given in this chapter are from the mass.mas03 database. This

gives values for mass excess, binding energy and atomic mass in ASCII format, that can

be converted to an EXCEL spreadsheet. The AMDC mass.mas03 and NUBASE tables

give different nuclide properties, but give the same nuclide mass excess values (Audi et

al., “NUBASE”, p.6). The NUCLEUS program has been used to create 2-D and 3-D views

of the the chart of nuclides.

The National Nuclear Data Center (NNDC), at the Brookhaven National Laboratory (BNL)

…http://www.nndc.bnl.gov/ provides a wealth of information, including the excellent

Interactive chart of nuclides, at…http://www.nndc.bnl.gov/chart/ and the Nudat database,

at… http://www.nndc.bnl.gov/nudat2/

“Qcalc”, for finding Q-values for nuclear reactions, is at…http://www.nndc.bnl.gov/qcalc/

Other sources of data on nuclides are…

The Particle Data Group, http://pdg.lbl.gov/ and

The National Institute of Standards and Technology http://physics.nist.gov/cuu/index.html

(All web-sites accessed on 6 April 2011)


http://pdg.lbl.gov/

http://www.nndc.bnl.gov/nudat2/

http://www.nndc.bnl.gov/amdc/nubase/nubtab03.asc

http://www.nndc.bnl.gov/amdc/web/nubase_en.html



Calculations

I've taken 1 amu = 931.494 MeV/c2, and the electron mass, me = 0.511 MeV/c2. Values

have been rounded, usually to 1 d.p., so there will be some rounding “errors”. Masses

should strictly be given in units of MeV2/c, but I shorten this to just MeV, for simplicity, and

as many others do.

The mass of a nucleus, that is without its electrons, can be calculated two ways...

1) via the binding energy…

mnucleus = mass of free protons + mass of free neutrons – nuclear binding energy.

So mdeuteron = p + n – BEdeuteron = 938.27 + 939.57 – 2.2 = 1875.6 MeV

2) or via the mass excess (or mass defect)…

matom = (number of nucleons x amu) + mass excess for that atom.

So matom = (2 x 931.494) + 13.1 = 1876.1 meV.

But this is for the deuterium atom, a deuteron nucleus plus its electron, and we need to

subtract the electron mass…

So mdeuteron = 1876.1 – 0.5 = 1875.6 MeV.

The two deuteron mass values agree.

the atomic mass unit (amu)

We’ve seen that nucleons lose mass when they combine into nuclei. So we need some

reference nucleus to provide a standard value for nuclear masses. That standard is the

carbon-12 atom, with a nucleus of 6 protons and 6 neutrons and containing also 6

electrons. The standard atomic mass unit (amu) is 1/12 th of the mass of a carbon-12 atom,

equal to 931.494 MeV/c2.

Unbound free nucleons have masses significantly bigger than 1 amu. We can get at these

from their mass excess values.

For a neutron, mn = 1 amu + neutron mass excess = 931.494 + 8.071 = 939.6 MeV (1d.p.)

We can find the mass of a hydrogen (H) atom, 1 proton plus 1 electron…

mH = 1 amu + hydrogen mass excess = 931.494 + 7.3 = 938.8 MeV

and then subtracting the electron mass gives us the mass of a proton…

mp = 938.8 – 0.5 = 938.3 MeV (1d.p.)/tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

McNeil, 23/04/11,

The mass excess (keV) is the difference between the atomic mass (amu) and the mass number - Audi et al, p.6, Clayton, p.292. Thus the nuclide helium-4 (2p,2n) has an atomic mass of 4.002603 amu and a mass number of 4 nucleons. So its mass excess is +0.002603 amu, and since 1 amu is equivalent to 931.5 MeV, this gives us helium-4’s mass excess as +2,425 keV (rounded figures), that appears in the AME2003 and NUBASE tables, and in Clayton, p.26.. The bnl chart of nuclides defines atomic mass and mass excess in the glossary, at… http://www.nndc.bnl.gov/chart/help/glossary.jsp

..., 23/04/11,

The masses of free protons and neutrons is given to 2 d.p. to avoid an apparent error due to rounding.

..., 23/04/11,

For example, “for brevity, one normally omits c2” – Bertulani, p.6.


It’s important to know if the data is for “naked” nuclei or for whole atoms (the “mas03”

table, for example, is for atoms). Nuclear properties are very accurately measured, and

the data is freely available to all – a wonderful resource. Readers are welcome to browse

the data and use it to calculate nuclear masses and other properties. In what follows, I'll

give nucleus masses and binding energies in MeV; the example of the deuteron nucleus

shows how these have been calculated.

mass-energy accounting

The particle data resources are the result of a huge amount of meticulous work by

thousands of physicists, that have been made freely available. The sheer amount of data

can be overwhelming, and clearly the full story is very complex. I will stick to the simple

picture we’ve seen with the deuteron nucleus – that the total mass-energy of a system of

nucleons remains the same whether they are separate or bound together. The mass loss

of the separate nucleons in forming the nucleus is the energy involved in their being

bound by the nuclear force. From this point of view, I will just be doing the mass-energy

accounting.

3.2.6 the simplest nuclide – the deuteron

We're now ready to try some mass-energy accounting for the proton-neutron pair, the

deuteron – figure 3.4.

Figure 3.4: The bound deuteron has less mass than its componentsThis simple example shows how the mass-energy accounting works (all figures in MeV).../tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

p+

p+

no

no937.8

938.3

939.6

mass of bound deuteron cluster = 1875.6 MeVso average mass of each bound nucleon = 937.8 MeV

Mass of separate nucleons= 939.57 + 938.27= 1877.8 MeVBinding Energy (BE)= mass loss of system= 1877.8 – 1875.6= 2.2 MeVSo BE/nucleon= 1.1 MeV

mas

s,M

eV


mass of bound nucleus + binding energy = total mass of separate nucleons = mp + mn

1875.6 + 2.2 = 1877.8 = 938.27 + 939.57

The mass “lost” when a proton and neutron are bound together is about 2.2 MeV – the

binding energy for this nucleus. In order to compare nuclei with different numbers of

nucleons, we’ll calculate the average binding energy per nucleon. For the deuteron it is

quite simply 2.2 / 2, which is 1.1 MeV/nucleon. This quantity, the binding energy per

nucleon, will be important when we look at the full range of nuclides.changing the proton-

neutron balance.

the proton-neutron balance

The deuteron (pn) is only one way to combine two nucleons - we can also have a pair of

protons, a diproton (pp), or a pair of neutrons, a dineutron (nn). We'll use these three

combinations in a hypothetical example see the effects of changing the proton-neutron

balance in the cluster. Figure 3.5 shows the mass-energies of the three nucleon pairs.


..., 23/04/11,

Bertulani describes these "three possibilities to build a system of two nucleons" - chapter 3. This is such a simple example of the conflicting factors of the protons’ electrical repulsion and the neutrons’ greater mass, that I want to use it if I can. I’m not sure if this approach is admissible, so I’ve called it hypothetical. To get an electrical pe greater than the mass difference between the proton and neutron, I’ve put the two protons only 1 fm apart – they are not this close in fact. Of the 3 combinations, only the deuteron is stable, but it’s not due to having the lowest mass-energy. However, we will see that real cluster-families (isobars) have mass-energies that follow similar valley profiles, and the stable nuclides are those at the valley bottom. This is a glimpse of the nuclear valley.

..., 23/04/11,

The deuteron is quite weakly bound, and so the prton and neutron have a significant probability of lying quite far apart. “This is where we catch another glimpse of how strange the quantum world is: the proton and neutron spend part of the time outside the range of the force that holds them together, yet they still remain bound together, unless the deuteron is hit by something like a powerful gamma ray” – Mackintosh et al, p. 53.


Figure 3.5: The proton-neutron pairing has the least mass-energy of the three possible pairings of two nucleonsSince a neutron is heavier than a proton, the pp pair should have the least mass and the

nn pair the most. However, we also need to consider the electric repulsion between the

protons in the pp pair. We have to force the protons together, thereby storing potential

energy in the system. This is just like a compressed spring - release the protons, and they

will fly apart. When two protons are 1 fm apart this electrical repulsion adds ~1.5 MeV to

the system’s mass-energy. Protons are lighter, but bring the energy of their mutual

electrical repulsion; neutrons have no repulsion, but are heavier. Between these extremes

there is a balance of protons and neutrons with least mass-energy, lying at the bottom of

a mass-energy “valley”.

cluster stability arises from conflicting factors

This simple example illustrates the conflicting factors that decide the stability of larger

nuclei. In general, a proton-rich nucleus has lighter constituents, but has a bigger total


Sys

tem

ene

rgy,

MeV

no no

no

proton-proton2mp + electrical p.e.

= 2x938.3 + 1.5 = 1878.1 MeV

p+ p+

p+

neutron-neutron2mn = 2x939.6= 1879.2 MeV

proton-neutronmp + mn = 938.3 + 939.6 = 1877.9 MeV

1879.2

1878.1

1877.9

change pn in electrical p.e.

is bigger than in mass-energy

change np in mass-energy

no change in electrical p.e. here

neutron-richproton-rich

..., 23/04/11,

This is the precursor of the nuclear valley that we will explore later.

..., 23/04/11,

This is from the standard formula for electrical potential. This repulsion is much weaker than the nuclear force, but has a greater range. We’ll see later how these differences set a limit on the size of stable nuclei, and also decide the stable proton/neutron ratio. PE calculation and graph have been deleted – 6 April 2011.


mass-energy due to the protons' mutual electrical repulsion. The transformation of a

proton to a neutron (I will use the shorthand pn) will reduce this mass-energy, if the loss

of electrical pe is bigger than the gain in mass. Conversely, a neutron-rich nucleus has

heavier constituents, but a smaller electrical pe contribution to the total mass-energy. The

transformation of a neutron to a proton (np) will reduce this mass-energy, if the loss in

mass is bigger than the gain in electrical pe. We can't see this with the nn pair, but we can

see it will apply to bigger clusters of nucleons, with more than one proton.

Figure 3.5 gives a hypothetical situation, arranged to show the importance of the proton-

neutron balance. We will see that for real nuclides, stability is found with the balance of

protons and neutrons that minimises the cluster’s mass-energy. Proton-rich clusters

reduce their mass-energy, and move towards stability by transforming protons to

neutrons; conversely, neutron-rich nuclei transform neutrons to protons.

Of the three combinations, only the pn pairing, the deuteron, is only just stable, and the

bound diproton and dineutron do not exist. This reveals the fundamental nature of the

particle-wave, that if it is confined, in effect “squeezing” its wave aspect, then it speeds up:

“a particle confined to a very small space must move very quickly, so a large force is

required to keep it from getting away”. The nuclear force, powerful as it is, can only just

confine a proton and a neutron in a deuteron. The potential energy well for the deuteron is

~100 MeV deep, but the proton and neutron rattle around so fast that “the deuteron

almost jumps out from the potential hole!” The deuteron is weakly bound system, that

needs only 2.2 MeV to overcome the small binding energy and separate the proton and

neutron - “It is almost unstable; it has no excited state, the smallest rotation or vibration

tears it apart”. It turns out that the nuclear force is strongest between unlike nucleons, and

so is just unable to confine pairs of neutrons or protons.

The universe’s first step in building nuclei is not a very firm one. If the strong force were a

little less strong then making the deuteron, the first step in creating the elements, would

not be possible. However, if the strong force were much stronger than it is, sufficient to

bind diprotons, then this would affect the sequence of nuclide formation in the early

universe.

why don’t diprotons and dineutrons exist? - a technical section

This is a technical section, that the reader can skip, that tries to answer these questions: if

the nuclear force is so strong, then (1) why does the deuteron, in a potential well ~100


..., 23/04/11,

Barrow and Tipler surmised that all hydrogen would be consumed as diprotons, and succinctly said “If the diproton existed, we would not!” – p.322. MacDonald and Mullan calculate that this extreme outcome would not occur.

..., 23/04/11,

Barrow and Tipler, p. 322.

..., 23/04/11,

This is the effect of nucleons’ spin – maybe something kept out of the general account.

..., 23/04/11,

Both quotes from Marx, p.378.

..., 23/04/11,

Hogan, p.29.

..., 23/04/11,

Bertulani, p.75.

..., 23/04/11,

Of the three combinations, the pn pair, the deuteron, is the only one that is stable, but the instability of the pp and nn pairs is not due to their greater mass-energies. The nuclear force is only just able to confine and bind the pn pair. The deuteron is close to breaking up. Difference in the nucleon spins in the pp and nn pairings weaken the nuclear force, so these pirs are unstable. The mutual repulsion of the protons in the pp pair add to its instability. See Bertulani, p.75, MacDonald and Mullan, and Barrow and Tipler, p.321, and http://en.wikipedia.org/wiki/Dineutron http://en.wikipedia.org/wiki/Isotopes_of_helium (both accessed 7 April 2011)

..., 23/04/11,

This is the weak interaction, which interchanges u and d quarks. We've seen the details of the reactions in the previous chapter on quarks. We'll meet them later in this chapter, as the radioactive beta decays, whereby unstable nuclei "seek" stability.


MeV deep, have a binding energy of only ~2 MeV, and (2) why are the diproton and

dineutron unstable?

The first part of the answer is to be found in the close confinement of the nucleons in the

nucleus. Quantum mechanics tells us that there are no objects but wave-packets; no

lumpen “things”, but constructs of energy, aggregates of universal action. Nothing can

ever be still, it is always on the move, and cannot be pinned down. The more you try to

confine a particle’s position, the less certain you are of how its moving. There is a

mathematical relationship between these uncertainties…

∆p x ∆x h

This tells us that the product of the uncertainties in momentum (p) and position (x) is

approximately equal to Planck’s constant. The more confined the particle, the faster it can

be moving. This can be rearranged as

∆p h/∆x

If the separation of the two nucleons in deuterium is ~2 fm, then that sets the uncertainty

in their positions, ∆x. The kinetic energy, 0.5mv2 can be written as 0.5m2v2/m and so as

0.5p2/m, where p is the momentum, mv. So for a particle with an uncertainty s in its

position, will have an uncertainty in its kinetic energy given by

∆Ek 0.5∆p2/m = 0.5h2/m∆x2

= 0.5 x 6.6 x 10-34 / 1.6 x 10-27 x (2.6 x 10-15)2 = 3.4 x 10-11 Joules = 212 MeV

The nucleons in the deuteron may be confined, but they are rattling around very fast in

their confinement, with kinetic energies that may be a couple of hundred MeV. This is

comparable with the depth of the deuteron potential energy well, about 100 MeV. The

powerful nuclear force is only just able to confine the proton and neutron within the

deuteron – “the deuteron almost jumps out from the potential hole!” The deuteron is

weakly bound system, that needs only a small energy input of 2.2 MeV to separate the

proton and neutron - “It is almost unstable; it has no excited state, the smallest rotation or

vibration tears it apart”.

The two protons in the diproton repel each other, but this adds only about 0.56 MeV,

which is much less than the deuteron’s 2.2 MeV binding energy. George Marx says that

“the simplest explanation is that the mild electrical repulsion destabilises the 2He

[diproton]”. We can see that there’s more to it than this. And also, why is the dineutron,

with no repulsion, unstable?


..., 23/04/11,

Marx, p.378.

..., 23/04/11,

MacDonald and Mullan, section III, give the proton separation as ~2.6 fm, which gives the electrical pe as 0.56 MeV.

..., 23/04/11,

Both quotes from Marx, p.378.

..., 23/04/11,

This is calculation is from George Marx. He used a position uncertainty of 1.5 fm, and got a kinetic energy of ~60 pJ, or 375 MeV, a much bigger energy than the depth of the deuteron energy well. Using ∆x~2 fm gives ~200 MeV, which is comparable with the energy well depth – though even this seems too big, but this is an approximation. Using h-bar, gives ke~5 MeV, too small.

..., 23/04/11,

Bertulani gives the “mean square radius” of the deuteron as 2.1 fm – p.37. MacDonald and Mullan model the deuteron as a square potential well with a radius of 2.02 fm – section III. So the separation of proton and neutron in the deuteron looks to be ~2 fm.

..., 23/04/11,

This is the standard expression for Heisenberg’s uncertainty principle, yet it varies in its presentation. Marx (p.378) and Hey and Walters (p.23) give Planck’s constant, h; Allday (p.79) and Bertulani (p.6) give h-bar, that is h/2pi; while Williams (p.19) gives h-bar/2.


The second part of the answer lies in the spin of the nucleons. The nuclear force is

stronger between nucleons with parallel spin than between nucleons with opposite spin.

The proton and neutron in the deuteron, being different fermions, can have parallel spin.

But the diproton and dineutron each comprise two identical particles, and so “the Pauli

exclusion principle requires that the nucleons … have opposite spin”. Consequently, the

nuclear force is weaker and not able to hold these particles together.

It is, however, a close-run thing, for an increase of only 9% in the strength of the strong

force would make the dineutron stable. A slightly bigger increase of 13% is needed for the

diproton, in order to overcome the protons’ repulsion. The diproton falls short of being

bound and stable by only ~92 keV.

So the extremely strong nuclear force, under the most favourable conditions, is only just

able to confine two differing nucleons, with parallel spins, into a stable configuration. If the

nucleons are the same, and have opposite spins, then no stable structure is possible. In

helium-3 (2p,1n), each nucleon is attracted to two others, and the binding energy per

nucleon (BE/A) is ~2.6 Mev, more than twice that of the deuteron.

3.2.7 Conclusion

We have a nuclear force that is very powerful but short range, that can bind nucleons

when they get as close as their own size, regardless of whether they are protons or

neutrons. We thus have a means of building nucleons into bigger structures. It looks fairly

simple, but we've seen that there are conflicting factors: protons are lighter but carry an

electrical charge, neutrons carry no charge but are heavier. We can use these simple

ideas to explore model nuclei.

3.3 Building nucleon clusters – a simple nuclear modelThis section describes the factors that decide the stability of nucleon clusters, and the

ways that unstable clusters rearrange themselves to become more stable. We’ll start with

a simple model of a cluster and see how this can start to explain how each of these

factors affects its stability. We’ll see that “the competition between gluonic [strong] and

electric forces (which tend to drive the nuclei apart) creates a rich arena of nuclear

phenomena and determines which stable chemical elements can exist in nature”.

Clusters of nucleons vary in two ways: (1) by their balance of protons to neutrons – the

p:n ratio, and (2) by their size. Thus we can think of the cluster-12 “family”, which


..., 23/04/11,

Hogan, p.10.

..., 23/04/11,

I think it’s worthwhile doing this, first, to show how even simple models can readily predict patterns of behaviour, and second, to illustrate how scientists often think. The model here is a 2D version of the liquid drop nuclear model. Williams, ch. 4 is a good example of a more rigorous, mathematical approach.

..., 23/04/11,

Stability, like uniqueness, is an absolute condition; a nucleon cluster is either stable, or it’s not. But it’s useful to talk of clusters being more or less stable, in the sense that a “less stable” cluster is one that is more likely to decay to a “more stable” configuration.

..., 23/04/11,

He-3 is the only stable nuclide having more protons than neutrons, except for the proton itself – Clayton, p.22.

..., 23/04/11,

And as a, perhaps fanciful, analogy parents and teachers will know that it can take a lot of energy to confine someone to a task they would rather not do!

..., 23/04/11,


..., 23/04/11,


..., 23/04/11,

MacDonald and Mullan – section III.

McNeil, 23/04/11,

The nuclear force is the “approximately independent of the charge state of the nucleons, that is, of nn, pp, or np”, but does depend on their spin – Bertulani, p.75.


comprises 12 nucleons in any combination, from 12 protons (12p,0n), through more

balanced ratios, like (6p,6n), to 12 neutrons (0p,12n). Each specific combination of

protons and neutrons has its own identity and characteristics, and is called a nuclide.

Thus the nuclide (6p,6n) is very different from its nearest “sibling”, (7p,5n), in the cluster-

12 family.

We know that nucleons are closely clustered into roughly spherical nuclei, rather like

marbles in a string bag. Imagine a simpler, two-dimensional analogue of this, where the

nucleons are represented by flat discs, closely packed together, like coins on a table. This

would be like taking a slice through the centre of a spherical cluster. We’ll use this to

explore the effects of varying the p:n ratio, in a cluster of constant size.

varying the proton:neutron ratio

Imagine a 2-D cluster of 14 nucleons, about 4 nucleons across, and with a 1p:1n ratio,

thus (7p,7n) – the middle cluster in figure 3.6. Changing one neutron into a proton gives

cluster (a), and doing the opposite gives cluster (c). We already know that protons are

lighter than neutrons and mutually repel. So the mass-energy of a cluster such as this is

the sum of the nucleon masses, as bound by the nuclear force, and the mutual repulsions

of the electrically charged protons.


Key: proton neutron electrical repulsion

neutron-richproton-rich

(c) cluster-14 (6p:8n)6 fewer p-p repulsions

mass increases by (mn-mp)

(b) cluster-14 (7p,7n)ratio 1p:1n

(a) cluster-14 (8p,6n)7 more p-p repulsions

mass decreases by (mn-mp)

..., 23/04/11,

The cluster size is the same, so the strength of the nuclear binding force will be more or less unchanged.

..., 23/04/11,

I think the general reader will be more at ease with a cluster-family rather than an isobar. This will provide the setting for the isobaric beta-decay reactions, whereby a nucleon cluster works towards stability. Like humans, nucleons keep a lot of things “within the family”.


Figure 3.6: (a) the arrows show the 7 extra p-p repulsions from changing one neutron to a proton (the dotted circle), (b) cluster-14 with a 1:1 proton:neutron ratio, (c) the arrows show the 6 p-p repulsions that disappear when a proton is changed to a neutron (the dotted circle)If we change one neutron into a proton, going (b) to (a), then there will be 7 more proton-

proton (p-p) repulsions to add to the cluster’s mass-energy, and a loss of mass (mn-mp).

If we go the other way and replace one of the middle cluster's protons by a neutron, then

there will be 6 fewer repulsions, and a gain in mass (mn-mp). Whichever way we go, there

is the same change in mass, but a different change in the number of p-p repulsions.

If we now imagine that the energy gain due to 7 extra p-p repulsions is just slightly bigger

(by less than the value of a single p-p repulsion) than the energy loss associated with the

mass difference (mn-mp), then the cluster (a) (8p,6n) will have slightly more mass-energy

than cluster (b) (7p,7n). Going the other way, the energy due to 6 p-p repulsions will be

slightly less than the energy associated with the mass difference (mn-mp), and so cluster

(c) (6p,8n) will also have slightly more mass-energy than cluster (b) (7p,7n). Thus the

mass-energy of cluster-14 is a minimum for the combination (7p,7n); shifting the p:n ratio

either way increases it.

The general argument above suggests that every cluster will have some p:n ratio that

gives it the minimum value of mass-energy. Too many protons, and the cluster mass

increases due to their repulsions, but too many of the heavier neutrons will also increase

the mass of the cluster. Since the “lost” mass is released as binding energy, the nucleons

will be most strongly bound in the cluster with the least mass. This is a general extension

of the pattern we saw with the deuteron in section 3.2.6.

We have seen how energy becomes incorporated in matter, according to Einstein’s

equation, E = mc2, and also how a particle is unstable if it can transform to another

particle with a smaller mass. Thus we have seen that only the generation I quarks, the

ones with least mass, are stable. The heavier quarks in generations II (charm and

strange) and III (top and bottom) all decay to the light up and down flavours. This fits with

the idea that each cluster family has only one stable p:n ratio, the one with least mass. It

is energetically favourable for the other heavier configurations in to “move” towards this

stable ratio, by interchanging protons and nucleons within the cluster.

varying the cluster size

Now we’ll keep the p:n ratio constant, and vary the cluster size. A cluster of protons and

neutrons is subject to two competing forces. The powerful nuclear force attracts all


McNeil, 23/04/11,

Mackintosh et al, p.98.

..., 23/04/11,

What about a highly unbalanced cluster achieving a more amenable p:n ratio simply by ejecting a nucleon? This sounds feasible, but in fact is energetically unfavourable, since nucleons have more mass outside the cluster than inside, due to the binding effect of the nuclear force. Thus a system of A nucleons would have less mass as a single cluster than as a cluster of (A-1) plus 1 free nucleon.

..., 23/04/11,

This will have been covered in chapter 2, on quarks and leptons.

..., 23/04/11,

This will have been covered in chapter 0, discussing the idea of energy being invested in matter, and thereby trying to explain why less mass means more stability.

..., 23/04/11,

We saw a pattern of behaviour like this with the 2-nucleon clusters - see section 3.5.5 on the deuteron.


nucleons, but is very short range, reaching little further than the next-neighbour nucleon.

The much weaker electrical repulsion affects only the positively charged protons, but is

long range, and can "reach" across a big cluster. Figure 3.7 shows a series of clusters of

increasing size, all with ratio close to 1p:1n. Because we're keeping the proton:neutron

ratio the same, we don’t have to consider the proton-neutron mass difference. If we

consider the forces acting on one proton at the edge, we can see that the nuclear

attractions from the adjacent nucleons compete with electrical repulsions from all other

protons in the cluster.

Figure 3.7: attractive and repulsive forces in small 2-D clusters with a 1:1 proton:neutron ratioWe can see that the number of next-neighbours very quickly builds up as the cluster size

increases. In this 2-D case the maximum number is three, and this is reached in cluster-4.

So while the total attractive forces quickly increase, they also quickly reach their

maximum. In contrast, the electrical repulsions build up slowly but steadily. They reach

right across the cluster, and are roughly proportional to the number of protons - doubling

the number of protons will about double the number of repulsions on the proton at the

edge. Figure 3.8 shows the shifting balance of these competing forces, as the cluster

grows.


Key: proton neutron nuclear force attraction electrical repulsion

(a) cluster-21p:1n

(b) cluster-42p:2n

(c) cluster-73p:4n

(d) cluster-147p:7n

..., 23/04/11,

This is a simplification, though I don't think it invalidates the result. The nuclear force, with a range of about 1-2 fm, will reach to the next-nearest nucleon. The electrical repulsion is longer range, though it does weaken with distance - see back to figure xXx. After thinking through this explanation, I found a similar approach in Mackintosh et al, p.98 – July ’10, and also Benjamin Crowell… http://www.lightandmatter.com/html_books/4em/ch02/ch02.html#Section2.5 (August 2010)


Figure 3.9: how the balance of competing nuclear and electrical forces varies with cluster size, keeping a 1:1 proton:neutron ratioThe attractive nuclear forces quickly build up a "lead" over the repulsive electrical forces,

establishing a range of stable clusters, with a maximum stability at the lower end of the

range. But after an initial fast start, the nuclear forces are soon approaching their limit,

and their curve flattens out. The total electrical force is roughly proportional to the number

of protons, and is represented here by a straight line. So there must come a cluster in

which the forces are balanced, and beyond which no cluster is stable. It's like the race

between the hare and the tortoise, where the hare starts off fast, but the slow patient

tortoise eventually catches up.

The model suggests that there is a maximum stable cluster size, beyond which clusters

will “down-size”, as a proton at the edge is ejected by the repulsions from all the other

protons. This is in contrast to the behaviour of clusters smaller than the stable maximum,

which look like they will stay the same size, and internally adjust their p:n ratio.

Summary


Forc

es o

n th

e ed

ge p

roto

n

maximum size

attractivenuclear forces, very short range, limited to adjacent nucleons...quickly build up… ...then slowly approach a limit

This cluster is most strongly bound, where the nuclear attractions have built up the biggest "lead" over the electrical repulsions

The electrical repulsions start slowly, and at first lag behind...

The nuclear force attractions are limited

by the maximum number of adjacent

nucleons. They can't get bigger than this

The strong and electrical forces are balanced. This is the biggest possible stable cluster.Clusters larger than this are unstable.

…but keep steadily increasing, and finally "catch up" the strong attractions

maximum stability

0 cluster size

electrical repulsions

are potentially limitless

McNeil, 23/04/11,

This is ok for now, but we’ll see later that this is rather simplistic, and that the “down-sizing” process must reduce the total mass of the nucleons.


The picture that emerges shows a nucleon cluster as being subject to conflicting internal

forces, and having a mass-energy that is sensitive to its p:n ratio. The cluster may be

unstable, and either fragment, or transform to a configuration with less mass-energy. We

can view a cluster as a transient configurations of nucleons, that will transform to a more

stable configuration if it gets the chance. A nuclide, a specific configuration of protons and

neutrons, is then a stage in this process.

Our brief look at a very simple nuclear model suggests that…

(1) a cluster of a constant total number of nucleons has a minimum mass-energy, and

is therefore stable, at some particular p:n ratio,

(2) clusters with a constant p:n ratio have an optimum size, at which they are most

tightly bound, and…

(3) … a maximum size, beyond which they are unstable, and will down-size by ejecting

a proton.

We’ll now look at the real nuclides and see if these inferences are valid.

3.4 Real nuclei3.4.1 what does a nucleus look like?

We're ready now to stand "outside" the nucleus, and look at the way the nucleons are

packed inside. We're now viewing the nucleons as simple spheres, like marbles gathered

together in a spherical cluster in a string bag. However, this simple picture is static, and

we will see that the nucleons in the cluster are in incessant motion: “Perceived from

outside, a stable atomic nucleus looks like a solid citizen of nature; but way inside there is

a boiling microcosm - a world of complex unstable hadron interactions”.

We’ll now look inside a nucleus, and try to get a “feel” of the tiny world to which nucleons

are confined.

particle scattering tells us the size

We probe very small objects like nuclei, by firing even smaller particles at them, and

seeing how these are scattered. When we use negatively charged electrons as probes,

they respond to the oppositely charged protons in the nucleus, and thereby give us an

idea how closely the nucleons are packed.

the nucleus has a fuzzy boundary


..., 23/04/11,

Electrons measure the “charge density”. Other particles, which respond to the nuclear forces, will tell you the "matter radius" of the nucleus. These two values are very similar (HyperPhysics, and Bertulani, p.99), and I've just used "charge density" as a simple indicator of the closeness of nucleon packing.

Andrew McNeil, 23/04/11,

Fritzsch, Elementary, p.86.

..., 23/04/11,

There are several models of nuclear clusters, each of which which explains some aspect of nuclear behaviour - Bertulani, ch. 5. Most nuclei have two basic shapes – spherical or prolate, like an American football, though there are other shapes possible – Mackintosh et al, p. 52. I’ll stick to the simple spherical case.


Such scattering experiments show that the nucleus is not a hard sphere, like a snooker

ball, but has a “fuzzy” edge, where its density gradually falls to zero. The nucleons are

packed tightly together in the central region, giving a constant density, and thin out at the

edge – figure 3.10.

Figure 3.10: upper – a pictorial representation of the protons and neutrons in a nucleus. The effective radius can be simply expressed as 1.2 times the cube root of the total number of nucleons (A) – se the text below.lower – a plot of the density of nucleon packing across the nucleus

some real nuclei

Figure 3.11 shows the charge density plots for several nuclei, from small helium, with only

2 protons, to massive lead, with 82. Here the density of electric charge on the protons is

used as a measure of how tightly the nucleons are packed.


central dense region of closely packed nucleons

the nucleus "skin" where the nucleons are thinning out and the

density is decreasing

the outermost nucleons, where the nucleus density has fallen to

zero

pictorial visualisation of the distribution of nucleons in a large nucleus

a plot of the density of nucleons in the nucleus

..., 23/04/11,

From a university physics course... http://www.kutl.kyushu-u.ac.jp/seminar/MicroWorld3_E/3Part2_E/3P22_E/charge_E.jpg This agrees with graphs in Mackintosh et al., p. 50, Williams, fig. 3.5, p.46, and Bertulani, fig. 4.2, whose graph is based on Hofstadter, fig.8, at… http://nobelprize.org/nobel_prizes/physics/laureates/1961/hofstadter-lecture.pdf (accessed 5 April 2011) Halo nuclei are a recently discovered class of unstable nuclei, mostly neutron-rich, which have anomalously large sizes – Mackintosh et al, p.53, Bertulani, p.390.

..., 23/04/11,

upper - from "Nucleus: A trip into the heart of matter", by Ray Mackintosh et al., p.48 accessed at… http://pntpm3.ulb.ac.be/pans-info/site/html/activities/CHAPT_4.PDF (accessed July 7 2010) lower - from HyperPhysics, at… http://hyperphysics.phy-astr.gsu.edu/HBASE/Nuclear/imgnuc/nucradius.gif

..., 23/04/11,

Williams, p.40.


Figure 3.11: charge distributions for a range of nuclei: helium, He (2p,2n); calcium, Ca (20p,20n); nickel, Ni (28p,30n); samarium, Sm (62p,90n) and lead, Pb (82p,126n). Only part of the plot for hydrogen, H (1p,0n) can be shown, since the charge is so concentrated.In the lead nucleus the nucleons are tightly packed together, out to about 5 fm, beyond

which is the "skin", where they are packed more and more loosely, with the outermost

nucleons being about 9 fm from the nucleus centre. Helium has only 2 protons, but is very

compact - about 3 fm to the outer nucleons - and so has a large central charge density.

The hydrogen nucleus, a single proton, has its single charge concentrated in a very small

region.

checking the empirical radius formula

The nuclear radius depends on the number of nucleons, and approximately follows the

empirical equation…

rnucleus = r0 A1/3

where A1/3 is the cube root of the number of nucleons, and the constant r0~1.2 fm.

The "effective" nuclear size is usually taken as the radius at a point about half way up the

charge density slope, as is shown for calcium (Ca) by the red arrow (labelled R) in figure

22. Using this, we can check the empirical formula for some of the nuclei.

For helium-4, He-4, rHe = 1.2 x 41/3 = 1.9 fm,

and for lead-208, Pb-208, rPb = 1.2 x 2081/3 = 7.1 fm.


if the lead nucleus was a hard sphere it would

have a charge distribution like this

the approximate plot for hydrogen, H(1p), goes way up beyond this graph

lead (Pb) nucleus dense central region skin edge

..., 23/04/11,

Bertulani, p.99, and Mackintosh et al, p.48. Nuclei behave in the opposite way to the atoms of which they are the centre. To a first approximation, nuclear material has a more or less constant density, so nuclei of different masses have different sizes. In contrast, atoms of different masses have similar sizes. This leads, for example, to lead and aluminium having very different densities – Mackintosh et al, p.49.

..., 23/04/11,

from http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/imgnuc/nucradius.gif If you set A=1, in effect, a hydrogen nucleus, then you get a very rough idea of the radius of a single proton as 1.2 fm.

..., 04/23/11,

Indeed - why include samarium? The isotopes of samarium (and probably of other elements too) have different shaped nuclei. The Sm-148 isotope is “a typical spherical nucleus”, but the Sm-152 isotope is “ deformed like an American football” – Mackintosh et al., p. 104. I don’t know which Sm isotope is shown in the graph.


In each case, the formula puts the effective nucleus boundary 1-2 fm inside the point

where the charge density fades to zero. The charge on a single proton is very

concentrated, so only part of its curve can be shown, and this suggests a radius a little

less than 1 fm. This agrees fairly well with the current value of the proton charge radius ,

which is given as 0.877 fm.

a simple sum

We can also do a simple sum, which might help us get more of a “feel” for the physical

presence of the nucleus. The dotted blue line shows the charge distribution in a lead

nucleus, if it was truly like a snooker ball, and had a uniform charge density and a sharp

boundary. In this form it’s a sphere with a radius close to 7 fm, and a charge density about

0.06 electron charges/fm3 (e/fm3). We can use this to estimate the number of proton

charges in the nucleus.

So the lead nucleus volume = 1.33 r3 = 1.33 x x 73 = 1437 fm3

With a uniform charge density of ~0.06 e/fm3, this will contain 1437 x 0.06 = 86 proton

charges (remembering that protons and electrons carry the same amount of charge, just

opposite in sign). This rough figure of 86 is satisfactorily close to the figure of 82 proton

charges the lead nucleus actually holds.

how big is a nucleon?

We can use the figures above to get an idea of how closely the nucleons are packed

inside a nucleus. A lead nucleus, with a volume of 1437 fm3 contains 208 nucleons. So

each nucleon occupies a volume of 1437/208 = 6.9 fm3. If we treat this as a cube,

enabling the nucleons to be closely packed in a 3-D cluster, then its side length is the

cube root of 6.9 = 1.9 fm. So the nucleons are about 1.9 fm apart; they have a "size" of a

bit under 2 fm. This makes the average nucleon radius a bit less than 1 fm, which fits with

the value given above for the proton charge radius, and with Bertulani - "The radius of

protons and neutrons that compose the nucleus is of the order of 1 fm."

So, we have a working grasp of the size of a nucleon – it has a radius of about 0.8-0.9 fm,

and so is bit less than 2 fm across. If we scale nucleons up so that 1 fm becomes 1 cm,

then a proton will sit on the tip of one finger. A helium nucleus is about 3-4 finger widths,

and a lead nucleus is about two hand breadths across.

Generally, protons and neutrons are described as having a size of “about a fermi”. In the

end, thankfully, precise sizes are not an issue for us here!


..., 23/04/11,

This section started with my trying to visualise a nucleus and wondering how big a nucleon is. It seemed a simple question! I wanted to be a bit clearer, and found this was not straightforward. The graphs and figures I've given seem to be consistent in saying that a nucleon has a radius of 1 fm, and so is 2 fm across. But this doesn't fit with the professional authors, who say a nucleon has a size of "about a fermi". I would take the word "size" as meaning the diameter, the distance across. So, there seems to be some inconsistency here. It may well be an error of mine. However, the 1 fermi size fits well with the nuclear force-distance graph that we meet soon. Smith, p.29, says the size is “0.8 fermi across”. For an example of someone else trying to find out the size of a proton, see… http://www.physicsforums.com/showthread.php?t=241465m (accessed 5 April 2011). Missing section:”How big is a human?” removed here – 5 April 2011.

..., 23/04/11,

Bertulani, p.98. A clear statement at last - May 2010.

..., 23/04/11,

An alternative approach goes... We can put the formula rnucleus = r0 A1/3 (remembering that A is the number of nucleons) into the formula for the volume of a sphere… Vnucleus = 1.33 ( rnucleus3 = 1.33 ( (r0 A1/3)3 = 1.33 ( r03 A = 7.2 A fm3 That means that, on average, each nucleon packed in the nucleus occupies 7.2 fm3. So in the tightly packed nucleus, the centres of the individual nucleons are about 1.9 fm apart (the cube root of 7.2), so the average nucleon radius is 0.8-0.9 fm. This is a more general, and better approach, but I think the reader will be more at ease with the concrete example I've given. We can calculate the saturation density of a nucleus… A/V = 1/7.2 ~ 0.139 nucleons/fm3. We get this value for any kind of nucleus – nuclear matter has an almost constant density. The average internucleon distance is then the cube root of 7.2, about 1.9 fm - close to the value calculated for the lead nucleus. http://www.kutl.kyushu-u.ac.jp/seminar/MicroWorld3_E/3Part2_E/3P22_E/nuclear_size_E.htm http://en.wikipedia.org/wiki/Nuclear_matter

..., 23/04/11,

Plots for several other nuclei can be found in Robert Hofstadter's 1961 Nobel prize address, from which the plot for hydrogen has been taken. http://nobelprize.org/nobel_prizes/physics/laureates/1961/hofstadter-lecture.pdf The value of the proton charge radius has been put at 0.877 fm… http://physics.nist.gov/cgi-bin/cuu/Value?rp|search_for=atomnuc! The sub-atomic particle properties have been rviewed by C. Amsler et al. (Particle Data Group), Physics Letters B667, 1 (2008)… http://pdg.lbl.gov/2009/listings/rpp2009-list-p.pdf You can get to it via reference 1 in…http://en.wikipedia.org/wiki/Proton). “Until recently the best estimate of the proton radius was 0.877 fm with an uncertainty of 0.007 fm”, but a recent very careful study has put the radius at “0.8418 fm with an uncertainty of 0.0007 fm – in striking disagreement with previous measurements.” The consequences of this small difference could be extensive, for "the finding could mean that physicists need to to rethink how they use quantum electrodynamics (QED) - or even consider a major overhaul of this theory." This may seem "picky", but the first validation of Einstein's special theory of relativity was the observation of a tiny deviation in the path of starlight - reffff "Putting the squeeze on proton size", Physics World, vol. 23, No. 8, p.4, August 2010, and Nature vol. 466, p.213. The mean square radius of the deuteron is 2.1 fm and 0.8 fm for the proton – Bertulani, p. 37.


A football as heavy as Everest

We’ve perhaps got a “feel” for a nucleus as something like a string bag full of marbles –

something we can imagine holding in our hand. It’s worth reminding ourselves that this is

not ordinary material. Nuclei are incredibly dense – “if a solid football were made of pure

nuclear matter, it would weigh as much as Mt Everest.”

3.4.2 the nuclide plot

Figure 3.12 is a standard plot of the known nucleon clusters - the known nuclides - in

“nucleon-space”.

Figure 3.12: Decay half lives for the known nuclei. The colour code follows the visible spectrum, with black for stable and very long-lived nuclei, then from purple through to red as half lives get shorter. Note that this makes bismuth-209, with a very long half-life, appear as the last stable nuclide.The neutron and proton numbers are plotted along the x- and y-axes, respectively. The

nuclides range in size from 1 proton up to massive clusters of about 180 neutrons and

120 protons. There are thus nearly 22,000 potential proton/neutron combinations, out of

which, about 3,200 are known, and of these a mere 260 or so are stable, the thin arc of

stable nuclides, shown as black cells running diagonally across the diagram. The coloured

cells represent known but unstable nuclides, and beyond these, in the grey area of the

plot, the clusters are so unstable and short-lived that they effectively don't exist. There is a


nuclides with p:n ratios further from the stable arc are more unstable

arc of stable and long-lived nuclides

(black cells)

increasingly proton-rich

increasingly neutron-richratio

about 1p:1n

large clusters have ratios of 1p:~1.5n

highly unstable nuclides, with half lives < 0.1 s (red cells)

this 45o diagonal line is for cluster-200 (Z+N=200). The p:n ratio varies along the line,

with one stable combination, at (80p,120n).

all nuclides on a vertical line have the

same number of neutrons (here N=20)

all nuclides on a horizontal line have the same number of protons (here Z=20)

the biggest stable nucleus - cluster-208

..., 23/04/11,

The limits are given by the “drip-lines”, “the point where the nucleus can no longer hold another particle” – Bertulani, p.385. We’ll look at these when we get to stellar nucleonsynthesis, later in this chapter.

..., 23/04/11,

Bertulani, p.115. The stable arc is also known as the beta stability line, because unstable nuclei “move” towards stability by a radioactive beta decay process.

..., 23/04/11,

This is the Segrè plot, or nuclide plot, and is the standard way to represent nuclides. This figure is from the Nucleus-Win application, “nucWxp3.exe”, available from… http://www.nndc.bnl.gov/amdc/web/nubdisp_en.html (accessed 20 April 2011), which can create 2-D or 3-D displays of the nuclides. I am grateful to Dr. G. Audi at CSNSM for his assistance with the 3-D version of the NUCLEUS program. This and other plots of all the nuclides have been obtained as screen captures from the Nucleus-Win application, which uses the nuclear database NUBASE 2003. Actual figures, for nuclide numbers are 3179 known nuclides, of which 257 are stable (NUBASE 2003). I’ve given approximate figures in the text because the number of known nuclides is growing, and some unstable nuclides with very long half-lives are classified as stable. I'm treating stability as a relative property, in that all unstable nuclides will decay, but the more unstable they are, the shorter are their half lives. This might not be strictly correct, but it seems helpful in a context where every unstable configuration of nucleons follows its particular journey towards stability. The biggest stable nuclide is the isotope lead-208 (82p,126n) (NUBASE 2003). Databases, such as NUBASE, may set a time limit, such as 1 billion years (1 Gy), beyond which a nuclide is shown as stable. This plot shows bismuth-209 as the biggest stable nuclide, though it is actually unstable, with a very long half-life of 19 x 10 18 years. It also shows two more very long-lived nuclides well beyond the stable limit of 208 nucleons. All configurations of the next biggest cluster, 210, are distinctly unstable, the longest half life being 22 years, for (82p,128n) – see the interactive chart of nuclides. The three natural decay series of big unstable nuclei all end at a stable isotope of lead - see Bertulani, p.174, and my old copy of Roger Harrison's "Book of Data", though now a bit out of date, gives helpful diagrams. It's as if the each of the unstable nuclei takes the most direct route to stability, rather like a drowning swimmer will make for the shore - but this is maybe a bit fanciful. Another, interactive, nuclide chart is at… http://www-nds.iaea.org/relnsd/vchart/index.html (accessed 20 July 2010.)

..., 23/04/11,

Mackintosh et al, p. 49. Nuclear material is about 3 x 1014 times denser than water – Mackintosh et al, p.106. We’ll revisit this when we encounter neutron stars.


limit to the size of a stable cluster - the biggest has 82 protons and 208 nucleons, and is

about 14 fm, roughly 7 nucleons, across. Bigger clusters than this are found naturally and

can be artificially made, but they are all unstable.

moving around in nucleon-space

We need to be able to “move around” comfortably in nucleon-space. Vertical lines mark

clusters with the same number of neutrons, while horizontal lines mark clusters with the

same number of protons. The 45o diagonals are important lines that mark clusters of the

same size, where the total of protons and neutrons (Z + N) is constant. So the nuclides on

a diagonal line all belong to the same cluster “family”, having the same number of

nucleons, but different p:n ratios. The diagram shows the diagonal for cluster-200. Of all

the possible p/n combinations in this cluster, only one is stable – the nuclide (80p,120n).

We could draw 208 diagonals for all the clusters up to the largest stable nuclide.

stability is sensitive to p:n ratio

Up to the maximum stable size of 208, all cluster families behave similarly, in that stability

is sensitive to the p:n ratio. Shifting a cluster’s p:n ratio along the diagonal away from the

stable value in either direction makes it increasingly unstable, and consequently have a

shorter half life. The outermost nuclides are very unstable, with half-lives less than 0.1

seconds. This far out, the p:n ratio is so extreme that a nuclide may eject a proton or

neutron, so quickly that “the nucleus may not have a distinct existence before the nucleon

is emitted”.

the shifting p:n ratio and the arc of stable nuclides

In small clusters the stable ratio is very close to 1p:1n, shifting to 1p:~1.5n in the biggest

stable clusters. Having worked with the simple model, we can understand this shift.

Decreasing the p:n ratio (that is, having more neutrons than protons) will tend to stabilise

larger clusters, since converting pn will reduce the number of proton repulsions within

the cluster. But this can only go so far, for there comes a point where the benefit of

removing another repulsive proton is outweighed, literally, by the gain in mass due to the

heavier neutron. We thus have the “arc of stable nuclides”, that makes a gentle curve in

nuclide space.

cluster size and stability

A nuclide plot, with the nuclides colour-coded to show the average binding energy for

each nucleon is shown in figure 3.13.


..., 23/04/11,

NUCLEUS displaying binding energies.

..., 23/04/11,

Williams, p. 76. There are two drip-lines that can be drawn on the nuclide chart, one each for protons and neutrons, beyond which it is energetically favourable for the nucleus to eject a nucleon – Williams, p. 65, and Mackintosh et al, p. 73. We’ll see later that the drip-lines are important in nucleosynthesis in stars.

..., 23/04/11,

Bertulani, p.115.

..., 23/04/11,

Physicists use Z and N for the proton and neutron numbers, respectively - "N" is ok, but "Z"? The technical term is isobar, one more technical term, so I’ll use the more accessible phrase cluster-family.

..., 23/04/11,

Isotones have the same number of neutrons (vertical lines), isotopes have the same number of protons (horizontal lines), and isobars have the same mass number (A = Z + N) - the same total of nucleons in the cluster (diagonal lines)...most nuclear physics texts, eg Bertulani, p.116. We don't get into isotones. Isotope is familiar, and relevant to our atomic world, and I briefly introduce them at the end of section 3.7. A small note: the letter”n” in isotones helps us remember that the neutron number that is constant. The letter “p” in isotopes tells us it is protons that are constant. We are very much concerned with isobars, but I avoid introducing another technical term, and talk about cluster families instead, eg cluster-12, that has all the p/n combinations that total 12.

..., 23/04/11,

We can use the empirical nucleus size formula to get an idea of size - see back to the formula and the graph in section 3.5.2. A simple but useful example is marbles stacked into cubes. A 2x2x2 stack will hold 22 = 8 marbles; a 3x3x3 stack holds 27, A 6x6x6 cube holds 216 marbles.

..., 23/04/11,

Some particular proton-neutron combinations are not stable, for example, clusters of 5 or 8 nucleons are not viable with any p:n combination. Also, there are no stable nuclides with 43 or 61 protons, or with 19, 35, 39, 45, 61, 89, 115 or 123 neutrons – NUBASE 2003, half-life data.


Figure 3.13: Binding energies per nucleon for all nuclei. B/A = binding energy/number of nucleons, colour coding gives values in keV/nucleonThe colour coding shows the nuclide binding energies to form a long narrow ridge in

nuclide space. The maximum binding energy occurs at a cluster size of about 60, with a

p:n ratio of about 1:1. The binding energies steadily decrease as you move away from this

size and ratio.

the binding energy ridge

If we view this as a contoured geographical map, we can see a narrow ridge, with its

summit near the south-west end. If we start at the south-west end and follow the spine

along its length, we climb steeply to the summit, then have a long gentle descent to the

biggest clusters. These still have quite large binding energies, but the steady

accumulation of protons and their growing mutual repulsions eventually make the clusters

unstable, as we saw in figure 3.9.

In walking along the ridge we are changing the cluster size, and keeping the

proton:neutron ratio pretty much the same. But what if we walk across the ridge? Now we

are keeping the cluster size the same, and changing the proton:neutron ratio. Wherever

we cross, we'll have a short climb to the spine, and then a similarly short descent; the

binding energy rapidly peaks and then decreases.

the simple model checks out


the biggest stable cluster, where the attractive and repulsive forces are just balanced. This holds 208 nucleons and

is about 14 fm across.

unstable clusters - repulsive forces are bigger than attractive forces

the clusters with the biggest binding energies hold about 60 nucleons (~9-10 fm across), with a p:n ratio close to 1:1

the colour coding shows binding energy decreasing outwards from the

central black region

the spine of the binding energy "ridge"

following a constant sized cluster across the “ridge”

McNeil, 04/23/11,

This section picks up from the previous graph, and relates to the plot of binding energies, that comes later.


These nuclide plots confirm the inferences of the simple 2-D model of nuclei. Cluster

stability is sensitive to p:n ratio, with the stable nuclides in each cluster-family having the

greatest binding energy, and therefore the minimum mass-energy (prediction 1). The

binding energy per nucleon is greatest for clusters of a certain size, ~60 nucleons

(prediction 2). There is a maximum stable cluster size, which is 208 nucleons (prediction

3).

stable and unstable

The nuclide chart shows us that the vast majority of the known proton/neutron

combinations are unstable, either because of their p:n ratio, or because they are too big.

The simple cluster model explains this, but questions remain: what do unstable nuclides

do? And what decides that a nuclide will be stable? We will see that an unstable nuclide

has a limited number of available options, that a nuclide’s stability depends on its

neighbours, and that most stable nuclides are not all that they seem.

3.4.3 the nuclear valley

We’ve seen that the mass of a set of nucleons is less when they are bound in a cluster,

than when they are free, and that this mass loss is released as the binding energy. The

greater the mass loss, the more binding energy is released, and the more tightly the

cluster is bound. Thus the average mass of the nucleons in a cluster is a measure of how

tightly they are bound. If we take the 2-D nuclide plot of figure xXx, and incorporate the

average nucleon mass on the vertical axis, we get the 3D plot shown in figure 3.14. This

is the “nuclear valley”, in which all the known nuclides reside.


..., 23/04/11,

NUCLEUS – in 3-D mode. The “nuclear valley”, or “valley of stability” - a well-established way of viewing the nuclides, for example, Ray Mackintosh et al, ch.6, and George Marx. An article in the CERN Courier, shows an image of the heavy elements in the valley, with clouds overhead and the super-heavy elements as mountains in the distance. The nuclides are referred to as species - “Of the thousands of known nuclear species, only about 300 are stable, that is they exist along the so-called "valley of stability". The unstable species forming the valley "walls" - those with an overabundance of protons or neutrons - tend to decay quickly, sometimes within milliseconds.” – CERN Courier, Feb 22, 2002, at… http://cerncourier.com/cws/article/cern/28587 (accessed 29 March 2011)

McNeil, 23/04/11,

I’m explaining things in terms of average nucleon mass because this fits in with concept of the “nuclear valley”, as used by Mackintosh and Marx. I think the reader will relate relative nuclide stabilities to the nuclear valley terrain, and nuclear decay to “falling” down the mass-energy slope. Marx fig.3 shows a plot of the nuclear valley in terms of binding energy/nucleon. The 3D nuclear valley is the nuclear habitat, within which nuclides co-exist and compete. Any “move” down the mass-energy slope will release binding energy and lead to a more stable configuration. But we must be careful with the values of average mass/nucleon. The lowest nuclide in the nuclear valley is not the most tightly bound. The nuclide iron-56 has the smallest average nucleon mass (930.175 MeV/nucleon), closely followed by nickel-62 (930.187 MeV/nucleon). However, it is nickel-62 that is most tightly bound, having the highest binding energy per nucleon (8.795 MeV), followed by iron-58 (8.792 MeV), and then by iron-56 (8.790 MeV). The reason for this apparent disparity in order is that Ni-62 (28p,34n) has a slighly higher proportion of heavier neutrons than does Fe-56 (26p,30n), and so even though it has a higher binding energy its average nucleon mass is greater than Fe-56. So the way to release the maximum energy per nucleon is to bring 28 protons and 34 neutrons together into a nickel-62 nuclide. However, we can use the terrain of the nuclear valley to see if a particular nuclide decay is energetically favourable. We can easily do the energy “accounting” using nuclide masses - if there is a reaction that will re-configure a system of nucleons with a loss of mass, then that reaction is energetically favourable. For any reaction, calculations using mass or binding energy per nucleon give the same results. This is the 3D extension of the 2D Nucleus program, available from… http://www.nndc.bnl.gov/amdc/web/nubdisp_en.html You need a reasonable 3D graphics card in your computer to display it. Controls are… Q/A to increase/decrease altitude V/B to “step” left or right Home/End to zoom in or out, though these give extreme movements Page Up/Page Down to “look” up or down Cursor keys (/( to move forwards or backwards, and this can be done with the mouse scroll wheel Cursor keys (/( to turn your “look” left or right, and this can be done by clicking and dragging the reticule.


Figure 3.14: A panoramic view of the nuclear valleyfinding our way around the nuclear valley

Plotting the average nucleon mass on the vertical axis has transformed the flat 2-D

nuclide plot into the 3-D nuclear valley. The entirety of the valley is best seen from the

large nuclide end, with the small nuclides in the far distance. The valley is long and

narrow, with steep sides at the small nuclides end, widening in the middle, and becoming

more of a broad slope for the very large nuclides at the near end. The 2-D and 3-D plots

both show the nuclides colour coded by half life, and can be directly compared. Every one

of the roughly 3,200 cells in the valley terrain represents a known nuclide, with the cell

height representing its average nucleon mass.

Of all the possible proton/neutron combinations, these coloured cells represent the few

nuclides that exist long enough to be known and named, and of these, only a small

minority are stable. The valley is bordered by the highly unstable, short-lived nuclides.,

beyond which the nuclides “survive for such a brief period, they can hardly be said to exist


neutrons, N

protonsZ

mass per

nucleon

The long shallow mass per nucleon curve for all stable

nuclides. (black cells)

highly unstable nuclides with very short half lives, <0.1 seconds (red cells)

nuclides further from the arc of stability have bigger values for

mass/nucleon, and are higher up the valley slopes

the biggest stable nuclide –

cluster-208

the largest nuclides,

and highly unstable

the smallest nuclidesThe most stable nuclides, with the smallest mass per nucleon, are at the lowest point

in the valley, around cluster-60

the beta-decay curve for one cluster family

(~115 nucleons)add more curves

mass per nucleon

..., 23/04/11,

Mackintosh et al, p. 72.

McNeil, 23/04/11,

The nuclear valley resembles the skeleton of a wooden boat. The long shallow curve of the central keel corresponds to the arc of stability. The set of ribs that intersect with the keel correspond to the set of transverse curves for each cluster-family.


at all”. The terrain of the nuclear valley is the “habitat”, in which the nuclides co-exist and

interact.

the nuclear valley is defined by two curves

The shape of the nuclear valley can be visualised in terms of two curves - or rather, one

long curve that is intersected by a set of transverse curves. This resembles the skeleton

of a wooden boat, in which the long curving keel is intersected by a set of transverse ribs

(figure 3.15).

Figure 3.15: The curves of the nuclear valley superimposed on the skeleton of a boat. The 2-D plot can be seen as a projection of the 3-D plot on to a flat surface.The 2-D nuclide plot is shown as a projection of the boat skeleton, helping us to relate the

two views. Thus the arc of stable nuclides can be seen as the “keel” of the nuclear valley,

and the set of cluster-family curves then are the “ribs”. All nuclides (up to the 208 stable

maximum) lie on a rib, and the keel connects the lowest point of each transverse rib.

the arc of stable nuclides follows the valley bottom

The 2-D plot (figure 3.12) showed the stable nuclides as the black cells, lying on a long

curving arc in nuclide space. We can now see that this stable arc runs along the bottom of


neutrons

protons

mass/nucleon

line for ratio 1p: 1n

the keel: the arc of stable nuclides

2-D nuclide plotdiagonal for a cluster-familycurving arc of stability

the ribs: individual curves for the cluster families

McNeil, 23/04/11,

Mackintosh et al, p.100, comment on how the “energy valley…has an overall smooth structure which can be summed up in just three curved lines”, which fit moderately well with the lines of a simple 3-D boat design. I’m using the two curves of the boat, and leaving out the curving arc of stability in the 2-D nuclide plot, though this is shown in the figure - it may help some readers relate the 2-D and 3-D plots. It was not easy finding a suitable photo, even with Google image search. Finally I found this computer construction, at… http://i69.photobucket.com/albums/i64/Orinoco_guy/Tall%20Ships/BuildinganOceanGoingJunk1.jpg permission… Is this image going too far? Maybe, but I’ll include it in this first version.


the nuclear valley (figure 3.14), dropping steeply from the single proton at the far end, and

then rising slowly to end at the largest stable nuclide, cluster-208. The very lowest point

on this stable arc, at a cluster size of about 60 nucleons, represents nuclides with the

smallest average nucleon mass.

the cluster families traverse the valley

In the 2-D plot each cluster family, containing a fixed number of nucleons, lay on a straight

diagonal line, that crossed the arc of stable nuclides. In the 3-D view, we can see that

each cluster family traverses the nuclear valley in a U-shaped curve, dropping down the

slope on one side, crossing the stable arc at the bottom, and continuing up the other side.

Just about every cluster family, up to 208 nucleons, has one, maybe two, stable nuclides,

situated at or near the bottom of its U-shaped mass-energy curve. The stable nuclide(s) in

every cluster are those with the lowest average nucleon mass, that release the maximum

binding energy.

We can now see that the further a nuclide deviates from its cluster’s stable p:n ratio, the

higher it is up the valley slope, and generally its half life is shorter. There are no stable

nuclides to be found on the valley slopes.

the valley’s lowest point is around cluster-60

The arc of stable nuclides follows the valley bottom, with the lowest point at around

cluster-60, where the nuclides have the smallest average mass per nucleon. All the other

stable nuclides have more than this minimum value, and are therefore less strongly

bound, and so, strictly speaking, they are unstable.

an inconsistency?

We seem to have an inconsistency. If we look at a transverse curve for a single cluster,

the only stable configuration is the nuclide with least mass per nucleon. However, if we

look down the length of the valley, there is an almost unbroken series of nuclides carrying

excess mass that are stable. It looks as if an individual cluster can readily rearrange itself

to the stable lowest mass configuration, but there are constraints to clusters acquiring or

ejecting nucleons, and thereby changing size. In terms of the terrain of the nuclear valley,

individual clusters can internally rearrange themselves to descend transversely towards

the valley bottom, but the stable nuclides can’t gain or lose nucleons, and move towards

cluster-60 at the lowest point of the valley.

3.4.4 The nuclear binding energy and mass curves/tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

..., 23/04/11,

The standard treatment is to describe how unstable nuclides decay. But I think the important questions are: why are there stable nuclides with other than 60 nucleons? What are the impediments to decay of these, striclty unstable, nuclides? I’ll address these questions later.

McNeil, 23/04/11,

I will stick to this form of words, and avoid phrases like “the most stable nuclide”.


If we now select only the stable nuclides, and plot their average values for binding

energy/nucleon and mass/nucleon, then we get the two graphs in figure 3.16.

Figure 3.16: Graphs of binding energies (upper) and mass (lower) per nucleon for the stable nuclides. The figures in the box for lead-208 appear inconsistent, but this is due to rounding to the nearest whole MeV.mirror image graphs

Plotting the average values per nucleon means that the graphs are independent of

nucleon identity and cluster size. We have recently been introduced to these curves: the

upper graph is the profile of the binding energy ridge (figure 3.13), and the lower graph is

the profile of the bottom of the nuclear valley (figure 3.14).


0

1

2

3

4

5

6

7

8

9

bind

ing

ener

gy/n

ucle

on (M

eV)

930

931

932

933

934

935

936

937

938

939

0 20 40 60 80 100 120 140 160 180 200 220cluster size

mas

s/nu

cleo

n (M

eV)

hydrogen: 1pmass 938.27binding energy 0total 938.27

carbon-12: 6p,6nnucleusmass: 12 x 931.24 = 11,175binding energy 12 x 7.68 = 92total 11,267

Separate nucleons6 protons: 6 x 938.27= 56306 neutrons: 6 x 939.57 = 5637

total 11,267

lead-208: 82p,126nnucleusmass 193,687binding energy 1,636total 195,323Separate nucleons82 protons 76,938126 neutrons 118,386total 195,324

iron-56: 26p,30nnucleusmass 52,090binding energy 492total 52,582Separate nucleons26 protons 24,39530 neutrons 28,187total 52,582

helium-4: 2p,2nnucleusmass: 4 x 931.85 = 3,727.4binding energy: 4 x 7.07= 28.3total 3,755.7Separate nucleons2 protons: 2 x 938.27 = 1876.52 neutrons: 2 x 939.57= 1879.1

total 3,755.7

oxygen-16 (8p,8n) neon-20 (10p,10n) magnesium-24 (12p,12n)

deuterium: 1p,1nmass: 2 x 937.8= 1875.6binding energy:2x1.1= 2.2total 1877.81 proton 938.27

1 neutron 939.57

total 1877.8

..., 23/04/11,

All figures from the mass.mas03 database, selecting that are indicated as stable in NUBASE.. Upper graph The plot of binding energy/nucleon for the stable nuclides, taken directly from the mass.mas03 table. The binding energy per nucleon has more or less reached its maximum by the time cluster size is ~30. This tells us that cluster-30 is the biggest that the nuclear force can reach across. "...the nucleon-nucleon forces have a range much smaller than the nuclear radius" of cluster-30 - Bertulani, p.103. The simple formula tells us the radius of a cluster of 30 nucleons is ~3.7 fm See back to the charge density plot for different nuclei – calcium, with 40 nucleons, has a radius a bit less than 4 fm. Recall Smith’s statement that the nuclear force has a range of 3-4 fm. Lower graph The plot of average mass/nucleon for the stable nuclides, calculated from the mass.mas03 table as explained below. This diagram describes the assembly of nuclides from separate nucleons - so there are no electrons involved. The mass.mas03 table gives atomic data - that is, nuclei plus electrons - so the electrons have to be "removed" to get figures for the bare nuclei. Thus... Nuclear mass = atomic mass - mass of electrons Then… mass/nucleon = nuclear mass / number of nucleons. For carbon-12 (6p,6n), the atom is 6 electrons = 6x0.51 = 3.06 MeV heavier than the bare nucleus. Not a lot compared with the nucleus mass of 11,175 MeV. I think that involving electrons at this point would be a distraction. They would have to be added in to each calculation, and explain that the electron binding energies are small enough to ignore. In the narrative of the book we haven’t explained what atoms are yet. "Note that in NuDat masses are given in energy units." - from bnl web-site. Bertulani, figure 4.3, gives the binding energy plot for all known nuclei.


In the upper graph we can see the binding energy start at zero for a single unbound

proton, rise rapidly to a maximum of nearly 9 MeV/nucleon around cluster-60, and then

slowly decrease as the cluster size increases to the biggest stable nuclide. The lower

graph, for the average mass per nucleon, is the mirror image of the upper, since the

binding energy of the cluster is "paid for" by the nucleons' loss of mass (see section 3.2.3

on the mass-energy bank accounts).

We’ll now see how the mass-energy accounting works with carbon-12.

carbon-12 – doing the accounting for mass and binding energy

In carbon-12 (6p,6n), the average nucleon mass is only 931.24 MeV, so the cluster of 12

nucleons has a mass of 11,175 MeV, much less than the mass of the separate nucleons.

The mass difference of 92 MeV is the binding energy. Thus "we see that [carbon-12] is

considerably less massive than the sum of its twelve building blocks. The disappearing

mass is caused by the negative energy of the binding of those twelve nucleons together.

Einstein's E=mc2 explains the amount exactly !"

The “negative energy of the binding” is due to the attraction of the nuclear force. If we

want to separate the nucleons in the carbon-12 nuclide, we have to supply the binding

energy in order to restore the missing mass. The nucleus of the hydrogen atom is a single

proton, so its binding energy is zero. This is the only nucleus that suffers no mass loss.

The example of deuterium, that introduced the concept of binding energy, is now shown in

context, and there are also figures for other nuclides, with the working shown for helium-4

and carbon-12.

The general pattern for the bound nuclei is...

mass of bound nucleus + binding energy = total mass of separate unbound nucleons.

The helium-4 (2p,2n) nuclide is a very stable cluster, with a large binding energy, and so it

shows as a peak on the upper graph. There are nuclides that are multiples of He-4, that

also have relatively large binding energies, and these show as smaller peaks on the rising

binding energy curve. We'll see later how alpha particles play an important part in the

build-up of small nuclides, and also the decay of large ones.

energy “invested” in matter

If we see a nuclide as the result of the conversion of energy into matter, then the lower

graph shows how the relative “efficiency” of this conversion varies with cluster size. For


..., 23/04/11,

Donald Clayton, chapter on carbon-12.


helium-4 the conversion rate is about 932 MeV/nucleon. For the slightly better rate of 931

MeV/nucleon you can assemble clusters of about 20 or about 180 nucleons. But the

nuclide iron-56 has the lowest mass per nucleon of all the nuclides – we get a cluster of

26 protons and 30 neutrons at the bargain rate of only 930.175 MeV/nucleon. No other

nuclide offers a better deal.

mass per nucleon and binding energy per nucleon – a technical section

We’ve got to be a bit careful here, with binding energies and masses.

Average mass/nucleon

Iron-56 (26p,30n), 930.175 MeV

Nickel-60 (28p,32n), 930.181 MeV

Nickel-62 (28p,34n), 930.187 MeV

Binding energy/nucleon Nickel-62, 8.795 MeV iron-58, 8.792 MeV Iron-56, 8.790 MeV

So, if you want to release the maximum binding energy per nucleon, bring together 28

protons and 34 neutrons into a nickel-62 nuclide. If you want to have the least mass per

nucleon, then assemble 26 protons and 30 neutrons into an iron-56 nuclide. So nickel-62

is the most strongly bound nuclide, but iron-56 has the least mass/nucleon. Iron-56 has

slightly less binding energy, but has a smaller proton:neutron ratio, and so fewer of the

heavier neutrons, and so has the least mass per nucleon. Similarly, if you choose the

smaller apples from the available selection, then your average price per apple is less.

So, the nuclide iron-56 (26p,30n) sets the very lowest point in the nuclear valley, but it is

not the most strongly bound. However, the values of mass/nucleon for the different

nuclides define the nuclear valley, and allow us to determine whether a nuclear reaction is

energetically favourable. If a set of nucleons can be rearranged with a loss of mass, then

there will be a release of binding energy, and the process will be energetically favourable.

Wallerstein gives a nice example of this for two nuclides in the cluster-36 family: argon-36

(18p,18n) and chlorine-36 (17p,19n). The chlorine nucleus has a slightly larger binding

energy, but is slightly heavier, because of the larger proportion of heavier neutrons. Argon

should decay to the more tightly bound chlorine, but does not do so, because this would

involve a mass increase.

Iron-56 has commonly been described as the most tightly bound nuclide, but in fact it

comes (a very close) third in the binding energy competition. Yet iron-56 is more abundant

in the universe than nickel-62. We shall see that the creation of the nuclides in stars was

influenced by the ongoing nuclear reactions as much as nuclide stability.


..., 23/04/11,

Rod Nave, at… http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin2.html#c1 (accessed 11 April 2011)

..., 23/04/11,

Wallerstein, p.1060.

..., 23/04/11,

using data from the mass.mas03 table

..., 23/04/11,

This is like the most fuel-efficient car having the lowest figure for litres/kilometre – using the continental format, rather than the UK or American format of miles/gallon.


Iron-56 is a very important nuclide, that marks a threshold in the lives of stars and the

creation of the chemical elements. I shall describe it as the nuclide with the smallest

average nucleon mass. I shall follow the example of Williams who states, “The binding

energy per nucleon of nuclei is greatest near A = 60”, where A is the physicists’ symbol

for cluster size, and this will be the basis for our understanding fusion and fission

processes.

3.5 Cluster-12 - an individual cluster seeks stabilityWe will now look at an individual cluster to start to learn what unstable nuclides do, and

what decides if a nuclide is stable. We’ll take the small cluster-12 as representative for all

the clusters up to 208, the biggest cluster capable of stability.

3.5.1 the cluster-12 family in nuclide-space

Cluster-12, the “family” of 12 nucleons, lies down at the bottom left corner of nuclide-

space (figure 3.12). Figure 3.17 shows a detail of this region of nuclide-space, with the

cluster-12 family of nuclides lying along the diagonal line.


..., 23/04/11,

This and other data have been obtained from the NUCLEUS program, and the National Nuclear Data Centre: information extracted from the Chart of Nuclides database, http://www.nndc.bnl.gov/chart/ The assistance of Dr. Alejandro Sonzogni in the interpretation of data in the Interactive Nuclide Chart is gratefully acknowledged. A choice was needed here: The Nucleus program is excellent for overall views of all the nuclides, and it also connects with the 3-D view of the nuclear valley. The bnl nuclide chart is in some ways better for showing selected regions of nuclide space (it gives the nucleon numbers and the cells are simpler), but the colour coding is different, as are some of the nuclide classifications – for example, one will show a nuclide as an alpha emitter, and the other as mainly beta-decay. So I have used the NUCLEUS program for all views of nuclide space. The zoom setting determines how much information is displayed. I’ve usually kept the cells blank, to avoid distracting and unnecessary detail, but sometimes the nuclide identities are revealed, for example, 16O, which in the text would be O-16. The red border just indicates the selected nuclide at the centre of the display – it has no other significance. The two sources are in agreement on the main properties of the nuclides - there might be minor dfferences in the half life or the balance of decay modes. Data checked on the Interactive nuclide chart and with “Nucleus” – 10 June 2010.

..., 23/04/11,

We are in the world of nucleons and the ways they behave in nuclei take no account of our atomic world. Thus, for example, cluster (4p,8n) is an isotope of the metal beryllium, cluster (6p,6n) is carbon-12, and cluster (7p,5n) is an isotope of the gas nitrogen. If we think of these as nucleon clusters, then we can focus on what the nucleons are doing, without being distracted by the element names.


Figure 3.17: The cluster-12 mass-energy valley. The decay modes are colour coded as shown. The inset graph plots the atomic masses of the


0 1 2 3 4 5 6 7 8 9 10 11 12 neutron number, N

(8p,4n)decays by

ejecting a protonT1/2 < 10 -15 s

(7p,5n)decays by pn

T1/2 = 11 ms

(3p,9n)decays by ejecting

a neutronT1/2 < 10 -8 s

(5p,7n)decays by np

T1/2 = 20 ms(6p,6n)stable

prot

on n

umbe

r, Z

All nuclides in the cluster-12 family lie on this diagonal line so that in every case Z + N = 12.

the atomic mass of each nuclide along the cluster-12 diagonal is plotted in the graph

neutron-rich

proton-rich

(4p,8n)decays by np

T1/2 = 21 ms

11,170

11,180

11,190

11,200

11,210

11,220

11,230

4 5 6 7 8 9neutron number

atom

ic m

ass

The (6p,6n) combination has the lowest mass of the cluster-12 family

12

11

10

9

8

7

6

5

4

3

2

1

0

Arc of stable

nuclides

proton emissionbeta-plus decay and electron capture (pn)stable nuclide

beta-minus decay (np)neutron emissionalpha-decay (ejection of (2p,2n)


The nuclides in the cluster-12 family run from the very proton-rich (8p,4n) to the very

neutron-rich (3p,9n), with only the nuclide (6p,6n) being stable, and the inset graph shows

that this has the least mass of all the nuclides in the cluster-12 family.

the weak interaction

In the last chapter we encountered the weak interaction, mediated by the W and Z0

bosons, that interchanges protons and neutrons. We can summarise these nuclear

reactions as follows…

beta-minus (np) decay: n0 p+ + e- +

beta-plus (pn) decay: p+ n0 + e+ +

We saw in chapter 2 that this reaction is energetically favourable if it reduces the atomic

mass by more than 2 electron masses (2 x me = 1.02 MeV). If this is not the case, then the

transformation can be achieved if the nuclide captures an electron.

electron capture (effectively pn): p+ + e- n0 +

In this chapter, we will see that these beta-decay reactions are the means for an unstable

configuration of nucleons to move towards stability.

the beta-decays of the unstable nuclides

Nuclides with extreme p:n ratios eject a nucleon, to become a smaller cluster, and nearer

to the arc of stable nuclides. In less extreme cases, nuclides undergo one form of beta-

decay, transforming one nucleon type to the other, and moving one diagonal step towards

the stable ratio. Thus, “beta decay is the process by which complex nuclei return towards

the line of stability by emitting electrons or positrons, or by electron capture.”

the cluster-12 family traverses the nuclear valley

The cluster-12 family traverses the nuclear valley, and so a plot of the nuclide masses is a

cross-section of the valley at that point. The nuclides in the cluster-12 family bear out the

predictions of the simple model. A cluster has an optimum p:n ratio that minimises its

mass: too many protons, and the mass increases due to their electrical repulsions, but too

many of the heavier neutrons also increases the mass. Nuclides on either side of this

optimum ratio are unstable due to their excess mass, and undergo beta-decay to

interchange protons and neutrons and reduce it. Nuclides further from the stable (6p,6n)

ratio have bigger masses and shorter half lives.

a sequence of decay steps


..., 23/04/11,

The inset graph plots the atomic masses of the nuclides in the cluster-12 family, lying along the diagonal line – data taken from the AMDC “mas03” table. Note that an excess mass of only 10 MeV for an atom of mass about 11,000 MeV, that is about 0.1%, is enough to make the cluster unstable. In this section we’re looking at the behaviour of nuclei, without the accompanying electrons that would make them atoms. It might seem odd to use atomic masses, but we have to consider whole atoms when we look at radioactive beta-decay, and I want to be consistent. You can get just the nuclide masses from the “mas03” table, and they show the same patterns as the atomic masses, just being ~3 MeV less.

..., 23/04/11,

Williams, p. 278.

..., 23/04/11,

These are the nuclides’ major decay modes. Many of the cluster-12 nuclides have another, minor, decay mode. For example, nuclide (5p,7n) mainly undergoes beta-plus (p(n) decay, but also occasionally follows this by emitting an alpha particle (2p,2n) - ref. NUCLEUS. Such “beta-delayed particle emission”, where a cluster emits a nucleon or nuclide straight after beta-decay is not uncommon - see, for example, the glossary to the BNL Interactive nuclide chart… http://www.nndc.bnl.gov/chart/help/glossary.jsp


Unstable nucleon combinations transform themselves in a sequence of decay steps to

become the stable minimum mass nuclide. Nuclide (4p,8n) becomes (5p,7n), which

becomes the stable (6p,6n), each decay “moving” the configuration one cell down the

slope of the nuclear valley. The beta decay reactions enable an individual cluster to

reduce its mass and work towards stability.

Clusters very far from stability take a more violent course and eject a proton or neutron,

thus becoming a cluster one nucleon smaller, which then follows a similar decay

sequence.

3.5.2 the cluster-12 family in the nuclear valley

Figure 3.18 shows the view of the nuclear valley as seen from the high peaks at the end

with the smallest nuclides. The arc of stable nuclides runs along the valley bottom into the

far distance, with the proton-rich nuclides on the left, and the neutron-rich on the right.

Figure 3.18: Looking down to the bottom of the nuclear valley from the high peaks of the small nuclides. The decay of the cluster-12 family of nuclides can be seen in the context of the bigger pattern. Nuclides are colour-coded by decay mode.


(3p,9n)

(8p,4n)

(7p,5n)

(6p,6n)

(5p,7n)

(4p,8n)

unstable neutron-rich nuclides move towards

stability by transforming np (blue cells)

unstable proton-rich nuclides move towards

stability by transforming pn (yellow cells)

the nuclides with the least mass per nucleon, at the bottom of the valley

the dotted line shows the nuclides

in the cluster-12 family

eject a proton (pink)

eject a neutron (lilac)

proton number

neutron number


The dotted line shows the atomic mass curve for the cluster-12 family (see back to the

last figure). The stable configuration of this cluster has the lowest mass-energy, and sits

at the bottom of the "valley of stability". The nuclides to either side are like “boulders

perched up the side of the valley”, and they decay so as to reduce their larger mass-

energies, "falling" down the energy slopes towards stability.

boulders roll, water flows, nuclides decay

Every cluster-family has its own mass-energy curve, that traverses the valley and

intersects the arc of stability. The unstable nuclides interchange protons and neutrons,

and move in a series of beta-decays downhill towards the arc of stability at the valley

bottom. Boulders roll, water flows, nucleon clusters decay – all seeking the state of lowest

energy.

energy invested in matter

We have seen that matter is created from energy, according to the equation…

E = mc2 or rather, m = E / c2

Thus energy is converted into matter, and in a sense, energy is incorporated or invested

in matter. The valley of stability for cluster-12 shows the “efficiency” of converting energy

to 12 nucleons of matter. The most efficient configuration of matter is as the cluster

(6p,6n), anything else has a greater mass. Thus the unstable cluster-12 nuclides decay to

reduce their mass, and thereby increase efficiency of the energy to matter conversion.

3.6 An inner structure to the nuclear clusterWe’ve seen that the average mass per nucleon of a cluster is determined by its p:n ratio

and its size, and in explaining this, we have regarded nucleons in the cluster as if they are

all mixed up, like marbles in a string bag, or like molecules in a drop of liquid water. But

there is evidence that indicates that nuclei have an inner organisation and structure, that

favours configurations with pairs and “magic” numbers of nucleons.

3.6.1 nucleon pairs - clusters-100 and -101

It has been found experimentally that like nucleons “pair up” in the nucleus, so two

protons are always more strongly bound than a single proton – and the same goes for

neutrons. The greater stability of nucleon pairs affects the stability of clusters bigger than

~20. We’ll look at two examples – cluster-101 and cluster-100.

cluster-101/tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

..., 23/04/11,

Williams, ch.5, uses these two examples. I’ve given only proton and neutron numbers, and omitted the usual way of naming these clusters, which is by naming the element. Thus the stable combination (44p,57n) is called ruthenium-101, or Ru-101. This would just add another level of difficulty for the reader, so I’ve left these element names out. I’ve taken nuclear data from the mas03 table, and plotted it in EXCEL. The atomic mass values differ slightly from those in Williams, probably because he the semi-empirical mass formula, rather than actual data. The decay details are from the “Nucleus” program working with the NUBASE data. As with C-12 I've plotted the values of masses of atoms rather than nuclides. Plotting nuclides would give very slightly smaller masses - ~22 MeV less - and show the same pattern, so I'll stay close to the atomic mass data in the mas03 table. Bertulani compares nuclei with odd and even nucleon numbers – p.122. Mackintosh et al show the mass-energy valley for cluster-137 – p. 72,101.

..., 23/04/11,

Williams, p.58.

..., 23/04/11,

For description of the different models of the nucleus see Bertulani ch. 5, and Williams, ch. 4 and 8. There are two types of models. The collective models treat the nucleons like water particles in a liquid drop, and describe the nucleus as a whole. The independent particle models, also called shell models, treat the nucleus as having an inner structure. Each type of model is useful in explaining particular nucleus properties – Bertulani, ch.5. Williams notes that “ the shell model and the liquid drop model are so unlike tha it is astonishing that they are models of the same system” – p.155. He also writes, “nuclei are complicated many-body objects held together by poorly understood forces” – p.157 – so it’s remarkable how much can be explained from these diverse models. I’m only going far enough to show the nucleus as more than an amorphous aggregation of nucleons, like marbles in a bag. Tiny as it is, the thing has an inner structure, that has subtle effects on the stability of the whole. The shell model of the nucleus relates to the shell model of electron orbitals in atoms, and we’ll look at this in ch. 4.

..., 23/04/11,

I’ve used the units MeV for amount of matter, though we know that the strictly correct unit is MeV/c2.

..., 23/04/11,

This will have been introduced in chapter 0. This is usually treated as the equivalence of energy and matter, but if we think of it more as energy being incorporated, or invested in matter, then it can perhaps help explain why the stable systems are those with the least mass, and that decay processes are those that reduce the system’s mass.

..., 23/04/11,


..., 23/04/11,

Mackintosh et al use the term “energy per nucleon” – p. 73/4. We’ll see later that there are cluster families that have two stable combinations. This is because the stability of the nucleus is sensitive to whether the proton and neutron numbers are even or odd - see Williams, p.73.


Figure 3.19 shows the mass of different p:n combinations in cluster-101.

Figure 3.19: The mass-energy curve for cluster-101 has only one minimum – one stable p:n ratioThe pattern is similar to what we saw for cluster-12. There is one stable p:n combination,

with the smallest mass. On either side of this the masses increase, due either to the

electrical repulsions between protons on the left, or to the heavier neutrons on the right.

Neutron-rich combinations decay by the beta-minus np reaction. Proton-rich

combinations can convert a proton to a neutron, either by the beta-plus pn reaction, or

by electron capture. We’ve seen that if the atomic mass difference is bigger than twice the

electron mass, then both of these reactions can occur, and this applies to nuclides

(47p,54n) and (46p,55n). However, nuclide (45p,56n) can only decay to (44p,57n) by

electron capture, since the atomic mass difference is only about 0.6 MeV.

Cluster-101 has an odd number of nucleons, so whatever the p:n combination, one of

them will be odd, and one even. These are shown as “oe” and “eo” under the graph. Thus

there is one unpaired nucleon in every combination.

cluster-100

Now compare this with cluster-100, whose atomic mass graph for the different p:n

combinations is quite different (figure 3.20).


93,992

93,993

93,99493,995

93,99693,997

93,998

93,99994,000

94,001

94,002

54 55 56 57 58 59 60neutron number

atom

ic m

ass

(MeV

)

Proton number 47 46 45 44 43 42 41odd or even? oe eo oe eo oe eo oe

Only one combination is stable

Decay by np

large mass losses, decay by electron capture and

pn reaction

small mass loss, decay only by

electron capture

all proton and neutron

numbers are odd-even

proton-richpn reaction or electron capture

neutron-rich

np decay


Figure 3.20: The mass-energy curve for cluster-100 has two minima - but only one truly stable p:n ratioThis cluster has an even number of nucleons, so the proton and neutron numbers are

either both odd, “oo”, or both even, “ee”. In the broad valley of stability the alternation

between “oo” and “ee” affects the cluster mass, so there are now two minima in the curve

– the main one at (44p,56n) and a second at (42p,58n). The first is truly stable - this is the

combination with the lowest mass-energy. The second must surely be stable too, after all,

the nuclides to either side have larger masses. But if this nuclide (42p,58n) could

somehow transform two neutrons to protons at the same time, then it could become the

nuclide (44p,56n). And this nuclide is in fact unstable, though it is often shown as stable,

having a very long half life of of 8.5 x 1018 years, and its decay mode is by a double np

decay.

Thus cluster-100 is only truly stable in one combination, (44p,56n), but effectively stable in

another, (42p,58n). The combination (43p,57n), in between these two, has two possible

decay reactions, one giving a much bigger reduction in mass than the other. The main

decay is by the np reaction, losing a neutron, and about 3.2 MeV of mass. But in about

about 2 in 1,000 decays it can “move” the other way and capture an electron, and thus

lose a proton, but now only lose about 0.2 MeV of mass.

clusters 100 and 101 in the nuclear valley


proton-richpn reaction or electron capture

neutron-rich

np decay

93,060

93,062

93,064

93,066

93,068

93,070

93,072

93,074

53 54 55 56 57 58 59 60neutron number

atom

ic m

ass

(MeV

)

Proton number 47 46 45 44 43 42 41 40Odd or even? oo ee oo ee oo ee oo ee

proton and neutron numbers are either

“oo” or “ee”

electron capture

only

two decay routes...by electron capture (~2 in 1,000 decays)or by np (~100%) not completely

stable...half life 8.5 x 1018 y, decay by double np

to (44p,56n)

the only truly stable combination

..., 23/04/11,

or 8.5 exayears, where 1 exayear is 1 x 1018 years. Bertulani gives the half life as 7 x 1018 years – close. This nuclide is sometimes shown as stable, for example, in NUBASE and the Interactive Nuclide Chart. Several nuclides undergo double beta decay – Bertulani, p.211.


If we position ourselves in the nuclear valley around cluster-90, looking in the direction of

the heavy nuclides, we see the view shown in figure 3.21.

Figure 3.21: The curves for clusters 100 and 101 nuclide families are in the foreground, with the mass peak for the biggest nuclides in the background. Nuclides are colour-coded by half life.In the middle distance the end of the black line marks the largest stable nuclide, and

beyond that is the summit of the largest known nuclides, marking the end of the valley. In

the foreground are the nuclides in clusters 100 and 101. The curve for cluster-101 has

one true minimum, and one truly stable nuclide. The curve for cluster-100 has a little

bump in it, with a minimum on either side. Thus there is one truly stable nuclide (44p,56n),

and the other has such a long half life that it is marked here as stable. Thus the slight

effects of nucleon pairings deep in the nucleus reveal themselves in the terrain of the

nuclear valley.

Patterns of stability

If we extend what we’ve seen with cluster-100, this suggests that clusters with an even

number of nucleons can have more than one stable combination. This is true, and we can

now see why there are more than 208 stable nuclides. Stable nuclei with even numbers of


the cluster-100 curve has two minima – one truly stable nuclide, and one with a very long half-life

the cluster-101 curve has only one minimum - one true stable nuclide

44p,56nstable

44p,57n42p,58n

very slow decay by double np reaction to 44p,56n

43p,57nhas 2 modes of decay to a

lower mass nuclide

the largest stable nuclide, lead-208

the peak of the largest nuclides, marking one

end of the valley

..., 23/04/11,

Bertulani, p.116, gives 284 stable nuclei, covering 83 elements – so he’s including bismuth.

..., 23/04/11,

Williams, p.74, and Bertulani, p. 121.


protons are more numerous than ones with odd numbers, and the same is true for

neutron numbers. Consequently, stable nuclides with even numbers of nucleons, “oo” and

“ee”, are more numerous, and the latter even more so. Conversely, it looks as if there may

be no stable odd-odd “oo” nuclides, and this is nearly true – the only exceptions are four

light nuclei, each with less than 20 nucleons: (1p,1n), (3p,3n), (5p,5n) and (7p,7n).

3.6.2 Magic numbers

Nucleon pairing is part of a bigger pattern, for it has been found that nuclides with certain

numbers of protons or neutrons, commonly called magic numbers, are more tightly bound

than neighbouring nuclides. The existence of these magic numbers can only be explained

by seeing the nucleus as having a comprehensively organised structure, where nucleons

arrange themselves in concentric shells, rather like the spherical layers of an onion, and

quite unlike a liquid water drop. The “magic numbers are a signature that the nucleons lie

in simple orbits”. The "magic" nucleon numbers which fill an inner shell and confer extra

stability on a cluster are:

2, 8, 20, 28, 50, 82 and 126

and they apply to both proton and neutron numbers separately (figure 3.22).


the largest truly stable nuclide, lead-

208 (82p,126n) is doubly magic

neutron number, N

prot

on n

umbe

r, Z

helium-4 (2p,2n) is doubly magic

the doubly magic nuclide (50p,82n)

there tend to be more stable nuclides with magic proton or neutron numbers

2 8 20 28 50 82 128

82

50

28

20

82

magic numbers: protons… neutrons

magic proton numbers

..., 23/04/11,

The 2D NUCLEUS chart, with the basic magic numbers added, according to Williams, p.131, and the bnl Nuclide chart.

McNeil, 04/23/11,

Williams summarises the evidence for magic numbers, and some of the consequences - ch.8. See also Mackintosh et al, p. 74 and 103. These numbers can be predicted using quantum mechanics - Frank Close, "The New Cosmic Onion", p. 43. See also… http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/shell2.html The bnl Interactive nuclide chart marks the magic numbers 8, 20, 28, 50, 82, 126. NUCLEUS gives a more thorough treatment, which goes further than I want to here, and gives a different set of magic numbers lines (see the NUCLEUS help file). It would be easy to use the bnl interactive nuclide chart, with its simple set of magic numbers (though it omits the number 2). But I want to keep to one nuclide chart for consistency, so I have added the magic number lines to the NUCLEUS chart. The nuclear magic numbers are analogous to the magic numbers of atomic physics (2, 10, 18, 36, 54, 80 and 86) which are due to the way electrons are arranged in “shells” in atoms. We’ll look at this in ch.4.

..., 23/04/11,


..., 23/04/11,

Bertulani, p.122.

..., 23/04/11,

Respectively, hydrogen-1, lithium-6, boron-10 and nitrogen-14 – Bertulani, p.116, and look at the NUBASE nuclide chart with different parity settings - the oo setting, for example, shows no stable nuclides at all beyond nuclide (7p,7n).

..., 23/04/11,

Bertulani, p.116.


Figure 3.22: The magic shell numbers in the chart of nuclidesA cluster with a full shell of either type of nucleon is more tightly bound, and the nuclide

chart shows that there tend to be more stable nuclides when the proton or neutron

number is magic. The tightly bound helium-4 (2p,2n) nuclide is doubly magic, as is the

largest stable nuclide lead-208 (82p,128n). Elements with magic numbers of protons or

neutrons are more abundant on Earth than elements with similar, but non-magic, nucleon

numbers.

magic numbers 50 and 82

The larger binding energies of nuclides with magic numbers of nucleons means that their

average nucleon masses are slightly smaller, and we can discern this in the 3-D nuclear

valley. Figure 3.23 shows a region of the nuclear valley where two magic numbers, 50

and 82, intersect at nuclide (50p,82n).

Figure 3.23: The intersection of the magic numbers 50 and 82 in the nuclear valley. Nuclides are colour coded by decay mode. Compare this with the 2-D nuclide plot.All the nuclides on the red dotted line have 50 protons, and those on the green line have

82 neutrons, and each of the intersecting rows of cells makes a slightly uneven step in the


nuclide (50p,82n)

proton number

neutron number

All clusters along this line have the magic number 50 protons. Hence, thay are more stable, have less mass/nucleon, and so

make a small step in the local slope.

all clusters along this line have the magic number 82 neutrons

arc of stable nuclides

82

50

..., 23/04/11,

It’s not easy to show the magic numbers. This is the best example I found.


overall valley slope. As with nucleon pairings, the presence of a closed shell of nucleons

in the magic number nuclides alters the topography of the nuclear valley.

3.6.3 Super-heavy clusters and the "Island of Stability"

At the end of the arc of stability is cluster-208, which is doubly "magic", containing 82

protons and 126 neutrons. The next "magic" number is 184, and so it is believed that

there may be an "island of stability" centred on a cluster with 184 neutrons and 114 or 126

protons - there is some uncertainty which of these is the "magic" number in clusters of this

size. Various clusters containing 114 protons have been produced, which are

comparatively stable for clusters this big. Such super-heavy clusters can be produced by

colliding selected smaller clusters together. For example…

cluster-48 (20p:28n) + cluster-244 (94p:150n) cluster-289 (114p:175n) + 3n ejected

The product of this collision had a half life of about 2.6 seconds, before it decayed by

emitting an alpha particle. This is clearly unstable, but the cluster is 9 neutrons short of

the magic 184. The doubly magic cluster-298, comprising 114p:184n, is predicted to be

more stable, with a half life of around 17 days. This kind of work takes patience: in some

experiments it's taken around 5 billion billion collisions to produce a single super-heavy

nuclide.

3.7 Fusion and fission – changing cluster sizeSo far, we have focussed on individual cluster-families, each lying on a transverse curve

that crosses the nuclear valley. We have seen how the nucleons are interchanged by

beta-decay reactions to minimise the cluster’s mass-energy, and arrive at a stable

configuration. Now we step back and widen our view to consider all the nuclides along the

length of the valley, and look at how clusters gain or lose nucleons in the search for

stability.

There are two basic nuclear processes that change cluster size. Fusion is cluster growth,

where two clusters merge into a single bigger cluster. Fission is the splitting of a cluster

into smaller fragments. We have seen how the nucleon mass curve shows that the most

energetically favourable cluster size is about 60 nucleons - figure 3.24.


McNeil, 23/04/11,

This is the basis of the generation of electrical power by nuclear fission of heavy nuclei like uranium, and by the fusion of light nuclei, like deuterium (1p,1n) and tritium (1p,2n). We’ll look at fusion processes, since these power the stars that create the elements of our atomic world. “The stars are machines which convert hydrogen into heavier elements” - Williams, p.128.

McNeil, 04/23/11,

Emsley: Element 114, p. 467.

McNeil, 04/23/11,

Wikipedia… http://en.wikipedia.org/wiki/Ununquadium

McNeil, 23/04/11,

http://en.wikipedia.org/wiki/Ununquadium (accessed 13 April 2011).

McNeil, 04/23/11,

(1) Philip Ball, "The Elements", p.115, (2) John Emsley, Element 114, p. 467, in the chapter "The transfermium elements", (3) Mackintosh et al, p.80, and (4) Wikipedia, more recent (Mar 2002), and has a nice diagram of the proposed "Island of Stability". http://en.wikipedia.org/wiki/Island_of_stability (accessed 13 April 2011). NUBASE 2003 shows two isotopes of element 116, way out on their own.

..., 23/04/11,

It seems that the standard magic numbers (2, 8, 20, 28, 50, 82, 126) may not apply to some unstable nuclei, and this could cast some doubt on the validity of the nuclear shell model… http://www.fynu.ucl.ac.be/librairie/ocde/rapportocde/lr_oecd_vs16_2.htm (accessed 13 April 2011).


Figure 3.24: Clusters of nucleons can reduce their average mass/nucleon by converging on the minimum at cluster size of 60. Small clusters grow by fusion; large clusters fragment by fission. The box gives the energy “savings” from the fusion of small clusters into bigger ones.For small clusters, less than about 60 nucleons, the average nucleon mass decreases as

the cluster grows, up to about 60 nucleons. So, it is energetically favourable for a small

cluster to undergo fusion and grow, since it will reduce its average nucleon mass, thereby

releasing binding energy. For clusters larger than about 60 nucleons, past the minimum

mass, the situation is reversed. The average nucleon mass is now increasing with cluster

size, so now it is energetically favourable for large clusters to be smaller, and to down-

size by fission.

We’ll now look at how the processes of fusion and fission operate, and see what are the

constraints on them.

3.7.1 fusion - the energetics of cluster growth

The inset box in figure 3.24 gives some specific examples of the energy savings in a

sequence of fusion reactions. Combining the three helium quartets into a single cluster of

a dozen “saves” 7 MeV of mass-energy, which is released as binding energy. Bringing

two such clusters of twelve together “saves” another 14 MeV. Fusion is energetically

favourable only up to a cluster size of about 60. Further enlargement beyond that

increases the average nucleon mass, thus requiring energy, rather than releasing it.


930

931

932

933

934

935

936

937

938

939

0 20 40 60 80 100 120 140 160 180 200 220cluster size

mas

s/nu

cleo

n (M

eV)

1 x helium-4(2p,2n) = 3727.4 MeV

3 x helium-4(2p,2n) = 11,182 MeV1 x carbon-12(6p,6n) = 11,175 MeV saving = 7 MeV

2 x carbon-12(6p,6n) = 22,350 MeV1 x magnesium-24(12p,12n = 22,336 MeV saving = 14 MeV

fusion of small clusters <~60 reduces the

average mass/nucleon

fragmentation of large

clusters >~60 reduces the

average mass/nucleon

..., 23/04/11,

We can take this simple approach as far as calcium-40 (20p,20n), the last stable nuclide with a p:n ratio of 1:1.


We will see shortly how a sequence of fusion reactions such as these power the stars,

and in so doing create many of the elements of our atomic world.

small clusters are isolated by their mutual repulsion

Clusters of less than ~60 nucleons are, strictly speaking, unstable, since it is energetically

favourable for them to enlarge, at least up to a size of about 60. Yet our world is largely

made of atoms with these small nuclei - such as carbon, oxygen, magnesium - that have

been stable for billions of years. Why have these small clusters not all aggregated into

clusters of 60 nucleons?

Nucleon clusters carry large positive electrical charges, due to the protons they contain,

and so they repel each other strongly. It is only at enormously high temperatures - tens of

millions of degrees and more – that small clusters have enough kinetic energy to

overcome this repulsion, and come together so their nucleons can mingle into a single

larger cluster. At lesser temperatures, anywhere outside the interior of a star or a particle

physics experiment, clusters never come together, but are forever isolated by their mutual

repulsion.

Thus nuclear fusion is a reaction that “wants” to happen, but can’t. What about nuclear

fission?

3.7.2 fission - the energetics of cluster splitting

the spectrum of nuclear fission reactions

There is only one way for nuclear clusters to merge and undergo fusion, but for fission,

there is a “spectrum of possibilities in which a heavy nucleus breaks into two (or more)

parts…at one end of the spectrum the result is one small part and one large part, at the

other it is two almost equal parts.” - figure 3.25.


..., 23/04/11,

Following Williams, p.78, I’ve presented the nuclear fission modes as a continuum of fragmentation,and added the ejection of a nucleon to the spectrum. However, they involve quite different mechanisms: alpha-decay and cluster-decay both involve quantum tunnelling, while spontaneous fission is usually described in terms of the liquid drop model. It’s not easy to find good diagrams of nuclei on-line. These nuclei have been extracted from diagrams in Mackintosh et al, available on-line at… http://pntpm3.ulb.ac.be/pans-info/site/html/activities/ (accessed 13 April 2011) I’ve tried to show the range of fission processes as simply as possible, so the diagram doesn’t show the recoil of the parent nucleus. This is spontaneous fission, not initiated by neutrons.

..., 23/04/11,

Williams, p.78.

..., 23/04/11,

Hydrogen “burning” starts around 10 million degrees, and helium nuclei will fuse at around 100 million degrees _ Williams, p.347/8. We’ll go into this later in the sections on stellar nucleosynthesis. The particles in cosmic rays may have enough energy to fuse with a nucleus.

..., 23/04/11,

The mass curve above, or more usually the binding energy curve, is commonly shown to explain how the iron-group nuclei, with ~60 nucleons, are the most tightly bound, and therefore the most stable. It was only in the autumn of 2010 that the thought occurred, to look at things the other way round, and ask the question: why haven’t the other nuclei decayed to become iron-group nuclei?


Figure 3.25: the range of nuclear fission processesUnstable nuclides can decay by ejecting a single nucleon (a), or a small cluster (b) and

(c), or by splitting into two clusters of comparable size (d). The ejection of a quartet of

nucleons, (2p,2n), is a common mode of decay among larger nuclides. This quartet is the

nuclide helium-4, also known as an alpha particle, so this is commonly called alpha-

decay. A few nuclides that undergo alpha-decay also eject larger nuclides, such as

carbon-14 (6p,8n) or neon-24 (10p,14n) - this rare mode is known as cluster decay.

the pattern of decay of the known nuclides

We’re ready now to view all these decay processes in the nuclide chart, now colour-coded

by decay mode (figure 3.26).


(b)…by ejecting a He-4 nucleus (2p,2n)

alpha-decay

(c)…or by ejecting a small nucleus like Ne-24, C-14

cluster decay

(d)…or by splitting into two smaller nuclei, and some free neutrons

spontaneous fission

a nucleus can decay

by fission…

(a)…by ejecting a single proton or

neutron…

McNeil, 23/04/11,

Cluster decay is not often mentioned – Williams mentions it on p.78 and 96, and Bertulani, Q.4 p.194. See… http://en.wikipedia.org/wiki/Cluster_decay (accessed 11 April 2011) …in which the data broadly agree with NUBASE 2003. The ejected nuclei, such as C-14 and Ne-24, don’t have to be stable. Cluster decays usually have very small branching ratios – they are very rare events compared to alpha-decay. For example, radium-223 emits one C-14 nucleus for every billion alpha particles (reference 1 in Wikipedia: “Cluster decay” and Williams, p.96). Perhaps this comparative rarity is because it is even less probable that larger numbers of nucleons come together in the right configuration to tunnel out of the parent nucleus. Before writing this section I knew only of alpha-decay and spontaneous fission. In trying to explain how the alpha particle’s large binding energy made it suitable for ejection, I realised that there were other even more tightly bound nuclei nearby (see Williams, p.78) A check with cluster-180, showed that it was energetically favourable for cluster-180 to down-size to cluster-168 by ejecting a carbon-12 nucleus. A check with NUBASE 2003 found several heavy nuclei that undergo “heavy cluster emission”. I wonder if a large nucleus is always surrounded by a cloud of smaller nuclei that have tunnelled through the potential energy well, but only those that can reduce the mass of the system are actually ejected. Can we imagine a nucleus surrounded by a cloud of single nucleons and small clusters, like He-4, C-12, up even to Mg-24, and all sorts of clusters in between. Only where the ejection brings a reduction in mass is the decay energetically allowed, and the particle can then "materialise" outside the parent nucleus. These particles would be a bit like the cloud of pions round a nucleon, except they are not created from the quantum vacuum, and not involved in binding interactions.


Figure 3.26: The major decay modes of the known nuclides, colour coded as shown in the diagram.We can see the arc of stable nuclides, curving through nucleon-space from a single

proton up to the maximum cluster size of 208 nucleons. On one side of this arc the

proton-rich nuclides seek stability by the beta-plus (pn) decay reaction; on the other

side, decay is by the beta-minus (np) reaction. The beta-decay reactions are important

decay modes for nearly all clusters, up to about 250 nucleons. Decay by the ejection of a

single nucleon is associated with highly unstable nuclides, with extreme proton/neutron

ratios, that are situated at the edge of the nuclear valley. Alpha-decay becomes a

significant decay mode for clusters bigger than ~145 nucleons, especially if they are

proton-rich. Spontaneous fission is confined to the massive nuclides, way beyond the

maximum stable cluster size. The one process that does not appear on this chart, of

course, is fusion.

This chart shows only the major decay modes of each nuclide. We shall see that many

unstable nuclides, especially the larger ones, decay by more than one reaction, as the

following example shows.


arc of stable nuclides (black cells)

beta-plus (pn) decay of unstable proton-rich nuclei

(orange cells)

beta-minus (np) decay of unstable neutron-rich nuclei (blue cells)

the diagonal for cluster-145

ejection of a neutron (purple cells)

ejection of a proton

(red cells)

alpha-decay (yellow cells)

spontaneous fission

(green cells)

largest truly stable cluster-208

stable nuclidebeta-plus decay and electron capture (pn)beta-minus decay (np)

proton emissionneutron emissionalpha-decayspontaneous fission

McNeil, 23/04/11,

How can there be nuclei on Earth bigger than the largest stable nuclide? The heaviest naturally occurring nuclide is uranium-238 (92p,146n). It is unstable, and hence radioactive, but it occurs naturally because its half life is about the same as the age of the Earth, 4,500 billion years, while heavier nuclides with shorter half lives have disappeared (Bertulani, p.395). Uranium half life: NUBASE2003.

..., 23/04/11,

Note that in this graph, the biggest stable nuclide is lead-208 (82p,126n), holding 208 nucleons. Bismuth-209 is shown as unstable being an alpha-emitter, though its half life is so long that when the nuclides are plotted by half life, it is colour-coded as stable.


the three decay modes of nuclide (77p,90n)

Figure 3.27 shows nuclide (77p,90n), situated high up on the proton-rich slope of the

nuclear valley.

Figure 3.27: the decay choices open to nuclidesThis highly unstable nuclide (it has a half life of only 35 ms) has three available modes of

decay: it can fission, by ejecting either (1) an alpha-particle or (2) a proton, or it can (3)

transform one proton to a neutron (beta-plus decay).

the energy accounting of fission

We have seen that fusion is energetically favourable if merging two smaller clusters into

one larger one reduces the total mass. Similarly, fission is energetically favoured if the

total mass of the fragments is less than the mass of the initial cluster - that is, the total

mass of all the nucleons is reduced by the rearrangement. The initial decaying nuclide is


(2) eject a protonmass-energy loss939 MeV > mass of free protonprobability: 32%

(3) convert p nmass-energy loss = 9 MeVprobability: 20%

(1) eject an alpha-particlemass-energy loss= 3734 MeV > mass of alpha particleprobability: 48%

the nuclide (77p,90n) has 3 modes of decay…

the nuclide (76p,90n) can only eject an alpha-particle (72%

probability) or convert pn (28%).It cannot eject a proton, since this would decrease its mass-energy by only 937 MeV, less than the

mass of a free proton.

87 88 89 90 91 92 neutrons

77

76

75

74

73

prot

ons

atomic masses (MeV):Ir-167 (77p,90n): 155,542Os-166 (76p,90n): 154,603Re-163 (75p,88n): 151,808Os-167 (76p,91n): 155,533Re-165 (75p,90n): 153,666

proton ejection: 32%Ir-167 - Os-166155,542.4 - 154,602.6 = 939.8

alpha decay: 48%Ir-167 - Re-163155,542 - 151,808 = 3734

beta-plus decay: 20%Ir-167 - Os-167155,542 - 155,533 = 9

Os-166…alpha .72…beta= 0.28…no proton NUBASE +nndc

..., 23/04/11,

Detail from the 2-D NUCLEUS nuclide chart, and mass data and decay probabilities from the mas03 database and the nuclide chart… http://www.nndc.bnl.gov/chart/ Nuclide (77p,90n) is, of course iridium, Ir-167. The elements are little known heavy metals – I don’t see how it would help the reader to have them named. I refer to nuclides in the text, but these decays are of atoms – nuclides plus their complement of acquired electrons, so the figures are for atomic masses. We’ve not yet reached the atomic level of the emergent hierarchy, so I write of nuclides, though this is not really correct.


commonly called the parent nuclide, and the smaller nuclide that is produced is called the

daughter.

Thus, proton ejection is energetically favourable if…

mass of parent nuclide > mass of daughter nuclide + mass of proton (938 MeV)

Similarly, for alpha-decay to be viable the condition is...

mass of parent nuclide > mass of daughter nuclide + mass of alpha particle (3,727 MeV)

So if the mass difference between the parent and daughter nuclides is bigger than the

mass of the fragment to be ejected, then that mode of decay is energetically favourable.

It is energetically favourable for nuclide (77p,90n) to undergo alpha-decay, because the

mass of the daughter nuclide is less by 3734 MeV, which exceeds the mass of an alpha

particle. Similarly, proton-decay is favourable because the the 939 MeV loss exceeds the

proton mass. And finally, the nuclide can undergo the familiar beta-plus decay,

transforming pn.

a balance of probabilities

The configuration of nucleons in the parent nuclide (77p,90n) thus has a choice of three

ways to reduce its total mass - three decay routes across the nuclear valley terrain, all

going downhill. The balance of probabilities is such that of 100 parent nuclides, about 48

will undergo alpha-decay, 32 will eject a proton, and the remaining 20 will undergo beta-

plus decay. If we piled one million parent nuclides on the cell (77p,90n) of nuclide space,

like casino chips, then within a second they would be all gone, and a million new daughter

nuclides would appear on the three nearby cells.

What then happens to the daughter nuclides? We’ll look at one example, the daughter

nuclide (76p,90n). The alpha and beta decay modes are available to this nuclide, but

since ejecting a proton would reduce its mass by only 937 MeV, just less than the mass of

a free proton, this mode of decay is not energetically viable (see figure 3.27).

competition between nucleon configurations

We’re discovering that a nuclide is not in itself inherently unstable. What makes a nuclide

unstable is the existence of an accessible configuration with a smaller mass. In a sense,

configurations compete for nucleons, and any configuration that can rearrange a set of

nucleons with a reduction in mass is energetically favoured. If no energetically viable

decay reaction is available, then the nuclide is, de facto, stable.


McNeil, 23/04/11,

The energy accounting is very strict; no special consideration, or rounding things up!

McNeil, 23/04/11,

The half life is 35 ms, and so there are 1000/35 = 28 half lives in 1 second, reducing the numbers by a factor of 0.528, enough for even 1 billion nuclides to be reduced to single figures.

McNeil, 23/04/11,

NUCLEUS gives the first two, totalling 80%, and NuDat gives these plus the 20% beta decay.

McNeil, 23/04/11,

Alternatively, if the fission process releases more binding energy, then it is energetically favourable. These decay reactions can be evaluated using binding energies instead of masses, and you get the same results.


is it enough for a decay reaction to be energetically favourable?

We’ve seen that for a decay reaction to be viable, it must be energetically favourable - it

must “pay its way”. But that does not mean that every decay mode that is energetically

viable will occur - the small stable nuclides of less then ~60 nucleons have shown us that.

It’s time for us to look at the fission of clusters larger than 60 nucleons.

3.7.3 Spontaneous fission

fission is energetically viable for clusters >~100 nucleons

Figure 3.28 is a close-up view of the minimum in the nucleon mass curve for the stable

nuclides. We can see the mass per nucleon values fall steeply from cluster size 40 to the

minimum at ~60, then begin their slow rise as the cluster gets bigger.

Figure 3.28: The minimum in the nuclide mass-energy curve. Fission only becomes energetically favourable for nuclides bigger than ~100 nucleonsWe can use the graph to get a rough idea when it is energetically favourable for a larger

cluster to split into two equal clusters. The average nucleon mass for a cluster of 80

nucleons is less than for a cluster of 40. Splitting a cluster of 80 nucleons into two clusters

of 40 would increase the total mass, so it’s not energetically favourable. However, the

graph shows that a cluster of around 95 or so nucleons can split into two equal clusters,

with no change in the average mass per nucleon. It is then energetically favourable for


930.15

930.20

930.25

930.30

930.35

930.40

40 50 60 70 80 90 100nucleons

mas

s/nu

cleo

n (M

eV)

fission increases the average nucleon mass

no change in the average nucleon mass

fission reduces the average nucleon mass

..., 23/04/11,

The data in the mass.mas03 table show that there are several ways that the protons and neutrons in the stable nuclide (44p,56n) can be distributed between two identical or very similar clusters with a loss of mass, and hence a release of binding energy. It is energetically favourable for smaller stable nuclides, down to 94 nucleons, to be split, but the range of viable daughter nuclides rapidly diminishes. Finally, there appears to be only one way for the two stable clusters of 92 nucleons to fission: if molybdenum-92 (42p,50n) fissions into calcium-46 (20p,26n) and titanium-46 (22p,24n) there is a tiny loss of mass of 0.4 MeV. All other splittings increase the mass of the system. Checking this might be a suitable pastime for a wet Sunday afternoon in November.

McNeil, 23/04/11,

Williams puts this succinctly, as befits a university textbook: “An examination of…the curve of binding energy per nucleon shows that it is energetically possible for a nucleus having A>100 to fission into two equal parts” - p.78. Since the stable p:n ratio changes as the cluster size increases, this is only a rough indication, though still useful. I’ve used the nucleon mass graph, to be consistent with the rest of this chapter.

McNeil, 23/04/11,

We can see the nuclides with the least mass/nucleon values: iron-56, followed by nickel-60, then nickel-62. Some clusters, eg cluster-70, have two stable nuclides (see the section on cluster-100).

..., 23/04/11,

Here we start to answer the question: why are there stable nuclides that have numbers of nucleons very different from 60, the most tightly bound?


clusters bigger than this to split in half, since the average nucleon mass is reduced, as the

cluster-100 example shows. The fission process will become even more favourable as the

cluster size increases further.

Yet the nuclide chart shows the arc of stable nuclides to be nearly continuous up to

cluster-208, and that spontaneous fission is the major mode of decay only for massive

nuclides. So what prevents the stable nuclides larger than 100 nucleons decaying?

spontaneous fission of very heavy nuclei

It is only the very heaviest nuclides, beyond uranium, that decay by spontaneous fission,

and usually in conjunction with other decay modes. Figure 3.29 shows a portion of the

nuclide chart beyond the stable cluster-208 limit.

Figure 3.29: Very large nuclides decay by a number of processes: spontaneous fission (green), alpha-decay (yellow) and beta-minus decay (blue). Some of the cells have an inset box; this gives information on other nuclear processes.We can see that spontaneous fission becomes a significant decay mode beyond cluster-

238. Cluster-250 decays by three processes, beta, alpha, and spontaneous fission. The

familiar beta pn reaction "moves" the cluster one cell along the cluster-250 diagonal,


neutron number, N

prot

on n

umbe

r, P

cluster-250 (96p, 154n)

has 3 modes of decay…

...beta-minus, np, ~8%

...alpha, ~18%

...spontaneous fission into 2 similar size fragments, ~74%

cluster-208, the end of the arc of stability

the cluster-238 family of nuclides

uranium-238 (92p,146n)

McNeil, 23/04/11,

Screen capture from the 2D NUCLEUS chart. Decay of nuclide (96p,154n) checked with NUCLEUS and NuDat. - September 2010.

..., 23/04/11,

The transuranic elements – Williams, p.94.


while the alpha-decay reaction, which ejects an alpha-quartet of nucleons (2p,2n), moves

the cluster two cells diagonally in nucleon-space towards stability. Compared to these two

small, precise decay reactions, spontaneous fission is like a hyper-space jump. You can't

predict what the fragments will be, though the usual outcome is that one has 90-100

nucleons, and the other has 130-140. Each decay mode has its own probability, as we

saw for nuclide (77p,90n).

the nucleus as a liquid drop

If spontaneous fission is energetically favourable for nuclides larger than ~100 nucleons,

why are there stable nuclides bigger than this? The reason is that in order for the

nucleons to get to the lower energy state, arranged in two smaller nuclei, they must go

through a higher energy configuration, that is, surmount an energy barrier.

We’re familiar with the way water drops, falling from a dripping tap, pull themselves into a

spherical shape as they fall. We might think of the surface tension, the attraction between

the water molecules in the surface, as pulling the drop into a sphere. However, it’s more

helpful to think of the surface tension giving the drop a surface energy, which is minimum

for a spherical shape. A sphere is the shape with the least surface area for its volume,

and any distortion in the drop’s shape increases its surface area, and hence its energy.

The nucleus is similarly bound by the attractive nuclear force, and behaves in some ways

like a liquid drop, and so nuclear fission then becomes analogous to a water drop splitting

(figure 3.30).


..., 23/04/11,

Williams describes spontaneous fission in terms of the nuclear liquid drop model – p.94. Also see Mackintosh et al, p.75, and Bertulani, p.319. However, quantum tunnelling also has some role to play in spontaneous fission – Bertulani, p.320, and Williams, p.95.

..., 23/04/11,

Bertulani, p. 319 and 321. The difference in mass number between the fragments is about 45 – Williams, p.78.

..., 23/04/11,

A simulation of neutron-induced fission of uranium-235 is available at… http://phet.colorado.edu/ This shows showing fission of a single atom, a chain reaction and the operation of a nuclear reactor (accessed 2 July 2010.)


Figure 3.30: A large nucleus fissions like a water drop splits in two.A large nucleus “wobbles” due to the incessant movements and collisions of its nucleons

(a), so its surface area gets bigger, and its energy increases (b). A large wobble distorts

the nucleus, so its shape is something between a large stretched nucleus and two smaller

nuclei very close together (c). At this point the nucleus energy is a maximum, and the

nucleus may relax back into the single cluster, or carry on (d) and split into two smaller

nuclei, with some free neutrons, and a release of energy from the loss of mass (e). We’ve

seen that bigger clusters need more neutrons per proton for stability, so when a big

cluster fissions into smaller fragments, the products are themselves unstable and there

are some free neutron “leftovers”.

The energy barrier opposing the splitting of a cluster of 238 nucleons is only 6-8 MeV

high, a tiny fraction of the cluster’s mass of more than 200,000 MeV, but this is enough to

make spontaneous fission nearly impossible. This energy barrier gets smaller as the

cluster size increases, and beyond a cluster size of ~238 nucleons spontaneous fission

becomes more common.

why there are stable nuclides >100 nucleons/tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

(a) A large nucleus, within which the nucleons are all randomly moving and colliding, making the nucleus “wobble”, and its shape change.

(b) Nucleon collisions have disorted the nucleus shape – its surface area and energy have increased.

(c) The nucleus shape has distorted further. Its energy is now a maximum, and it can either go back to the spherical shape (a), or distort further…

(d) …into two smaller nuclei, with a neck developing between them.

(e) Two smaller nuclei, of comparable size, with free neutrons and the conversion of the lost mass into energy.

..., 23/04/11,

Strictly, it’s the fissionability parameter, Z2/A, where Z=number of protons, and A=number of nucleons – Bertulani, p321 and Williams, p.95.

..., 23/04/11,

We use these to initiate fission in nuclei like uranium-235, to sustain the chain reaction in a nuclear power station or detonate an atomic bomb – see for example, Mackintosh et al, p.75 and 91.


So it is the existence of a small but insuperable energy barrier that prevents clusters

bigger than ~100 nucleons undergoing spontaneous fission. Spontaneous fission might

be energetically favourable, but impossible in practice. It’s perhaps like having a winning

lottery ticket, but you have to post it off to claim your prize, and you can’t afford the price

of a stamp.

3.7.4 alpha-decay

We’ve calculated that alpha-decay, the ejection of a helium-4 (2p,2n) nuclide, becomes

energetically favourable for the stable clusters bigger than ~145, and this is confirmed by

the nuclide chart (see back to figure 3.26). We can see that it’s the proton-rich nuclides

that favour alpha-decay, since this moves them down the valley slope towards the arc of

stable nuclides.

why is alpha-decay so common?

The nuclide chart (figure 3.26) shows that by far the most common mode of fission is the

ejection of an alpha-particle (2p,2n). Why this particular combination of nucleons?

We have seen how for a large nuclide to fission, the mass difference between parent and

daughter nuclides must exceed the mass of the ejected fragment. Consequently, the more

tightly bound is the fragment, the smaller is its average nucleon mass, and the more likely

that its ejection will be a viable mode of decay. Hence, the tightly bound helium-4 nuclide,

with its small mass per nucleon, is a very suitable candidate for ejection. For example, we

saw in section 3.7.2 that the nuclide (76p,90n) can eject an alpha particle quartet, with the

small average nucleon mass of 931.9 MeV/nucleon, but not a proton with a large unbound

mass of 938.3 MeV.

alpha-decay is viable for clusters >145 nucleons

We’ve seen that alpha-decay becomes viable if the mass of the daughter nuclide plus

alpha particle is less than the parent, or to put this another way…

mparent - mdaughter > malpha (3,727 MeV)

The parent nuclide can undergo alpha-decay if down-sizing to the smaller daughter

reduces its mass by more than the mass of a free alpha particle, 3,727 MeV.

The nucleon mass curve tells us that as a cluster increases in size beyond about 60, at

some point alpha decay should become energetically favourable. But at what size would


McNeil, 23/04/11,

Helium-4 has an average nucleon mass of 931.9 MeV/nucleon.

McNeil, 23/04/11,

See Williams, ch.6 and Bertulani, ch.7.


this occur? Some simple mass-energy accounting for the alpha-decay of nuclide

(60p,85n), the stable nuclide in the cluster-145 family, will give us an idea - figure 3.31.

Figure 3.31: the mass-energy accounts for the alpha-decay of cluster-145When nuclide (60p,85n) ejects the (2p,2n) quartet its mass is reduced in two ways…

1) Simply losing the 4 nucleons (2p,2n) reduces the mass of cluster-120 by 4 times the

average nucleon mass, that is, 4 x 930.7 = 3,722.8 MeV. This is less than the alpha-

particle mass, so this is not enough in itself.

2) The daughter nuclide is now a little closer to the minimum in the nucleon mass curve,

and so its average nucleon mass is slightly less. The slope of this straight part of the

nucleon mass graph tells us that losing 1 nucleon reduces the average mass/nucleon by

about 0.008 MeV. So, in losing 4 nucleons, the remaining 141 nucleons each have about

0.032 MeV less mass. This reduces the cluster’s total mass by a further 141 x 0.032 = 4.5

MeV

Adding these two gives the total mass difference parent - daughter as 3,727.3 MeV,

almost exactly the same as the mass of a free alpha particle. So there is no energy

advantage in cluster-145 undergoing alpha-decay. But what about clusters smaller or

larger than this?


930

931

932

933

934

935

936

937

938

939

0 20 40 60 80 100 120 140 160 180 200 220cluster size

mas

s/nu

cleo

n (M

eV)

free alpha particle: He-4 (2p,2n)mass, malpha = 4 x 931.85 = 3,727.4 MeV

Above a cluster size of ~100 the graph is nearly straight.Losing one nucleon reduces the mass per nucleon by about 0.008 MeV.So losing 4 nucleons (2p,2n) reduces the mass per nucleon by about 0.032 MeV.

cluster-145: average nucleon mass = 930.7 MeVmass loss…due to ejecting (2p,2n) = 4x930.7 = 3,722.8 MeVof the remaining 141 nucleons = 141 x 0.032 = 4.5 MeVTotal mass loss = 3,727.3 MeV

straight line fit to mass/nucleon graph

alpha-decay energetically unfavourable

alpha-decay energetically favourable

..., 23/04/11,

This plots the values of average mass/nucleon of the nuclides only. The difference with atoms is tiny, for example, the mass of a helium atom is 3,728.4, different by the mass of 2 electrons.


The ejection of 4 nucleons reduces the mass of each nucleon remaining in the cluster by

0.032 Mev, regardless of the cluster size, so the bigger the cluster, the bigger its mass

loss. Cluster-145 is a break-even size, where the mass loss of the fragmenting cluster is

almost exactly equal to the mass of a free alpha particle. But for clusters larger than 145,

the ejection of four nucleons will reduce the cluster’s mass by more than a free alpha

mass; for smaller clusters, the reduction will be less. Thus, the nucleon mass curve tells

us that it is energetically favourable for stable clusters bigger than about 145 nucleons to

decay by ejecting an alpha particle.

The nuclide chart (figure 3.26) indeed shows that alpha-decay becomes a significant

decay mode for nuclides with more than about 145 nucleons, in agreement with the

calculation above. So, if alpha-decay is energetically favourable for nuclides of more than

~145 nucleons, then why is there a nearly continuous arc of stable nuclides all the way up

to 208 nucleons?

why there are stable nuclides >145 nucleons

One of the features of alpha-decay is that the the rate of decay is very sensitive to the

energy released with the ejected alpha particle. The bigger the alpha particle energy, the

faster the decay rate, and the shorter the nuclide’s half life. Nuclides that undergo alpha-

decay have an enormous range of decay rates, with half lives ranging from microseconds

to millions of years.

In the cluster size range 144-206, there are seven unstable alpha-emitting nuclides that

can be found naturally because they have very long half lives, comparable to the age of

the Earth. The very low energies of their emitted alpha particles mean that they can only

decay very slowly. Williams concludes, “it is therefore certain that although most nuclei in

this range on the line of stability may be energetically able to decay by alpha-emission,

they do not do so at a detectable level because the [decay] rate is too small.”

Thus, alpha-decay is energetically favourable for the nuclides on the stable arc in the size

range 145-208, but for most of them the decay rates are negligible, and so these nuclides

are effectively stable.

alpha particles in a potential well

We have so far seen alpha-decay simply as the ejection of an alpha particle (2p,2n) from

the nuclide, and we’ve done the energy accounting for the process. But how does an

alpha particle manage to escape from the immensely powerful grip of the nuclear force?


McNeil, 23/04/11,

Williams, p.76, who refers to it as the transition rate.

..., 23/04/11,

Williams, p.76.

McNeil, 23/04/11,

Williams, p. 82.

..., 23/04/11,

This empirical relation is known as the Geiger-Nuttall rule – Williams, ch.6.

..., 23/04/11,

Also called the transition rate - the probability that a cluster will make a transition to another state in a time period of 1 second - in nuclear physics, and also called the decay constant - Williams, p.21. I’ll stick to calling it the decay rate.

McNeil, 23/04/11,

Alpha decay becomes energetically possible for clusters bigger than ~151 - Williams, p. 76, Bertulani, p.122, or with more than 50 protons – NNDC, at. http://www.nndc.bnl.gov/chart/help/alphadecay.jsp Why do all these calculations? Because I wanted to show the reader that broad patterns of behaviour arise out of the shifting balance of simple factors, and that these patterns can sometimes be predicted with fairly simple calculations. Williams gives an elegant calculation, using the semi-empirical mass formula, to derive the break-even cluster size as 151 - pages 61 and 76. I’ve done a much simpler energy accounting exercise for one specific nuclide.


Alpha decay is an example of a process known as “quantum tunnelling”. We have seen

waves and particle-waves can pass through barriers that would appear to be

impenetrable, and we will see that alpha particles can behave in the same way.

Alpha-decay differs from beta-decay, in that the nucleons making the alpha particle

already exist inside the nucleus, whereas the electron is first created by the weak

interaction and then ejected from the nucleus. So our first question is how any nucleons,

let alone a quartet, elude the grasp of the nuclear force. Nuclei which undergo alpha-

decay have a huge range of half lives – from less than a microsecond to nearly a

thousands of years. A second question, then, is why is there such an enormous range of

rates for what is “essentially the same process”?

The nucleons are packed together, “jostling about in a very small volume”, confined in the

nucleus by the strong nuclear force. So “when a nucleon approaches the surface and tries

to fly off the nucleus, it suffers an attractive force by the nucleons that are left behind,

forcing it to return toward the interior. Inside the nucleus it feels the attraction forces of all

the nucleons that are around it, resulting in a net force approximately equal to zero. We

can imagine the nucleus as a balloon, inside of which the nucleons move freely, but

occupying states of different energy”.

The nucleons are confined in a potential energy “well”, since energy must be supplied to

remove them. Figure 3.32 shows how the potential energy of an alpha-particle (2p,2n)

varies with distance from the centre of a typical heavy nucleus of about 236 nucleons.


..., 23/04/11,

This universally accepted way of describing the nucleus is very effective in explaining a lot of nuclear properties. The graph is plotted for an alpha particle and a nucleus of ~236 nucleons, (90p,146n), based on figure 6.2, Williams, p.84 (see also Bertulani, fig. 7.3, and Hey and Walters, p.85). The graph was plotted in EXCEL, using standard values for electron charge and the permittivity of free space, and taking Williams’s value of about 9.5 fm for the nuclear radius.

..., 23/04/11,

This is the basis of the Fermi Gas model of the nucleus, which explains many nuclear properties – Bertulani, p.124.

..., 23/04/11,

Hey and Walters, p. 87.

..., 23/04/11,

Williams, p.82.

..., 23/04/11,

Williams, ch.6, Bertulani, ch.7, and HyperPhysics… http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radact.html#c2

..., 23/04/11,

Bertulani, p.195.

..., 23/04/11,

The diagrams for this were made using the program “Quantum tunnelling and wave packets”, version 1.11, PhET Interactive Simulations, University of Colorado.http://phet.colorado.edu (accessed late July 2010). The program is written to show the quantum tunnelling of plane waves and electron particle-waves, through potential barriers of various heights and widths. To create the wave functions shown here, the barrier width was set to roughly equal the electron wavelength, and the height adjusted to give a transmission probability of 0.01; that is, 1 in 100 collisions with the barrier succeeds in tunnelling. The diagrams show screenshots from the simulation. I’m assuming that the mathematical wave functions for an electron are the same as for an alpha particle, only the dimensions will differ. The wave functions produced by the PhET program for the plane wave entering both the infinite and the narrow barrier are the same as those shown by Rae, “Quantum Physics”, p.52, and Williams, p.85, and in http://en.wikipedia.org/wiki/Quantum_tunneling


Figure 3.32: For a big nuclide to undergo alpha-decay a (2p,2n) quartet must break out of a the nuclide’s deep potential energy well.

inside the nuclear well

Nucleons fill the central potential well inside the nucleus, up to an energy of about -6 MeV,

so that it takes about 6 MeV to remove one nucleon (a). The constantly jostling nucleons

can briefly gain energy from random collisions, and so there are always excited nucleons

higher up in the potential well (b). This jostling also means that there are brief local

groupings within the nucleus, such as the alpha particle quartet (c), though all sorts of

groupings will occur. Whilst we should not think of alpha particles (2p,2n) as having a

permanent existence inside a larger nucleus, it is very likely that at any moment there is

somewhere in the nucleus an alpha-like grouping of nucleons. So, while “the simple

intuitive picture of alpha particles bouncing around in nuclei, each with its own unique set

of nucleons, is not quite accurate … alpha particle-like structures do occur in the nucleus

and sooner or later these appear ouside the nucleus as alpha particles”.

outside the nucleus


-20

-15

-10

-5

0

5

10

15

20

25

30

-100 -80 -60 -40 -20 0 20 40 60 80 100

distance from centre of nucleus (fm)

pote

ntia

l ene

rgy

(MeV

)

(e) …yet alpha-particles are emitted with a typical kinetic

energy of ~4 MeV

(d) an incoming alpha-particle with 9 Mev kinetic energy only gets as close

as this to the nucleus

(a) Nucleons filling the potential well from the bottom, up to an energy of ~ - 6 MeV

(c) some of these excited nucleons can briefly come

together as a quartet.

(b) some nucleons gain extra energy from collisions

the energy peak is ~27 MeV

9 MeV4

MeV - 6 MeV

the potential well inside the nucleus,

radius ~10 fm

outside the nucleus, beyond ~10 fm, an alpha particle is electrically repelled

..., 23/04/11,

Mackintosh et al, p.68, who explain how “the fact that all protons are identical in the special quantum mechanical sense means that every proton is part of every [alpha-like] cluster.” The alpha particle’s stability means that they “often have an independent existence, of a sort, within many nuclei”. - Mackintosh et al, p.68. Some nuclei, for example carbon-12 and oxygen-16, behave as if they were clusters of alpha-particles – Bertulani, p.186.

..., 23/04/11,

Williams gives the nucleon separation energy for heavy nuclei as about 6 MeV – p. 83. Clayton gives a similar figure, 8 MeV – p.292. The diagram shows nucleons stacked in the potential well – see Bertulani, p. 187 and 127. I’ve tried to present a simple picture of nucleons filling up the well from the bottom (and not getting into details such as energy levels and spin), and with a potential well depth of ~45 MeV, with the highest occupied energy ~ -8 MeV – Bertulani, p.127.


Beyond the very strong but short range attraction of the nuclear force an alpha particle,

carrying an electrical charge of +2, is electrically repelled by the protons in the nucleus. It

“falls” down the potential energy slope, towards zero potential energy, effectively out of

range of the repulsion. An alpha particle needs about 27 MeV of kinetic energy to

overcome this electrical repulsion and enter the nucleus. An alpha-particle with, say, only

9 MeV kinetic energy never gets near the nucleus, but runs part way up the potential

energy “hill”, and then rolls back down again (d). Similarly, an alpha particle would need

27 MeV to surmount the barrier and escape from the nucleus, but it would then be ejected

with 27 MeV of kinetic energy. Yet alpha particles are emitted from decaying nuclei with

only 4 MeV of kinetic energy (e). It is as if they somehow “emerge” part-way up the

potential energy hill at 4 MeV, and from there roll away from the nucleus.

tumbling and tunnelling

How do any nucleons manage to escape from the nuclear attraction? And why is it an

alpha particle particle, a quartet of nucleons, that emerges? The answer is that the alpha

particles escape the nucleus by quantum tunnelling. We know that, like any particle, an

alpha particle should be regarded as a particle-wave, and so in colliding with an

impenetrable barrier there is the probability that it will pass through.

The jostling nucleons, randomly coming together in alpha-quartets, are confined in the

nucleus by the strong nuclear attraction. We can see this as alpha particles colliding with

a potential barrier, and knowing that an alpha particle behaves as a particle-wave, we can

see that its wave function will penetrate some way into the barrier, and has a finite chance

of emerging on the other side. Figure 3.33 illustrates the sequence.


..., 23/04/11,

I’ve taken the electron particle-waves from the diagram of the narrow barrier and used them in the nucleus, having shrunk them to fit, and split them so the transmitted wave “appears” on the curve for electrical repulsion ( following Williams, fig. 6.2 and Bertulani, fig. 7.3), though not at 4 MeV as Williams gives – this is a qualitative depiction, after all. This diagram is very similar to the alpha-decay of polonium-212 at… http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/alptun2.html#c1 (accessed 13 Jan 2011). I estimate a 4 MeV alpha particle to have a de Broglie wavelength of ~7 fm, treating it as a particle-wave, so one can imagine the wave function extending outside the potential ‘hill’, and therefore giving the alpha particle a tiny probability of 'appearing' outside the decaying nucleus. Is this another useful way of treating alpha-decay? I've not seen it in reading around - 13 Jan 2011.

..., 23/04/11,

The usual approach is to treat alpha-decay as a probabilistic quantum tunnelling process. The starting point is to assume that the chance of decay is the probability of crossing the barrier multiplied by the number of attempts the particle makes to cross it (Bertulani, p.187). This simple idea leads in a (comparatively) straightforward way to a reasonably accurate prediction of half-life, and an explanation for the empirical Geiger-Nuttall rule, that relates alpha paricle energy to half life – Bertulani, ch. 7, Williams, ch.6. I’ve not found any analysis of alpha-decay in terms of the uncertainty in the alpha particle’s position, using the well established relation… (x(p(ħ/2. A 4 MeV alpha particle has a speed of ~1.4x107 m/s, and so a de Broglie wavelength of ~7 fm…maybe one can take this further?

..., 23/04/11,

Alpha particle energies range from about 4-8 MeV – HyperPhysics… http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radact.html#c2 I’ve followed Williams, figure 6.2, who shows an alpha particle emitted with 4 MeV energy.

..., 23/04/11,

Found by Earnest Rutherford in early experiments on alpha-decay – Hey and Walters, p. 86.


Figure 3.33: Alpha-particles collide with the walls of the energy well, and have a finite probability of tunnelling through.Nucleons jostle around in the energy well (a), and some gain extra energy from collisions

(b). A quartet of excited nucleons, corresponding to an alpha particle with about 4 MeV

energy, collides with the barrier (c), tunnels through (d), and appears on the other side (e).

The alpha particle is free of the nuclear attraction, is subject only to the electrical

repulsion, and so shoots away from the nucleus with 4 MeV kinetic energy (a speed of

~1.4 x 107 m/s, 14 million metres/second).At an energy of 4 MeV the barrier is 50 fm wide

(60 -10 = 50 fm, see the graph). But the range of the nuclear attraction is only a few fm at

the most – so why must the alpha particle tunnel the full 50 fm? The reason is that energy

must be conserved. The alpha particle goes into the barrier with 4 MeV, so it must emerge

with the same energy. If it emerged anywhere else on the potential energy hill, say at 20

MeV, then it would have somehow acquired another 16 MeV for free. So in the figure the

dotted line representing the tunnelling process must be horizontal – no forbidden energy

changes!

If the alpha particle has only a little more energy, say 6 MeV, then the barrier width is

much less, only about 30 fm (40 -10 fm). We can now qualitatively explain the huge

variation in half life for alpha-decay. An alpha particle with more energy, not only has a

better chance of penetrating a barrier of the same width, but also finds that the barrier is


-20

-10

0

10

20

30

-20 0 20 40 60 80 100distance (fm)

pote

ntia

l ene

rgy

(MeV

)

(e) … appear as an alpha particle outside the

nucleus, with 4 MeV potential energy

(d) …tunnel through …

(b)…and some gain extra energy from collisions with other nucleons.

(a) nucleons “jostle” around at the bottom of the

potential energy well…

(c) a quartet of excited nucleons (2p,2n) collide with the walls of the energy well…

4 MeV

..., 23/04/11,

The kinetic energy, 0.5mv2 = 4 x 1o6 eV. Knowing that the mass of an alpha particle is about 4 amu, 3700 MeV/c2, we can calculate the speed as 1.4 x 107 m/s. Whilst it’s accepted to write of masses in MeV, in the calculation we must remember that the mass unit is MeV/c2, from Einstein’s equation, E=mc2. There’s a simulation of alpha-decay, with half life and quantum tunnelling at…http://phet.colorado.edu/


narrower. So the probability of tunnelling is extremely sensitive to the alpha particle

energy.

a “cloud” of possibilities?

We’ve seen that some nuclei emit larger nuclear clusters as well as alpha particles – for

example, radium-223 emits one C-14 nucleus for every billion alpha particles.

Presumably, bigger clusters are brought together less often by random jostlings of

nucleons. But very occasionally this happens, and if it is energetically favourable, then the

large cluster is emitted by the nucleus.

If C-14 clusters can be emitted, then we must infer that all possible nucleon groupings can

form inside the nucleus, collide with the potential barrier, and have some small but finite

probability of tunnelling through it. We can perhaps envisage a sort of “cloud” of possible

particles around the heavy nucleus. But the only permitted outcomes are those that lead

to a reduction in the mass of the system – that is, the mass of the daughter plus emitted

nucleus must be less than the parent nucleus. We’re reminded of Kaufmann and

Freedman’s aphorism: everything is possible, unless it is forbidden. There’s a host of

potential possibilities, but only those that are energetically allowed can actually occur.

3.7.5 Why there is an arc of stable nuclides

We have seen that a nuclear cluster of ~60 nucleons has the least possible mass per

nucleon, and the greatest binding energy. Thus it is energetically favourable for all other

clusters to converge on this optimum size. We can now explain why this does not occur,

and why there is a long arc of stable nuclides running through the nuclear valley.

motive, means and opportunity

As in the best murder mysteries, successful nuclide decay depends on motive (a

reduction in average nucleon mass, and therefore the release of binding energy), means

(a viable nuclear reaction), and opportunity (a significant probability of this reaction

occurring). Figure 3.34 summarises this for the stable nuclides.


..., 23/04/11,

Reference 1 in the Wikipedia article on Cluster decay. See also the isotopes of radium in the NUCLEUS database.

..., 23/04/11,

For example… plutonium-239: decay energy 5.3 MeV, half life 24,000 years actinium-218: decay energy 9.4 MeV (less than double), half life 1 microsecond (~1020 times shorter) – Bertulani, p.191, and NUBASE 2003. We’ve already seen how varying the electrical potentials on a flash memory cell can either confine an electron for >10 years, or allow it to tunnel out in <1 ms, a factor of 1012 quicker.


Figure 3.34: The decay options for the stable nuclides that are larger and smaller than the ~60 nucleon optimum size.Small nuclides, <~60 nucleons, would readily merge if they had enough kinetic energy to

overcome their mutual repulsions. They have the motive (a reduction in mass), and the

means (a viable nuclear reaction), but not the opportunity - they can’t get close enough to

undergo fusion.

Middle-sized nuclides, ~60 - 100 nucleons, have the motive of mass reduction, but there

is no rearrangement that will reduce it - they have no means.

Larger nuclides, >~100 nucleons, have the motive and the means, via a fission reaction,

but the insurmountable energy barrier denies them the opportunity.

Even larger nuclides >~145 nucleons have the motive and the means, via alpha particle

ejection, but the opportunities for decay are extremely rare.

a surprising conclusion?

Thus we have the rather surprising, and maybe unsettling, conclusion that our physical

world of ”stable” nuclides is not what it seems. It is energetically favourable for stable

nuclides that are smaller or larger than ~60 nucleons to decay, but they are “trapped”

because the options for decay are either unavailable, or too slow. These “stable” nuclides,

the lowest mass nuclides of each cluster-family, will in time become the nuclei of the

chemical elements, and create our atomic world.


930

931

932

933

934

935

936

937

938

939

0 20 40 60 80 100 120 140 160 180 200 220cluster size

mas

s/nu

cleo

n (M

eV)

small nuclides <~60Fusion: motivation means opportunity

nuclides ~60 motivation

nuclides ~60 to ~100fission… motivation means

nuclides >~100fission… motivation means opportunity

nuclides >~145alpha-decay… motivation means? opportunity


3.8 Nuclear reactionsThe collision of two clusters can lead to a nuclear reaction, rearranging the nucleons into

new nuclides. For example, if we fire protons at an aluminium-27 (13p,14n) nucleus, there

are a number of possible reactions (figure 3.35).

Figure 3.35: A nuclear reaction can have a range of possible outcomesThe colliding particles combine to make an excited silicon-28 nucleus, indicated by the

asterisk (*), which then decays in a number of ways. It may remain as it is, and just lose

its excess energy by emitting a photon of gamma radiation. This is rather like a wet dog

shaking the water off its fur, except that to properly mimic the nucleus, the dog would

have to shake all its water off in one single amount, not a large number of small drops.

Alternatively, it may fragment into smaller nuclei and protons and neutrons.

In certain situations three nuclei can fuse…

3 helium-4 (2p,2n) carbon-12 (6p,6n)

and we’ll soon look at this important reaction occurring in the interiors of stars.

A nuclear reaction doesn’t need two colliding nuclei, it can be initiated by a photon of

radiation…

1) + copper-63 (29p,34n) nickel-62 (28p,34n) + p and…

2) + uranium-233(92p,141n) rubidium-90(37p,53n) + caesium-141(55p,86n) + 2n

In the first reaction a gamma ray photon knocks a proton off a nucleus, and in the second

it induces a giant nucleus to fission into two smaller nuclei and a pair of neutrons.

nuclear Lego


p + aluminium-27 (13p,14n)

silicon-28* (14p,14n)

- an excited intermediate

state

silicon-28* (14p,14n) +

silicon-27 (14p,13n) + p

magnesium-24 (12p,12n) + helium-4 (2p,2n)

sodium-24 (11p,13n) + 3p + n

..., 23/04/11,

This is of course, radioactive gamma decay, whereby a nucleus in a state of internal excitation loses its excess energy through the emission of a single gamma ray photon. This is one of the trio of radioactive decay mechanisms – alpha, beta and gamma. I deal with only the first two, because only these two change the particle make-up of the nucleus. It seems odd to make no mention of gamma decay, and to deal with alpha and beta decay processes are in different chapters. But then this is not a book about “science”, but about trying to find a route from “there” to “here”. References: Bertulani, ch.9,

..., 23/04/11,

The reactions in this section have been taken from Bertulani, ch. 10 and 11, and see also Williams, ch.7. Nuclear reactions are subject to a number of conservation “laws”. For low energy nuclear interactions, below ~140 MeV, the threshold for the production of mesons, there are no nuclear processes capable of transforming a proton into a neutron. Also the weak interaction, that interchanges protons and neutrons, is very slow compared to the times for nuclear reactions - ~10-22 to 10-16 seconds. Thus we see the conservation of protons and neutrons, so they balance on either side of a reaction – hence we can think of rearranging Lego bricks. Bertulani describes the behaviour of the compound nucleus (the silicon-28* nucleus in the first set of reactions), that is the excited nuclear intermediate between the reactants and products in low-energy nuclear reactions. We’ll see later that there is an excited intermediate state in chemical reactions. The shorthand notation for simple collisions: a + A ( b + B Particle a is fired at target nucleus A, giving two products, a lighter one b and a heavier one A. This is written as A(a,b)B. (Williams, ch.7) Williams notes that although some reactions are forbidden, such as those that do not conserve charge, “anything not forbidden can and may happen” – p. 129.


We can take a simple view of all these reactions as analogous to rearranging Lego bricks.

The number and type of bricks are constant, and they can be rearranged in any

combinations, as long as they are energetically allowed. Thus, we see the same number

of protons and neutrons on either side of the reaction, they are just differently arranged.

We’ll see shortly how the hot dense interiors of stars force all sorts of nuclear reactions to

occur.

nuclear and chemical reactions

There is a similarity between nuclear and chemical reactions. The former rearrange

nucleons in nuclear clusters, and the latter rearrange atoms in molecules. Both types of

reaction can release energy, but nuclear reactions typically yield about a million times

more energy than chemical, because the forces between nucleons are so much greater.

3.9 Life in the nuclear valley3.9.1 a balance of conflicting factors creates the nuclear valley

We have seen how nucleons are bound into a cluster by the strong nuclear force, with

virtual pions “fluttering” to and fro, constantly interchanging protons and neutrons. The

mass-energy, and hence the stability, of a cluster of a fixed number of nucleons is

decided by its proton/neutron ratio. But the nucleons in the cluster form relationships,

albeit very limited ones; thus, like nucleons form pairs and magic numbers of them form

closed shells, with slight effects on the cluster’s mass and binding energy. ”The nucleons

are the bricks, and the nuclear forces provide the mortar, while everything is under the

control of rigorously enforced planning regulations provided by the quantum rules”.

Because the nuclear attraction force is stronger than the electrical repulsion, stable

nuclear clusters are possible. Because the electrical repulsion has the longer range, there

is a limit on stable cluster size. Because of the protons’ mutual repulsion and the

neutron’s greater mass, each cluster has its minimum mass-energy at only one or two

precise p:n ratios. Each stable cluster lies at the bottom of its own transverse section of

the nuclear valley. The stable nuclides in all the clusters then form the stable arc that runs

along the valley bottom.

3.9.2 The pathways towards stability

the limited options for nuclide decay/tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

..., 23/04/11,


McNeil, 04/23/11,

the first draft of this section, drawing the nuclear physics together, done 27 Sept 2010. The section “Taking a Walk in Nucleon-space” preceded this, and was deleted 13 April 2011. Also the section on radioactive decay chains for the elements beyond lead has been deleted. We usually associate decay series only with the 4 natural decay series: the thorium, uranium and actinium series ending in lead, and the neptunium series ending in bismuth. But the bigger picture is that the entire nuclear valley is criss-crossed with decay pathways, whereby the unstable nuclides transform themselves to a stable configuration. So, even though there’s another nice 3-D diagram, this doesn’t add to the narrative – so it’s gone – 13 April 2011.

..., 23/04/11,


..., 23/04/11,

Chemical reactions are the everyday stuff of life. Nuclear reactions will only occur where the colliding nuclei have enough kinetic energy to overcome the mutual repulsion and to make contact. On Earth this has happened naturally for as long as the planet has been in existence when cosmic rays, high energy particles, mostly protons, come from space. In the last hundred years or so, particle accelerators have creating specific nuclear reactions. “…unlike a chemical reaction, the products resulting from a nuclear reactionare not determined unequivocally: starting from two or more reactants there can be dozens of possibilities of composition of final products…” – Bertulani, p.259. I’m not sure about this – think about burning fuel in a car enegine, where you get a mix of reaction products; CO, CO2, and a range of oxides of nitrogen – all from octane and oxygen and nitrogen – Jan 2011.


A configuration of nucleons is unstable if there is a viable decay reaction that can

rearrange them with a reduction in total mass, and also at a reasonable rate. We have

imagined a nuclide as a boulder perched on the slope of the nuclear valley. There may be

many nuclides lower down the slope, all with lower values of average nucleon mass, and

all closer to the arc of stability, down in the valley bottom. However, this unstable parent

nuclide has only a very limited range of nuclear reactions, whereby it can decay to a

daughter nuclide, and reduce its average nucleon mass, thereby releasing more binding

energy (figure 3.36).

Figure 3.36: The limited decay options available to unstable nuclidesOn one side of the nuclear valley, proton-rich nuclides have only three options: (1)

transform a proton to a neutron by the beta-plus decay reaction, (2) eject an alpha particle

or, in the case of an extreme proton excess, (3) eject a proton. On the opposite valley

slope, neutron-rich nuclides have the “mirror-images” of only the first and last options:

beta-minus decay or neutron ejection. Alpha-decay is not available, for it would not

“move” the nuclide closer to the arc of stability.

average mass/nucleon is absolute, but stability is relative


neutron number, N

prot

on n

umbe

r, Z

eject n

alpha-decay (eject (2p,2n)

beta-plus decay (pn)

parent nuclides

daughter nuclides

arc of stable nuclides along the nuclear

valley bottom

beta-minus decay (np)

eject p

proton-rich parent nuclide

neutron-rich parent nuclide

McNeil, 23/04/11,

I’m omitting spontaneous fission and cluster-decay, which are seen only in very heavy nuclei.


A nucleon configuration may be unstable in principle, but the constraints on the viable

nuclear reactions may make decay impossible, or negligibly slow. The balance of

competing reactions depends solely on the relative properties of the parent and daughter

nuclides, and takes no account of the origins of the parent nuclide, or of the decay modes

of the daughter nuclides. Thus, while the average nucleon mass in a cluster is a fixed

quantity, the cluster’s stability, and its mode of decay are relative, and are determined

only by the parent and daughter nuclides.

decay pathways

The nuclear valley is criss-crossed by decay pathways, as the unstable nuclides “move”

downhill towards the arc of stability running along the valley bottom. Figure 3.37 shows

the final sections of three decay series.

Figure 3.37: Decay pathways in the nuclear valley; proton-rich nuclides have alpha-decay and the pn reaction, while neutron-rich nuclides have only the np reaction. Decay pathways can branch on the proton-rich side, so an unstable


nuclide Y, (63p,90n)europium-153

nuclide Z, (60p,85n)neodymium-145

nuclide X, (69p,84n) thulium-153

alpha-decay beta-plus (pn)decay beta-minus (np) decay

The thickness of the arrow gives an indication of the relative decay probabilities.

decay series A

decay series C

decay series B

..., 23/04/11,

A 3-D view of the nuclear valley, and decay data from the NUCLEUS program. I took 2 views of the valley, one with each cell named, and one without. The decay pathways were put in on the first view, which was then replaced with the second view. I’ve omitted the names of the nuclides; it doesn’t help, and spoils the view. I’ve named the 3 nuclides that start and end the decay sequences; the interested reader can follow things up from there. I’ve chosen a part of the valley around cluster-145, where alpha-decay makes its first appearance. This makes for branching decay paths on the proton-rich side, which are more “interesting” than the simple beta-minus (n(p) decays on the neutron-rich side.


nuclide can produce two stable nuclides. The nuclides are colour-coded by their major decay mode.Nuclide X, part way up the proton-rich side of the valley, is the start of a cascade of

decays that run down the slope to terminate in nuclides Y and Z. A minority of nuclides

have two modes of decay, so the pathway on the proton-rich side branches in places.

Nuclides Y and Z are the end-points of two other decay series, B and C. We have seen

that neutron-rich nuclides can only decay by the beta-minus (np) reaction, so these

decay paths do not branch.

Nuclides Y and Z share a common “ancestor” in nuclide X on the proton-rich side, and

different ancestors on the neutron-rich side. However, nuclide X is not the source of decay

series A, for it is itself a decay product of nuclides even higher up the valley slope. The

entire nuclear valley terrain is criss-crossed by decay pathways, that run down the slopes

and end at a nuclide on the stable arc.

nuclide “ancestors” and their “descendents”

Thus every unstable configuration of nucleons is on a journey to stability, following a

decay path down the slope of the nuclear valley, with each nuclide being a stage on that

journey. Every step on the path is downhill, releasing more binding energy, with the

nuclides getting closer to stability, and with increasing “efficiency” of conversion of energy

to matter. Every stable nuclide is the last “descendent” of a series of unstable nuclide

“ancestors”, that no longer exist. Its nucleons were once part of different clusters, and

what is now, say, a proton was maybe once a neutron. We can see a nuclide not so much

as a fixed thing, but rather as a transient configuration of nucleons. The nuclear valley,

then, is the environment, or habitat, in which the different nucleon configurations co-exist

and compete for existence.

decay pathways are determined by local topography

We’ve seen that the balance of the competing decay reactions is determined by the

relationships between the parent and daughter nuclides. So, while the decay pathway

must always be downhill in the nuclear valley, each step is decided solely by local factors,

and the decay series has no overall aim or preferred end-point. Similarly, water on a slope

will run downhill, always reducing its potential energy, but the precise path it takes is

determined solely by the local terrain it encounters, and not by the overall topography.

3.9.3 a rain of nucleons


McNeil, 23/04/11,

Some writers refer to the nuclides as nuclear species - “Of the thousands of known nuclear species, only about 300 are stable, that is they exist along the so-called "valley of stability". The unstable species forming the valley "walls" - those with an overabundance of protons or neutrons - tend to decay quickly, sometimes within milliseconds.” – CERN Courier, Feb 22, 2002, at… http://cerncourier.com/cws/article/cern/28587

McNeil, 23/04/11,

The pattern of decays of the unstable nuclides seems to bear parallels with the pattern of evolution of biological species (9 Sept 2010). The common features seem to be… (1) descent down a continuous ancestral line Every organism is descended from ancestral species, which are extinct, if you go back far enough. Every organism has an unbroken ancestry of creatures that have successfully reproduced. Every nuclide is at the end of an unbroken decay chain of unstable ancestral nuclides. The pathways of evolving species form a network running through biological space. Nuclide decay pathways form a network in nuclide-space. (2) modification Every offspring differs from its parents (with sexual reproduction). Competition with other organisms means that only the fitter organisms survive. Every daughter nuclide has a smaller average nucleon mass than its parent, and is thus one step closer to stability. We could perhaps see this as being more fit in the competitive nuclear environment. (3) local factors There is no evolutionary overall aim or purpose. The course of evolution is decided solely by local factors. Species evolve to deal with pressures acting on them here and now. The path of a nuclide decay chain is decided by the local options available to each nuclide.

..., 23/04/11,

Much as rainfall has carved channels on hillsides in arid regions on Earth.


We have seen all the possible combinations of protons and neutrons laid out like a chess

board, so that every possible nucleon combination has its own location in nuclide-space.

Now imagine a rain of nucleons, so that every cell is occupied by its nuclide. The great

majority of clusters break up almost as soon as they are formed; they are so unstable that

they hardly even exist. Only the clusters within a narrow diagonal region of nuclide-space

last long enough to exist as independent entities, and these define the nuclear valley -

(figure 3.38).

Figure 3.38: A rain of nucleons on the nuclear valley. Z and N are the proton and neutron numbers, respectively. The nuclides are colour coded by their major decay modes.All over the terrain of the nuclear valley, the nuclides are decaying as the clusters

rearrange themselves. The decay pathways show the pattern of nuclide “migration”,

always downhill, towards the arc of stability running along the valley bottom.

draining the nuclear valley

If we watch an unstable nuclide decay, we see the cell it occupied become empty, and the

transformed nucleon configuration reappear as the daughter nuclide in a nearby cell.

Nuclides further up the slopes are further from stability and decay faster, so we see the


very big nuclides undergo alpha and beta decay, and ‘move’ towards the largest stable nuclide, containing 82 protons

proton-rich nuclides decay by the beta-plus (pn) reaction and electron capture

towards the arc of stability

neutron-rich nuclides decay by the beta-minus (np) reaction towards

the arc of stability

some nuclides spontaneously fission

into smaller nuclei

a rain of nucleons

the lowest point in the

valley, at ~60 nucleons

Z

N


outer cells in the valley empty first. Almost all the moves are very short, either to an

adjacent or nearby cell by beta or alpha-decay. A few very big nuclides make great jumps

right down the valley, splitting into two smaller nuclides by spontaneous fission.

The region of occupied cells shrinks, like water draining out of a bathtub. Finally , all that

are left are the stable nuclides, occupying the cells along the valley bottom. If the stable

nuclides could flow like water, they would run in two streams downhill and meet at the

lowest point, around cluster-60. If you wanted to drain the nuclear valley completely, that

is where you would fit the plug.

3.10 The emergent nuclideWe have seen the emergence of a new structure – the nuclide, a community of protons

and neutrons - figure 3.39.

Figure 3.39: The emergent nuclear cluster is created and sustained by continuous activity within each nucleon.The continual exchange of pions fluttering between protons and neutrons creates “an

invisible, evanescent web … binding them together”. Thus a nuclide emerges from the

continual interactions between all the nucleons, mediated by pions. This nuclear force that

binds nucleons together, is an extension of the colour force operating inside each

nucleon. Nuclides are massive particles, carrying as many units of positive electric charge

as they have protons.

goodbye to quarks and the strong colour force

The quarks have effectively withdrawn, gathered into protons and neutrons by the strong

colour force. The residual of that force, the nuclear force, binds protons and neutrons into


…create and sustain…

… a nuclide – a cluster of protons and neutrons p+

p+p+n0

n0

n0n0

p+ n0

Continuous interactions between protons and neutrons, mediated by pions…

A stable nuclide has emerged, with a large mass, and carrying as many units of positive charge as it contains protons

free protons are stable, but are isolated by their mutual repulsions, and unstable neutrons cannot exist in isolation.We don’t consider the quarks inside protons and neutrons.

..., 23/04/11,

Mackintosh et al talk of “a river flowing along a valley floor to the lowest point” – p.74.

..., 23/04/11,

This might take a while. Some nuclides near the stability line have very long half lives, longer than the age of the Earth.


nuclei. Like a building’s foundations, the quarks and colour force are “underneath”,

holding everything up. It’s hard now even to distinguish between protons and neutrons –

they are just nucleons bound into a cluster by the nuclear force. Inside the nucleus the

weak interaction interchanges protons and neutrons, enabling unstable clusters to “move”

towards stability.

nucleosynthesis – making nuclei

We have now surveyed the array of known nuclides, and seen which are viable and why,

and how they transform from one to another. We know quite a lot about how real nuclides

behave, but nothing as yet about how they have been made.

There have been two phases of nucleus-building – nucleosynthesis – in the evolution of

the universe. The first ended when the universe was about quarter of an hour old, and the

second has been going on for the last 14 or so billion years.

We're now just emerging into a level of the physical universe that we can recognise. It is

made of only 4 particles: two quarks - up and down, and two leptons - electrons and their

neutrinos. We'll see that anti-matter, as positrons (anti-electrons) and anti-neutrinos, plays

a crucial röle in the creation of our world of atomic matter. The u and d quarks make

protons and neutrons, but only the proton is stable. On its own a neutron has a life time of

about 15 minutes, yet with the proton it has sustained our physical universe for about 14

billion years. Out of all the potential particles, these are all that are viable. How did the

Universe cook up its rich material banquet with only these few ingredients in the larder?

Almost everything we see when we look up into the night sky is the result of nuclear

reactions.

Ray Mackintosh et al, p.94

Every atom of carbon and oxygen on Earth (and in us) was forged inside stars that died

before our solar system formed. We are stardust: or less romantically, the nuclear waste

from stars.

Martin Rees, in Craig Hogan, p.viii

3.11 Nucleosynthesis 1 - The first quarter of an hourHere we will outline the events of the first 15 minutes or so after the big bang, when the

foundations of the material Universe were being laid. As Lawrence Krauss, writing about


McNeil, 04/23/11,

Krauss, p.24.

McNeil, 04/23/11,

I’m deliberately omitting dark matter and dark energy. While their presence is established, their natures and actions are not. So I’ll stick with “ordinary” matter and energy. While this picture is incomplete, it’s well established. I believe that dark matter and energy will not contradict this story, but add to it, perhaps in the same way that Relativity extended, not contradicted, Classical physics. I’ve covered the very early stages of this in chapter 2, on quarks. And see CERN slideshow on the evolution of the Universe, at… http://aliceinfo.cern.ch/static/Documents/outreach/animated_event/evolution.swf (accessed 9 April 2011) References for this section: Kaufmann, chs.28 and 29, Coughlan et al., ch. 45, Williams, ch.14, Close, "Particle Odyssey", Delsemme, ch.2, Allday, ch.12, John Gribbin, "The Universe", ch.4, Joseph Silk, Steven Weinberg. Earlier times covered in ch. 0,


The weak interaction that changes quark flavours, and thereby interchanges protons and neutrons, involves positrons and anti-neutrinos.


oxygen, puts it, “these are the conditions when the gist of our oxygen atom came to be,

when nothing became something”.

3.11.1 A time-line – linking temperature, time and energy

Figure 3.40 gives an outline of events, and links four important parameters – temperature,

time, energy and particle mass.

Figure 3.40:A time-line for the universe, linking temperature, time and energy. The temperatures are given in scientific notation, so “1.E+05” is 1 with 5 zeros, 100,000 degrees, and “1.E+09” is 1 billion degrees.The relation between the temperature of the universe and its age is well established, so

“the cosmic temperature can be used as a sort of clock, cooling instead of ticking as the

universe expands.”

We’ll now follow the series of temperature thresholds in the early “fireball” universe, with

each heading giving the temperature (T) in Kelvin (K), followed by the energy (E) in

electronVolts (eV), and then the time (t) - all values being approximate. In doing this we

will pass through levels in the universe’s hierarchy, that we have looed at in previous

chapters.


1.E+03

1.E+05

1.E+07

1.E+09

1.E+11

1.E+13

1.E+15

1.E-10 1.E-06 1.E-02 1.E+02 1.E+06 1.E+10 1.E+14

time (seconds)

tem

pera

ture

(K) Particle

masses10 GeV

1 GeV

100 MeV

10 MeV

1 MeV

100 keV

10 keV

1 keV

100 eV

10 eV

1 eV

300,000 years 1 year 1 day1 minute

..., 23/04/11,

The early universe, in which particle reactions occurred, is often called the primordial fireball – for example, Silk, p.56.

..., 23/04/11,

Kaufmann and Freeman, box 29-1. Steven Weinberg describes the first three minutes of the universe in a series of time frames, starting at t~1/100th second and T~1011K, with the temperature dropping by a factor of about 3 from one to the next. HyperPhysics summarises this sequence, with some comments … http://hyperphysics.phy-astr.gsu.edu/hbase/astro/bbcloc.html#c1 (accessed 9 April 2011) There’s a useful outline of the events in the fireball at… http://rst.gsfc.nasa.gov//Sect20/A1.html A timeline, with lots of links is at… http://en.wikipedia.org/wiki/Timeline_of_the_Big_Bang#Matter_domination:_70.2C000_years (accessed 9 April 2011) Hogan organises the sequence into epochs – ch.2.

..., 23/04/11,

Weinberg, p.86.

..., 23/04/11,

I’ve taken data on temperatures and times from a number of sources: Allday, chapter 12, Coughlan et al, ch. 45, Williams, ch. 14, Gribbin “Universe”, ch. 4, Weinberg, ch. 5, and Kaufmann and Freedman, figures 29.6 and 29.12. I took all their values of temperatures and times, and plotted them in the graph (figure 23). They are in good agreement, with each other and with other sources, see, for example, … http://rst.gsfc.nasa.gov/Sect20/A1.html (accessed 9 April 2011). This graph may be removed from future drafts, or modified - it's here to collate all the different sources.


3.11.2 Before the first threshold – temperature, T > 1015 K (energy, E > 100 GeV, time, t < 10-10 seconds)

At these enormous temperatures, "equivalent to energies far higher than anything

achieved at an accelerator on Earth, fundamental particles of matter and antimatter

emerged and annihilated continuously. The universe was an expanding froth of quarks,

antiquarks, leptons, antileptons, photons, W particles, Z particles, gluons, and maybe

other particles as yet unknown to experiment or undreamed of by theorists ." Individual

protons and neutrons can't yet exist, for "any quarks that did temporarily bind together

would be easily blasted apart again by collisions with the high energy photons".

This stage is sometimes referred to as the quark plasma, with a tiny excess of quarks

over anti-quarks. We can imagine pairs of photons and particles interchanging continually,

with particle-antiparticle pairs continually appearing out of the "sea" of background

radiation, and then disappearing back into it.

3.11.3 Threshold for creating W/Z particles (mass ~80GeV), T ~ 1015 K (E ~ 100 GeV, t ~ 10-10 seconds)

This first threshold temperature is for the heavy W/Z particles, with mass-energies around

80GeV. As the temperature drops, fewer and fewer photons have enough energy to

create these, for example…

+ Z0 + Z0 and the same for the W+ and W-

The W/Z particles annihilate back to photons, or decay to lighter particles, for example…

W- e- + e and the neutral Z0 e- + e+

After this threshold was passed, “the W and Z adopted their role of carrying the weak

interactions between particles, and had no independent existence except where they were

produced (briefly) in high-energy events involving collisions between particles, either

naturally or in particle accelerators designed for the purpose”. The quarks now "came into

their own” and started binding into hadrons – both mesons (pairs) and baryons (trios),

especially the light protons and neutrons.

The temperature is high enough for photons directly to create protons and neutrons in

particle-anti-particle pairs...

+ p + p and + n + n


..., 23/04/11,

Quotes from Gribbin, “The Universe”, p.68.

..., 23/04/11,

Generally, (1) W( decay to a lepton and a neutrino, one of these being an anti-particle, (2) Z0 decays to a lepton-anti-lepton pair. Williams, p.288, 371 and Coughlan, ch.24.

..., 23/04/11,

I’ll use “W/Z” as a simple way to refer to the W( and Z0 particles.

..., 23/04/11,

This is the threshold of currently laboratory-tested physics - Hogan, p.18. Particle accelerators capable of reaching energies of hundreds of GeV have been used for many years, so the events in the early universe after this first threshold are well established - Gribbin, "The Universe", ch. 4.

..., 23/04/11,

Allday, p.261. This is sometimes called the Quark Era – Nicholas SHort http://rst.gsfc.nasa.gov//Sect20/A1.html (accessed 16 April 2011). The tiny excess of matter over anti-matter will be dealt with in the first chapter.

..., 23/04/11,

Kaufmann, p.743, quote from Allday, p.261. Had there not been this intense background of radiation, a large proportion of the hydrogen would have been quickly ‘cooked’ into heavier elements – Weinberg, p.50.

..., 23/04/11,

Close, "Particle Odyssey", p.187.

..., 23/04/11,

Significant events occurred before even this early time, that established deep foundations of the physical universe – these will be covered in the first chapter.


The tiny quark excess carries over to the same excess of protons and neutrons over their

anti-matter counterparts. A slightly heavier neutron needs more energy for its creation

than a proton, but at this temperature the radiation photons easily have enough energy to

produce both in more or less equal numbers.

At the end of this phase the universe comprises photons of radiation and particles of both

both matter and antimatter: quarks that are free, or bound in protons and neutrons, and

electrons and neutrinos. Because the temperature is well above the thresholds for the

formation of all these particles, they are present in their matter and anti-matter forms in

almost equal numbers.

3.11.4 Threshold for creating protons and neutrons (mass ~940 MeV), T ~ 1013 K (E ~ 1000 MeV, t ~ 10-6 seconds)

The temperature has fallen to the point where the free quarks no longer have enough

energy to resist the strong force, and they become bound into pairs (mesons) and triplets

(baryons), mostly the light protons and neutrons. We have seen that the nature of the

strong colour force is such that it increases as quarks are pulled apart. The result is that

quarks will be confined in these particles for the next 14 billion years, up to the present

time - except for the few that find themselves at the core of a large star or in a particle

accelerator. So, by the time the universe is a millisecond old (t~10-3 s, T~3 x 1011K), single

quarks have effectively disappeared, “hiding exclusively inside protons and neutrons”, and

will never be seen free again.

Photon energies have decreased until they can only create the two lightest baryons -

protons and neutrons. Finally, the threshold for their creation comes at T~1013 K, and

these particles and their anti-matter counterparts annihilate each other back to radiation…

p + p + ...and the same for neutrons.

The vast numbers of protons and neutrons and their antimatter counterparts annihilate

each other, leaving a tiny residue of protons and neutrons. Never again will anti-protons

and anti-neutrons be viable constituents of the material universe; they will be only the

ghostly, fleeting products of nuclear processes in stars and particle experiments.

The tiny matter excess now reveals itself, and we’re left with 1 proton or neutron to about

109 photons of radiation, a ratio we can measure in the universe today. We can

summarise it like this…


..., 23/04/11,

Kaufmann, p.708 and p.735 and Weiberg ch.5. Delsemme gives "3 billion photons per particle of matter" - p.21. John Gribbbin explains how spectroscopic measurements of the proportion of helium in very old stars supports a photon to baryon ratio of close to a billion to one - "The Universe", p.77. Weinberg explains how the temperature of the cosmic microwave background (CMB) radiation at 3K tells us that the universe contains about 550,000 photons/litre, leading to a ratio of about 1 billion photons per nucleon – p.73. “It is truly impressive that with the plausible choice of a single free parameter, the ratio of atomic particles to photons, it is possible to account for the observed present abundances not only of ordinary hydrogen and helium (H1 and He4) but also of the isotopes H2 (deuterium), He3, and Li7. This is not only the most important quantitative success of modern cosmological theory, but the strongest evidence that we really do understand something about the history of the universe back to the first few minutes” – Weinberg, p.183.

..., 23/04/11,

The universe appears to be exclusively made of matter – “no one has seen signs of appreciable amounts of anti-matter anywhere in the universe. The cosmic rays that enter our earth’s upper atmosphere are believed to come from in part great distances in our galaxy, and perhaps in part from outside our galaxy as well … are overwhelmingly matter rather than anti-matter – in fact, no one has yet observed an anti-proton or an anti-nucleus in the cosmic rays.” – Weinberg, p.96.

..., 23/04/11,

Kaufmann, p.735.

..., 23/04/11,

Hogan, p.18.

..., 23/04/11,

John Gribbin, "The Universe", p.69.

McNeil, 04/23/11,

http://aliceinfo.cern.ch/Public/en/Chapter1/Chap1Physics-en.html (accessed 21 April 2011).

..., 23/04/11,

John Gribbin, "The Universe", p.69. Coughlan, p.220. Williams puts this at T~2 x 1012K – p.371. This is sometimes called the Hadron Era, lasting from ~10-5 to ~10-4 seconds… http://rst.gsfc.nasa.gov//Sect20/A1.html


(109 + 1) protons + 109 anti-protons 109 radiation photons + 1 matter proton

The residue of protons and neutrons remain vastly outnumbered in a universe awash with

radiation photons. Because the temperature is still well above the thresholds for the

creation of electrons and neutrinos, these particles and their antiparticles are still present

in almost equal numbers. The numbers of particles were decided by the balance between

the processes of creation and annihilation, and the “number and average energy of the

photons was about the same as for electrons, positrons and neutrinos.”

neutrino reactions

Neutrinos interact so little with ordinary matter that they can pass through an entire planet

like the Earth, and barely notice. However, the universe at this time is so dense, with the

particles pressed so close together, that the neutrinos readily interchange protons and

neutrons, via the weak interaction…

n0 + e- + p+ and p+ + n0 + e+

When the universe is a bit less than 1 second old the protons and neutrons are changing

their identities in this way about 10 times every second. This at first keeps the numbers of

protons and neutrons about equal, but as the temperature continues to fall, and the

average neutrino energy decreases, the neutrons’ slightly greater mass becomes an

increasing barrier to their creation, and they start to be outnumbered by protons.

These protons and neutrons are not yet bound into nuclei. The energy required to break

up a nucleus is 6-8 MeV per nuclear particle, and the thermal energy at ~1011K is more

than this, so any aggregates of protons and neutrons are destroyed as fast as they form.

3.11.5 Threshold for creating electrons (mass ~ 0.5 MeV), T ~ 6 x 109 K (E ~ 0.5 MeV, t ~ 1 second)

The temperature has now fallen below the threshold for electron creation, so the

reaction…

+ e- + e+

can no longer occur. Electrons and positrons (anti-electrons) annihilate, to leave a

remnant of electrons.


..., 23/04/11,

Coughlan, p.220. See Gribbin, Universe, ch. 4.

..., 23/04/11,

Kaufmann, p.735.

..., 23/04/11,

Weinberg, p.105.

..., 23/04/11,


McNeil, 04/23/11,

Krauss, p.74.

..., 23/04/11,

Williams, p.371, and Allday, p.262, Weinberg, p.105.

..., 23/04/11,

Weinberg, p. 103, gives the density at ~1011K as equivalent to a mass density about four thousand million times the density of tap water.

..., 23/04/11,

Weinberg, p.6.

..., 23/04/11,

Coughlan, p.220. This is sometimes called the Lepton Era… http://rst.gsfc.nasa.gov//Sect20/A1.html


This is where the original “anti-universe” finally takes its leave. We’re left with the universe

of matter – baryons (protons and neutrons) and leptons (electrons and neutrinos) – awash

with radiation photons – about 1 billion photons for each baryon.

This is the last phase of the creation of matter from photons. As the universe has

expanded and cooled the photons’ “spending power” has decreased, and the electron is

the last particle species to be created. The situation can be likened to a lottery winner in a

time of rampant inflation. Initially his winnings would enable him to buy a stately home, but

as time passes their value depreciates, and he can afford less and less – a penthouse

apartment, then a detached house, then a small terrace, and finally he can only afford a

garden shed.

the end of neutron creation

The universe has now expanded to be about 1 light-year across, not far off the distance

from Earth to the nearest star. All particles are flying apart, and the average distance

between a proton and the nearest electron is now about 1000 proton diameters. The

neutrinos are becoming increasingly isolated, and effectively cease interacting with

matter, so that the production of neutrons ends at T~3 x 109 K (t~13s), when the

proton:neutron ratio has shifted to about 83p:17n. The neutron’s greater mass has led to

its being outnumbered by protons in the cooling universe. We’ve seen that free neutrons

are unstable, and now that their creation has ceased, their numbers slowly fall as they

decay to protons.

Neutrinos remain in the universe – there are about 550 neutrinos in every cubic

centimetre, about the tip of your little finger – but they hardly react with matter, and fly

through the Earth and our finger tips at close to the speed of light, almost undetectable.

one tick of the clock

The universe is about 1 second old - one tick of the digital clock on my desk. So much

seems to have happened in so little time. But to get a clearer picture we need a different

kind of clock, one that ticks each time one particle interacts with another. Lawrence

Krauss calculates that in the universe's first second the particles in a volume of 1 cm 3

(roughly the tip of your little finger) experience about 1089 interactions - a stupendous

number. He goes on…”For comparison, during its 5 billion years of burning, in each cubic

centimetre in the fiery core of the sun a total of about 1055 interactions have taken place.

This is about 10 million billion billion billion times fewer collisions than occurred in the


McNeil, 04/23/11,

Atom, p. 62.

..., 23/04/11,

Kaufmann, p.737, or 550 million in a cubic metre. Close, "Particle Odyssey", p. 188, puts it at 300 neutrinos/cm3. Neutrinos are incredibly light. Their mass is uncertain, but only a few eV. But there are so many that their combined mass could be “10 times greater than all the matter in planets, stars and galaxies combined.” – Kaufmann, p.738.

..., 23/04/11,

Weinberg, p.109. Different writers all tell the same story, but with slightly different figures. Krauss, for example gives 82p:18n.

..., 23/04/11,

Values are approximate. I’ve used Williams, p.371, and others give similar figures.

..., 23/04/11,

Kaufmann, p.737, John Gribbin, "The Universe", p.71.

..., 23/04/11,

Krauss, p. 72-74. Weinberg estimates the circumference of the universe at a temperature of 101K as about 4 light years, but notes that to ask the size of the universe at this time may not even be a meaningful question – p.105.


same volume in the universe's first second. The number of collisions of atoms in this

volume of air during the 4 billion year history of life on Earth is about 1045, about 10 billion

times smaller still!”.

So the clock time is deceptive. The universe may appear very young, but the enormous

temperatures mean that the particles have experienced huge numbers of collisions, and

opportunities for interactions. In 1 second of clock time, two universes of matter have

been created, each capable of independent existence. One has been totally destroyed,

and the other very nearly so, such that only one billionth part of it remains, awash in a sea

of radiation.

energy budgets and lifespans

There is a link between this and the link between time and size in animals. The total

number of heartbeats in a lifetime is about the same for a hummingbird as for an elephant

or a whale. If you divide the life span by the number of heart beats you get about the

same number regardless of the size of the animal - "the total budgets for their actions are

the same". Metabolism, the rate of chemical reactions that consume biological fuel,

decreases as animal size increases; small animals "live faster". Our universe's

"metabolism", the rate of its nuclear reactions, increases with temperature and density, so

our small and hot young universe "lived faster".

3.11.6 Protons and neutrons start to combine, T ~ 109 K (E ~ 100 keV, t ~ 200 seconds) – the first nuclei

As we’ve watched the universe expand and cool we’ve crossed three thresholds – and

seen the end of the creation process of three particles; first of the massive W/Z, then of

protons and neutrons, and finally of light electrons. We have seen an entire universe of

anti-matter annihilated, to leave a tiny residual universe of matter. “For every billion

particles of matter and anti-matter, one was left behind. … The little particle that was left

behind, for every billion that were annihilated, is what makes galaxies, stars, planets and

people”. It is this “small seasoning of leftover electrons and nuclear particles” which are

“the main constituents of the author and the reader”.

the deuteron

So far, the universe has contained only isolated protons and neutrons. Even the strong

nuclear force can’t withstand the disruptive energy of very hot photons, and any nucleon


..., 23/04/11,

Weinberg, p.89.

..., 23/04/11,

Carlos Frenk in Jim AL-Khalili’s “Nothing”, ~57 minutes.

..., 23/04/11,

Bonner, p.116 et seq. Quote from p.117. The mathematical relation is... log(metabolic rate) (0.75 x log(mass) This allometric relation reappears in ch.5, on the basic constraints of biological life.


clusters were immediately broken up. However, as the universe expands, the radiation

photons are all the time being being stretched and cooled, and we now pass another

threshold. This one is marked, not by the end of production of an existing particle species,

but by the start of production of a new one. Now that photons have insufficient energy to

prevent it, we see protons and neutrons combining to make the first composite particle,

the first nuclear cluster – the deuteron,

p + n D(p,n)

This reaction starts when the temperature has fallen below about 1 billion degrees, and

the universe is about 3 minutes old.

neutrons are saved by deuteron formation

We’ve seen that a free neutron will decay into a proton by the weak interaction…

n0 p+ + e- + + 0.782 MeV

A free neutron's lifetime, its average "survival" time, is 886 seconds, about 15 minutes. In

the 3 minutes or so since neutron production ended they have been decaying, further

shifting the proton:neutron ratio, from 83p:17n to about 87p:13n. If things were to continue

like this all the neutrons would be lost, but instead, the remaining free neutrons are

gathered up into deuteron clusters. This is the first of a series of nuclear reactions that will

build up ever-bigger clusters of protons and neutrons. It is the first step in the long and

arduous journey towards our atomic world.

A free proton can’t decay to a neutron, because the proton is the lighter particle. But there

is another nuclear reaction that can consume protons – electron capture to create a

neutron and a neutrino

e- + p+ n0 +

0.5 + 938.3 < 939.6 ~0 MeV - the neutron is heavier by about 0.8 MeV.

The stability of protons is essential to the existence of the physical universe. Fortunately,

neither of these proton-consuming reactions can occur, because of mass difference

between the proton and neutron, but “it is a very small margin on which our existence

depends”.

We have seen that the strength of the nuclear force is a finely balanced thing: “a few

percent stronger and two protons (or two neutrons) would bind together. Nuclei consisting

of just two protons do not exist, but if they did all the hydrogen would have been


..., 23/04/11,

Mackintosh et al, p. 99. We looked this in the earlier section on the deuteron.

..., 23/04/11,

Williams, p. 71.

McNeil, 23/04/11,

Allday, p.263.and Croswell, p. 220, Williams, p.346.

McNeil, 23/04/11,

John Gribbin, New Scientist, says that if the neutrons had such a short lifetime that all were lost in 3 minutes or so, then there would be no deuterium, no helium, and no heavy elements in the universe today.

..., 23/04/11,

These two ratios from Weinberg, p. 110 – though on the previuos page he gives 86p,14n. Other writers give these figures, or something very close. You can use the exponential decay equation to check that about 3 of these 17 neutrons decay in the 3 minutes or so before the deuterium reaction starts.

McNeil, 23/04/11,

http://en.wikipedia.org/wiki/Neutron gives the mean lifetime as 885.7 seconds. A free particle's mean lifetime is a measure of how long it remains undecayed. Strictly, the lifetime is the time for the number of particles to decay to 1/e of their original number, where e is a "natural number" with a value of about 2.718. So in the lifetime the number of particles decays to about 37% of the original value. The half life (which can be written T1/2) is the time taken for a group of free neutrons to decay to half the original number. The two are related… Half Life = lifetime x ln2 (0.693) So the neutron's half life is… 885.7 x 0.693 = 614 seconds. John Gribbin, "The Universe", p.71, gives the neutron "a half life of 10.3 minutes". He also (New Scientist) describes the increasing precision of measurements of the neutron's lifetime, with the modern value as very close to 889 seconds, giving a half life of 616 seconds. The Brookhaven National Laboratory gives free neutrons " a half-life of 613.9 seconds". http://www.nndc.bnl.gov/chart/help/glossary.jsp#neutron The two terms seem to get interchanged quite a lot. For example, Krauss, p.77, gives "the average lifetime of a free neutron…about 600 seconds". In a scholarly work, Kris Heyde, p.133-5, gives the "half life" as around 890 seconds, but he also writes "…the measured lifetime (T1/2) for a free neutron…". He quotes work by J. Byrne et al. (Physics Review Letters, vol. 65, p.289, 1990) as giving the half life as 893.6 seconds. But these authors, in their paper, say this value is for the neutron's lifetime. Hmmmm. Robert Turnbull wrote in 1979 that "the half-life of the neutron is still one of the least accurately determined fundamental quantities in physics", and gave the then accepted value as 637 seconds - Turnbull, p.156.

..., 23/04/11,

Coughlan puts this at 3-4 minutes, T~8 x 108 K – p.221. Williams puts it at 225 seconds and 9 x 108 K – p.371. Weinberg puts t at t= 3 min 2 s, and T ~1 x 109 K – p.109. This is the start of what is sometimes called the Nucleosynthesis Epoch… http://rst.gsfc.nasa.gov//Sect20/A1.html


consumed in the Big Bang leaving none to power ordinary stars, including the sun….A

few percent weaker and the deuteron would not be bound”. In both cases, the

development of the physical universe would have been quite different, and the evolution

of life would be impossible.

All the nuclear reactions

The full set of 12 nuclear reactions in hydrogen-burning…

cluster size

1 2 3 4 5 6 7

p n H-2p,n

H-3p,2n

He-32p,n

He-42p,2n

Li-73p,4n

The first clusters of protons and neutrons

We have seen that each element is defined by the number of protons in the nuclei of its

atoms, while the number of neutrons can vary. There are three "versions" of hydrogen

atoms, and two of helium, and these are summarised in table 3.1.

element hydrogen hydrogen-2or deuterium

hydrogen-3or tritium helium-3 helium-4

chemical symbol H-1 H-2 H-3 He-3 He-4

number of nucleons in the nucleus 1 2 3 3 4


2 H-2 H-3 + p

2 H-2 He-4

H-2 + He-3 He-4 + p

H-2 + H-3 He-4 + n

H-3 + He-4 Li-7

2 He-3 He-4 + 2p

n + He-3 He-4

p + H-3 He-4

2 H-2 He-3 + n

2 H-2 H-3 + p

p + H-2 He-3

n + H-2 H-3

p + n H-2

McNeil, 23/04/11,

This is just for this first draft. There is a challenge to find a way to represent the nuclear reactions that will be meaningful to the non-technical reader. After all the conceptually difficult stuff on virtual particles, these reactions are as straightforward as building LEGO models. Yet the nuclear equations, while being correct, don’t convey this – not to me anyway. So, I’ve tried a couple of ways to show them visually for the monitor or the printed page. References: Croswell, p. 220, gives a total of 11, and Allday, p. 263, gives the 5 main ones that produce helium-4. This table is superfluous, it’s not going to help the reader, but it records most of the rich pattern of reactions, and I’ll keep it for this draft. It perhaps resembles the array of reactions in a living cell: anabolic – building things up, and catabolic – breaking them down.


protons and neutrons in the cluster (1p) (1p,1n) (1p,2n) (2p,1n) (2p,2n)

Table 3.1: The different atomic “versions” of hydrogen and helium.Thus, hydrogen-3, for example, specifies 1 proton in the nucleus, and a total of 3

nucleons, so there must be 2 neutrons.

This is the point where our atomic universe starts to come into view, with names that are

familiar. It's perhaps like the return from a foreign holiday - you know the town names, you

can read the advertisements, and the radio plays pop songs you know.

the creation of helium-4

We have seen that the nucleus of deuterium (1p,1n) is loosely bound – it is broken apart

at temperatures above about one billion degrees (1 x 109 K). Once the temperature is low

enough for deuterium to be stable, further reactions quickly occur that build up nuclei of

helium-4 (2p,2n). Figure 3.41 shows a particle view of the main nuclear reactions in the

fireball.

Figure 3.41: The sequence of nuclear particle reactions producing helium-4 (2p,2n) – to save space protons and neutrons are given as p and n respectivelyWe can imagine particles flying at high speed through space, colliding, fusing,

fragmenting - continually rearranging themselves, but steadily building up larger clusters,


p

n

p n

p and n collide and create a pn pair, hydrogen (H-2), deuterium...

…then two pn pairs collide and create trios of p and n, either He-3 or H-3…

p n

or...

p

ppn

n

nnp

+

+

p n

p n

pnnp

n np p

p

n

+

+

…and these trios collide with more pn pairs and create helium-4 (He-4) quartets.

H-2 or deuterium

H-3 or tritium

He-3 or helium-3

He-4 or helium-4

McNeil, 23/04/11,

There is a rich tapestry of nuclear reactions; Croswell, p. 220, gives a total of 11, and Allday, p. 263, gives the 5 main ones that produce helium-4. I only give a representative selection; the full set would be excessive. I've tried to present these in an accessible way for the reader, as particles and as moves in “nucleon-space”. And I've tried to write the reactions themselves in a more accessible way. They're usually written like this, for example… p + n ( 2H and 3H + 2H ( 4He + n

..., 23/04/11,

This is the ‘deutrium bottleneck’, Kaufmann and Freeman, p.737, and Weinberg, p.109.

..., 23/04/11,

Hmmm...this may not be true for all the generations in the family!


mainly helium-4. This sequence proceeds in 3 stages, with protons and neutrons

produced in the intermediate stages being fed back into the mix. We can think of these

nucleons behaving rather like Lego bricks, "clicking" together to make a variety of larger

clusters.

Nuclear “snakes and ladders”

However, this diagram can only show a few of the many reactions, and doesn't make

clear the systematic build-up of larger clusters. An alternative is to plot the nuclear

reactions in a graphical format in what we might call "nucleon-space" (figure 3.42). Here

the proton and neutron contents of each cluster are plotted on the y- and x-axes

respectively; adding protons moves upwards, adding neutrons moves to the right. We

thus have a 2-dimensional view of nuclear reactions as moves on a sort of nucleon

chessboard.

Figure 3.42: The nuclear reactions in the fireball plotted as a series of moves in "nucleon-space”. Follow the arrows, add up the nucleons, see where you land. The coloured squares represent unstable nuclides- we’re not concerned with these yet.


protons, p

lithium-Li, 3

helium-He, 2

Hydrogen-H, 1

0

+ p

+ n

He-4 + H-3 Li-7(2p,2n)+ (1p,2n) (3p,4n)

H-1(p)

H-3(p,2n)

H-2p,n

He-3 (2p,n) He-4 (2p,2n)

Li-6 (3p,3n) Li-73p,4n

+ p

0 1 2 3 4 neutrons, n

freeneutron, n

freeproton, p

+n

+ p

add

1 prot

on

add 1 neutron

He-3+H-2He-4 + n

2H-2He-3 + p

+ n

the neutron is fed back into the mix

the proton is fed back into the mix

Li-7 + p 2 He-4(3p,4n) + p 2(2p,2n)

..., 23/04/11,

Also called Segrè space or nuclide-space, and is a common way to represent the different nuclides. This is a screenshot detail, taken 10 June 2010, of the Interactive Nuclide chart, available at… http://www.nndc.bnl.gov/chart/

McNeil, 23/04/11,

I emailed Melissa Wallace, LEGO’s PR and Promotions manager, about using their name. She forwarded their lawyer’s reply (email received, 23 Feb. 2009)… “Yes, Melissa, we cannot prevent him, - so therefore, we should just ask for correct trademark use, i.e.: Always write “LEGO” in all uppercase letters, Use the ® at least once (first time) in a paragraph, and in headlines, Use a descriptive noun after the LEGO trademark “LEGO bricks” – and Finally, include a legal line: LEGO is a trademark of the LEGO Group, here used with permission. If pictures of LEGO bricks are used, - a “©2009 The LEGO Group” must be used right below the picture. Please stress that we do not allow the use of the LEGO logo. Furthermore we do not allow use of the LEGO trademark, photos etc on the cover of the book. Med venlig hilsen/Kind regards, Helle Rams Nørgård “


This figure shows the reactions given previously in figure 3.41, and a few more, including

the one producing lithium-7, where cluster (2p,2n) acquires cluster (1p,2n) and so moves

2 squares along and one up to arrive on the (3p,4n) square. If this cluster is hit by a

proton it can fragment into two helium-4 clusters. There is thus a rich tapestry of

interactions; small clusters fusing into larger ones, larger clusters being broken down into

smaller ones, and leftovers being recycled - all can be plotted as "moves" in nucleon

space.

The nuclear reactions in the rest of this chapter will be shown as moves in nucleon-space.

This will allow us to see the details of reactions and also the inexorable progress to the

right and upwards, as bigger nuclei are formed.

3.11.7 The end of nucleus formation – T ~ 3 x 108K (E ~ 30 keV, t ~ 13 minutes)

In order for small nuclei to fuse into bigger clusters they have to approach close enough

for the short range nuclear force to bind them together. This means the background

temperature must be high enough so they have enough speed to overcome the mutual

repulsion of their positive charges. When the universe is about 13 minutes old the

temperature has dropped so far that the nuclei are moving too slowly to make contact,

and the nuclear reactions cease. We left the universe at t ~ 3 minutes, with the ratio

87p:13n. Out of every 200 particles the 26 neutrons will combine with 26 protons to make

13 helium-4 clusters, each four times the mass of a proton. This gives a mass ratio of

helium as 13 x 4/200 = 26%, which is in good agreement with measurements, and is one

of the several experimental confirmations of the big bang theory. This is an impressive

example where calculations based on the behaviour of sub-atomic particles in a particle

accelerator give a result that agrees with measurements made on the entire visible

universe. So far, all the stars in the universe have burned very little (about 4%) of their

hydrogen to helium, so the current ~24 - 28% cosmic abundance of helium is well

explained. “It is a profound fact that the universe is made almost entirely of hydrogen and

helium, and this is because of what happened early in the Big Bang”.

After 13 minutes, about the time to the first ad-break in a commercial tv programme, the

universe looks a bit like this (figure 3.43).


McNeil, 23/04/11,

Diagram based on Kaufmann and Freedman, figure 28-10, p.710. No 5-nucleon cluster is stable. Be-7(4p,3n) is unstable and decays to Li-7(3p,4n) - Clayton p.42. The way the fractions of the nucleon clusters change in the cooling fireball is well established and is shown in a nice graph, figure 29-6 in Kaufmann and Freedman, and also available at… http://rst.gsfc.nasa.gov/Sect20/A1.html (accessed 11 April 2011) Observations of very old stars, formed from the primordial mix of nuclides tells us that Li-7 was present in tiny proportions – about 1 Li-7 nucleus for 5 billion protons – Clayton, p.37, and Hogan, p.114.

..., 23/04/11,

Hogan, p.96. The relative abundances of hydrogen, helium and deuterium are strong confirmation of the big bang model of the early universe – Hogan, p. 105.

McNeil, 04/23/11,

Delsemme, p.22.

McNeil, 23/04/11,

See, for example, Clayton, p.276, Silk, ch.5, Simon Singh, and Allday, ch.12, and Weinberg, ch,5. The big bang model correctly presicts the abundances of deuterium and lithium-7, using a ratio of several baryons per 10 billion photons – Hogan, p.102.

McNeil, 04/23/11,

John Gribbin gives 7p:1n, which scales up to 87.5p:12.5n, and gives 25% by mass of helium - "The Universe", p.73. Kaufmnann and Freeman give 1 He-4 nucleus:10 protons - p.737. Jonathan Allday gives 87p:13n, which leads to 26% helium – p.263. Hogan gives the 23-24% He – p.104. Weinberg puts the helium fraction at about 26%, and not exceeding 28% - p.110. Rod Nave gives 26% helium… http://hyperphysics.phy-astr.gsu.edu/hbase/astro/imgast/ndecay.gif Authors give slightly different figures; I've used figures here that are consistent with section 3.6.6. The maths... In a mix of 13 neutrons and 87 protons, the 13 neutrons will combine with 13 protons to make 6.5 He-4 nuclei, with a mass fraction of (6.5x4)/100 = 26%. So in the text it’s scaled up to 200 nucleons.

..., 23/04/11,

Williams puts it at about 30 min – p. 346.

..., 23/04/11,

This is not the usual way this is described, but it helps the reader see the Lego brick pattern of nucleus building. It takes familiarity with the nuclear physics to follow the statement "a He-4 nucleus acquires a H-2 nucleus to make a Li-7 nucleus". I think this would be distracting for the general reader.


Figure 3.43: Nucleons and electrons in the cooling fireball; protons and neutrons are gathered into small clusters, photons strongly interact with the charged nuclei, so the universe shimmers with scattered lightThe intense but brief heat of the early universe has fused all the surviving neutrons and

some of the protons into small clusters; pairs (H-2, deuterium) and trios (H-3 and He-3),

but mostly quartets of helium-4, and also a few septets of lithium-7, the biggest cluster

made.

The electrons and the oppositely charged protons and nucleon clusters are mutually

attracted, but as soon as they combine as neutral atoms “another photon would collide

with the atom and knock the electron free. With 1 billion energetic photons per electron,

the cards [are] stacked against the latter”.

Photons interact strongly with electrically charged particles, so they don't travel much

further than the nearest electron or nuclear cluster, "following a zig-zag path through

space like a high-speed ball in a crazy pinball machine." Thus “the universe was

completely filled with a shimmering expanse of high-energy photons colliding vigorously

with protons and electrons. This state of matter, called a plasma, is opaque, just like the

glowing gases inside … a neon advertising sign.” This would be like being in a fog, where

the light is scattered by the tiny drops of water suspended in the air, so that the whole fog

glows and you can’t see your way.

3.11.8 The threshold for ionisation - T ~ 3,000 K (E ~ 0.3 eV, t ~ 300,000 years) - the first neutral atoms


..

.neutrino - hardly interacts with matter -

-

-

-

-

-

-

--

p+

protons or hydrogen-1~ 93 particles in 100 (74% by mass)

p+

p+

n0

n0

helium-4 (2p,2n)~6-7 in every 100 particles (26% by mass)

n0p+

hydrogen-2 (p,n)or deuteriumabout 0.01% by mass

p+

n0 n0

hydrogen-3(p,2n)or tritium, about 0.0001% by mass

p+

p+p+n0

n0

n0n0

lithium-7(3p,4n) about 10-7 % by mass, 1 particle in about 5 billion

-electron1 for every proton n0

p+ p+

helium-3 (2p,n)about 0.001% by mass

..., 23/04/11,

Kaufmann, p. 709. This time, up to recombination, at ~300,000 years, is sometimes called the Radiation Era… http://rst.gsfc.nasa.gov//Sect20/A1.html

..., 23/04/11,


..., 23/04/11,

Krauss, p.90.

..., 23/04/11,

“The deuterium in existence today is primordial – made in the first few minutes after the Big Bang. This is known because any new deuterium made by reactions within stars is immediately destroyed by further reactions.” – Mackintosh et al, p. 130. See also Clayton, p. 16 – “The nucleosynthesis of [deuterium] is seen to have been even more strange than mankind had imagined. It is a nuclear ash of the fireball that began our universe…Stars are destroyers of deuterium.”


After this burst of activity, the “rush of creation began to subside. Physical processes

slowed … minutes turned to hours, hours to days, days to years, years to millenia .”

"Nothing much happens now for the next 300,000 years or so”. The fireball universe

expands and cools, the particles gradually slow down, the wavelengths of the photons

steadily increase. After about 300,000 years the temperature has fallen to about 3,000 K,

close to the boiling point of iron, and the average photon energy is now only ~0.26 eV.

The photons have a range of energies, however, and a tiny proportion of them have the

full 13.6 eV needed to remove an electron from the proton to which it is bound. But even

with 1 billion photons for each proton, this becomes insufficient, as the average energy

decreases. As the temperature keeps falling, the balance keeps shifting in favour of the

electrons, which settle in greater numbers on the protons and the nucleon clusters.

“Suddenly the face of matter was ready to change” and the universe now looks like this

(figure 3.44).

Figure 3.44: The separation of matter and radiation after the fireball. The nuclei are the same, and in the same places, but now they are combined with electrons, and the photons fly through without interacting. The universe is now transparent.

Space becomes transparent

No new particles have been created; things are merely rearranged. Each nucleon cluster

has gathered a number of orbiting electrons to make an electrically neutral atom. The

photons of radiation rarely interact with these neutral atoms, and fly unimpeded through

space. The "fog" of charged particles has cleared and space becomes the transparent


.

..

hydrogen-1

helium-4

hydrogen-2,deuterium

hydrogen-3,tritium

lithium-7

helium-3

--

n0

p+ p+

p+ --

n0p+

- p+

n0 n0

-

-

p+

p+

n0

n0

--

-p+

p+p+n0

n0

n0n0

neutrino

neutrino

..., 23/04/11,

The Hubble space telescope can see some of the first galaxies to form, that emitted their light when the universe was one tenth of its present age – so the photons have been travelling freely through space for around 12 billion years- Hogan, p.128.

..., 23/04/11,

“The raw materials were there, but not the architecture” – Krauss, p.89.


Diagram based on Kaufmann and Freedman, figure 28-10, p.710.

..., 23/04/11,

Krauss, p.89.

..., 23/04/11,

Krauss gives the average photon energy at 3,000K as ~0.6eV, and writes that ~1 in 10 million photons have 20x this energy, enough to ionise hydrogen - Krauss, “Atom”, p.91. However, using the equation E=kT (Weinberg, p.81) gives an energy of 0.26 eV at 3,000 K, and this agrees with HyperPhysics… http://hyperphysics.phy-astr.gsu.edu/hbase/astro/transp.html#c1 So, while Krauss paints an engaging scenario, I can’t be sure of his figures – Jan 2011.

McNeil, 23/04/11,

Krauss, p.89.

McNeil, 23/04/11,

Different sources give different times for this event. In many books it is given as 300,000 years (eg. Jonathan Allday, Marcus Chown, George Smoot, Armand Delsemme, Lawrence Krauss, and Joseph Silk). But other sources put the time at 700,000 years, for example, George Smoot, at http://aether.lbl.gov//www/tour/elements/early/early_a.html, and Rod Nave, at http://hyperphysics.phy-astr.gsu.edu/hbase/astro/bbcloc.html#c8 who bases his figure on Weinberg, p.112. Others put it at close to 400,000 years… http://www.damtp.cam.ac.uk/user/gr/public/cos_home.html http://map.gsfc.nasa.gov/resources/animconcepts.html (all accessed 11 April 2011) Williams gives 1 million years and 2,000 K – p.347, and Hogan gives half a million years – p.21 and 92. The currently accepted figure seems to be 379,000 (8,000 years: email from Paul Butterworth (18 March 2009). ([email protected]) I've rounded this up to 380,000 years. I'm not really concerned with the precise time, and in the absence of a clear consensus, I'll go with the majority, and the nice round number of 300,000 years. All seem to agree on a temperature of about 3,000K. What's important is that the universe cooled to a point where radiation and matter went their separate ways. The universe became the transparent place we know today, and gravity was able to start its patient, inexorable work on neutral matter. This temperature “marks the transition between a ‘radiation-dominated’ era, in which most of the energy in the universe was in the form of radiation, and the present ‘matter-dominated’ era, in which most of the energy is in the masses of the nuclear particles” – Weinberg, p.76.

..., 23/04/11,

Kaufmann, ch. 28. This account deliberately very brief, and focusses solely on the creation of baryonic matter, the "seeds" of our atomic world. We'll return to this in more detail later in this chapter, in describing the formation of stars. What if all the free neutrons had decayed? What if no nucleon clusters formed, and after the universe's first quarter of an hour there were only protons? This is one of the questions I can't answer (yet).

McNeil, 23/04/11,

Allday, p.264.

..., 23/04/11,

Krauss, p.81.


medium we're familiar with. The neutrinos continue as if nothing had changed – and they

still do!

We can imagine electrons settling around the nuclei rather like falling snow gently settles

on a landscape - on trees, fences, houses, cars. We know it’s the same solid, hard-edged

world underneath, but the shapes and patterns of things have changed and softened. So

it is with the electrons – they enfold the nuclei and enable a new level of subtle and

intricate particle interactions. All we've seen is the universal "fog" clear, and the light shine

freely, but the universe is transformed, and ready to emerge at a new level of complexity,

as we shall see in the next chapter.

Atoms of matter and photons of radiation now follow a quite different paths. We now have

separate matter and light - radiation has decoupled from matter. There are about 1 billion

photons for each proton or neutron, “a large number because matter when it was created

was only a trace contaminant born of the light”. The tidal wave of energy has subsided,

leaving small composite matter particles washed up on the shore of existence. So

"nothing" has become "something". What is this something?

The first atoms

For the first time there are atoms of substances we can recognise. There is hydrogen: the

flammable gas that lifted the first airships; the gas produced when we drop zinc into acid,

or if we over-charge a car battery. There is helium: the gas used in party balloons, that

also makes our voice go squeaky. And there is lithium: the metal used in batteries and

also to treat bipolar disorder. There the list stops - where are the familiar elements such

as carbon, oxygen, iron and copper? But if nucleons can gather in clusters of up to 7, and

containing up to 3 protons, then maybe they can go further.

3.11.9 review

From radiation-dominated to matter-dominated

We’ve gone from a universe dominated by radiation, where photons directly created

matter, to a universe that is now dominated by that matter. In our current universe the

average density of matter is tiny, equivalent to a few hydrogen atoms in each cubic metre.

In contrast, there are on average about 550 million radiation photons in every cubic metre,

with all possible wavelengths from radio through visible light to gamma-rays. “In other

words, the photons in space outnumber atoms by roughly a billion to one. In terms of total

number of particles, the universe thus consists almost entirely of microwave photons.”


..., 23/04/11,

Kaufmann, ch. 28-5, p708. Figures and quote from p.708 (and figure agrees with Weinberg, p.73) The densities of radiation and of matter have both fallen since the big bang. The formal transition from radiation-dominated to matter-dominated is calculated to have occurred about 2500 years after the big bang, when the density of radiation became less than the density of matter. At the time of the transition, a typical wavelength was about 40 nm, in the ultraviolet part of the spectrum. Now the typical wavelength is about 1 mm, in the microwave region. The figure for the matter density is the average value, if matter had not clumped into stars and planets, as solids, liquids and gases. For example, there are about 5 x 1025 atoms in each cubic metre of air we breathe. The density of matter is important for the ultimate fate of the universe – Kaufmann, ch. 28. I’ll not consider the nature of dark matter here.

McNeil, 23/04/11,

http://en.wikipedia.org/wiki/Lithium (accessed 11 April 2011)

..., 23/04/11,

…and they are extremely old. The protons in just about every hydrogen atom were created within a fraction of a second after the big bang; nuclei of deuterium, helium and lithium we encounter now may well be only a quarter of an hour less old. See the comment in the section “The evolutionary eras after the first minute”, in http://rst.gsfc.nasa.gov//Sect20/A1.html (accessed 11 April 2011)

McNeil, 23/04/11,

paraphrasing Krauss, p.24.

..., 23/04/11,

Hogan, p.93.

McNeil, 23/04/11,

This very brief and selective account focusses solely on the creation of baryonic matter, the nuclei of our atomic world. I come back to this. The temperatures of radiation and matter can now be very different. The present-day background radiation has a very uniform temperature of 2.726 K, whereas matter exists at a huge range of temperatures, from a few Kelvin in deep space to hundreds of millions of Kelvin in big stars – Kaufmann, p.710.

..., 23/04/11,

This formation of neutral atoms is called the ‘Era of Recombination’, referring to electrons “recombining “ to form atoms. Kaufmann notes that “the name is a bit misleading, because the electrons and protons had never before combined into atoms.” – p. 709. Weinberg: “a singularly inappropriate term, for … the nuclei and electrons had never in the previous history of the universe been combined into atoms!” – p.64.


However, despite their overwhelming numbers, these photons have little influence over

matter, since their wavelengths have been stretched, and their energies reduced, by the

universe’s expansion. In the matter-dominated universe we will see matter go its own

way, more or less regardless of the huge numbers of photons passing by. Steven

Weinberg gives a perspective; “well before the contents of the universe became

transparent, the universe could be regarded as composed chiefly of radiation, with only a

small contamination of matter. The enormous energy density of radiation in the early

universe has been lost by the shift of photon wavelengths to the red as the universe

expanded, leaving the contamination of nuclear particles and electrons to grow into the

stars and rocks and living beings of the present universe”.

the first ad break

The first phase of nucleosynthesis – nucleus-building - has lasted about thirteen minutes,

roughly the time to the first ad break in a TV programme. We’ve seen how nuclear

reactions can only occur at modest temperatures, high enough to bring the nuclei

together, but not high enough to break them up. But the temperature window for this is

small – between 100 and 1,000 million Kelvin (108 – 109K), and the universe cooled

through this window in a few minutes, enough time for only three elements to appear.

The next phase of nucleosynthesis has been running for the last 13 billion years or so,

starting when the universe was about a billion years old. Nuclei are built up in stars, which

recreate the heat and pressure of the early universe, and yet this phase is powered by the

weakest of the universal forces – gravity. We'll look at this in section 3.12, but before this

we need to understand the principles of nucleon clustering.

The physical universe now comprises:

nuclei – that is, protons and their clusters with neutrons, up to cluster-7, each

carrying a positive electric charge

electrons, carrying negative electric charge, in numbers that equal the protons

neutrinos, with no charge, that barely acknowledge matter’s existence

photons of radiation, about 1 billion for each nucleon, that interact weakly with

matter

So there are just two “players” left on the universal stage: negatively charged electrons

and nuclei, nuggets of highly dense matter, and as far as the electrons are concerned,

differing only by their positive charge. The weak and strong interactions confine their busy


..., 23/04/11,

One of these living beings, Steven Weinberg, wrote this towards the end of his book “The First Three Minutes”: “As I write this I happen to be in an airplane at 30,000 feet, flying over Wyoming en route home from San Francisco to Boston. Below, the earth looks very soft and comfortable – fluffy clouds here and there, snow turning pink as the sun sets, roads stretching straight across the country from one town to another. It is very hard to realize that this is all just a tiny part of anoverwhelmingly hostile universe. It is even harder to realise that this prent universe evolved from an unspeakably unfamiliar earlier condition, and faces a future extinction of endless cold or intolerable heat. The more the universe seems comprehensible, the more it also seems pointless. …The effort to understand the universe is one of the very few things that lifts human life a little above the level of farce, and gives it some of the grace of tragedy.” He seems to be saying that because conditions amenable to living things in the universe are highly localised both temporally and spatially, this means that it is “pointless”?

..., 23/04/11,

Weinberg, p.76.


activities in the nuclei. The much weaker electric interaction binds the oppositely charged

nuclei and electrons into neutral atoms. And the even weaker gravity can now act on the

atoms because they have neither attract nor repel, and they have mass. All the time the

neutrinos and radiation photons pass by, effectively indifferent to the doings of the neutral

atoms. Matter is now subject to gravity, by far the weakest of all the forces, but the only

one remaining.

3.12 Nucleosynthesis 2 - the next 14 billion yearsThe universe, having cooled to below 3,000K, now comprises almost exclusively neutral

atoms. This simple change has two enormous consequences: (1) matter and radiation

now go their separate ways, and radiation photons are free to stream without restraint

through the entire universe; (2) the electrical interaction is effectively confined into neutral

atoms, so the universe is a uniformly neutral medium, with no long range electrical

attarctions or repulsions. The weak and strong interactions are fundamentally very short-

range, and don’t extend outside the nuclei. This leaves the force of gravity, so much

weaker than the other three interactions, free to work its infinitely patient attraction on

these neutral atoms that now fill space. We’ll look now at what follows from these two

changes.

3.12.1 The cosmic microwave background

matter and radiation go separate ways

The fog of scattered light cleared, and our universe became transparent - but what was

there to see? There came a moment when each photon interacted with a charged particle

for the last time, after which it travelled freely through space, uninfluenced by neutral

atoms. When we look all around us out into space we see this radiation as the cosmic

microwave background, from the last time photons were scattered by matter at an

temperature of about 3,000 K.

photons are stretched by expanding space

What does an object at 3,000 K look like? The temperature of the tungsten filament in a

incandescent light bulb is 2,000 – 3,300 K; the visible surface of the sun is about 6,000 K.

So we can imagine the high energy photons, corresponding to the white-hot matter from

which they were scattered. But the universe has expanded to be about 1,000 times bigger

since then. Photons with energies corresponding to 3,000 K have had their wavelengths


McNeil, 23/04/11,

http://en.wikipedia.org/wiki/Incandescent_light_bulb (accessed 13 Jan 2011). Tungsten’s melting point is ~3,700 K. Hogan gives the sun’s surface temperature as 5,800 K – p.89.

..., 23/04/11,

Starlight from the farthest known galaxies has been travelling steadily to reach us for more than twice the age of the Earth – that is about 9 billion years – Adams and Laughlin, p.35.


stretched by a factor of about 1,000, and consequently their energies reduced by the

same factor (figure 3.45).

Figure 3.45: The source of the cosmic microwave background radiationWe are now surrounded by photons that have been travelling unimpeded through space,

that now have a temperature of about 2.7 K. “The sky is not actually dark; it is just like the

surface of the sun, only 2,000 times cooler”. We can no longer see these photons with the

naked eye, for they are microwaves, but if we look out into space with the right detector, in

every direction we see this cosmic microwave background (CMB) – figure 3.46.

Figure 3.46: Left: The cosmic microwave background (CMB) radiation from the universe all around us. The temperature is almost uniform, but there are tiny /tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

Time: 0 ~300,000 y about 14x109 y

Temperature: 3,000 K 3 K

separate nuclei and electrons

Radiation decouples from matter…

The big bang singularity

…and arrive at the here and now.

…and photons can now travel freely through space, with their energy reducing as the universe expands…

the temperature has fallen by a factor of ~1,000

+

--

-

+

+expanding space stretches the photons

the universe has expanded by a factor of ~1,000

..., 23/04/11,

The left image is from http://aether.lbl.gov/Images/wmap2.jpg It's based on 5 years of data from the Wilkison Microwave Anisotropy probe (WMAP) satellite, which built on the Cosmic Background Explorer (COBE) satellite data. The WMAP site is at http://map.gsfc.nasa.gov/news/index.html Right image from… http://aether.lbl.gov/www/projects/cobe/COBE_Home/DMR_Images.html (and is figure 14.13 in Allday) George Smoot and Marcus Chown tell the stories of the COBE project. Simon Singh, ch.5, tells the story of the discovery of the CMB, and its relevance to big bang theory. The excellent nasa web-site with animations and explanations… http://map.gsfc.nasa.gov/ A thorough article, with links… http://en.wikipedia.org/wiki/Cosmic_microwave_background

..., 23/04/11,

Hogan, p.79.

..., 23/04/11,

All are agreed on a temperature of ~2.7 K, but there seem to be inconsistencies in other respects. Wien’s Displacement Law states that for a black body spectrum, the wavelength for the peak intensity and the temperature are related by… (peakT = 3 mmK. So a temperature of 2.7 K should give a radiation intensity peak at (peak =3/2.7=1.1 x 10-3 =1.1mm. This agrees with Adams and Laughlin (p.19), and HyperPhysics (http://hyperphysics.phy-astr.gsu.edu/hbase/bkg3k.html#c1). These authors give frequencies in mm. But others show black body spectra with peaks at ( very close to 2 mm, for example, Barrow (p.14), Silk (p.51), and Hogan (p.87), and Smoot (fig. 8, Nobel Prize address).The graphs of Weinberg (p.59) and Smoot (p.86) have peaks at a wavelength between 1 and 2 mm. This latter group of writers give frequency in units of cycles per centimetre (cm-1). Silk gives the Wien’s law calculation in the text, coming out with (peak = 1 mm, but his graph has its peak at ~5 waves/cm, so (peak ~2mm. There seem to be inconsistencies, but I’m reluctant to conclude that experienced scientists are making simple mistakes; maybe there’s some subtlety I’m missing. The black body spectrum is a detail the reader doesn’t need here. Fortunately, all seem agreed on the temperature of ~2.7 K! So, we’ll move on – 12 Jan 2011.

..., 23/04/11,

This is based on the diagram given by Silk, p.55, and at… http://aether.lbl.gov/cmb.html

..., 23/04/11,

Hogan puts it at 1,100 – p.92. He points out that “the notion of ‘stretching’ is the most accurate verbal metaphor for the mathematical description of the cosmic redshift”. Explaining it as Doppler shift, due to the xpansion velocity, is “less scrupulously accurate” – p. 60.


“ripples”, with cooler (blue) and warmer (red) regions differing by a few millionths of a degree from the 2.73 K average. Even the smallest details in the inset picture are big enough to develop into clusters of galaxies (100 μK is 100 millionths of a degree).Right: The colour coding gives an idea of the size of the tiny temperature variations across the universe. (1 K = 1 millionth of a degree)a wall of light

The microwave background, this “wall of light” from the last scattering of radiation photons

off the charged nuclei at ~3,000 K, shows us the entire universe at an age of about

300,000 years. We can’t see further back than this, the dotted line in the diagram, for the

universe was then an impenetrable fog of scattered light. The average temperature is 2.73

K, just above absolute zero, and is almost, but not completely uniform. The false colours

suggest big temperature variations, but the differences are tiny - only millionths of a

degree between the coolest blue and the warmest red regions.

ripples in the microwave background

It's thought that these temperature ripples originated in tiny quantum fluctuations in the

energy density of space in the very early universe, and were stretched by the universe's

subsequent expansion. “Tiny temperature differences are the scars left by the quantum

vacuum on our universe. These irregularities, created in the first moments of existence by

the teeming quantum vacuum, meant that the matter of the universe didn’t spread out

completely evenly. Rather, it formed vast clumps, that would evolve into the galaxies and

clusters of galaxies that make up the universe today. … It now appears as if the quantum

world, the place we once thought of as empty nothingness, has actually shaped

everything we see around us.”

The temperature "ripples" show that the density of matter in the early universe was very

slightly uneven. The scale of these ripples was huge; even the smallest features in the

inset picture would expand to be as big as a whole cluster of galaxies. These density

fluctuations were the seeds that later formed the enormous chains and clusters of

galaxies that we see today. Our own galaxy is the result of one such quantum fluctuation.

“The idea that an object with billions of stars, like the Milky Way, began life as a quantum

fluctuation … of the vacuum, an object of sub-microscopic scale, is mind-boggling.”

3.12.2 Collapsing gas clouds

gravity takes over the show


..., 23/04/11,

Carlos Frenk, in Jim Al-Khalili, “Nothing”, ~54 minutes.

..., 23/04/11,

Kaufmann, ch.29-5. Adams and Laughlin, p.20, and p.34.

..., 23/04/11,

Jim Al-Khalili, “Nothing”, ~50 minutes. The quantum vacuum will have been described in the first chapter.

McNeil, 23/04/11,

I'm deliberately leaving the Inflationary phase of the universe out of this. Can be found in… Silk, p.181,

McNeil, 23/04/11,

Universe of California at Berkeley… http://universeadventure.org/eras/era1-consequences.htm Adams and Laughlin, p. xxv, Delsemme p.27, and Hogan, p.156.

..., 23/04/11,

Also called the surface of last scattering - George Smoot's web-site, on the cosmic microwave background http://aether.lbl.gov/cmb.html (accessed 12 Jan 2011). Hogan gives a helpful summary of the microwave background – ch.5.


This brings us to the second consequence of radiation decoupling from matter - the work

of gravity. Once there were no long range repulsions between charged nuclei, "gravity

completely took over the show", and started to work on the almost imperceptible clumps

of matter. Gravity can't create clumps of matter in a perfectly uniform gas, it can only

amplify pre-existing clumps. As the universe expanded, these clumps expanded very

slightly less than the surrounding regions; every time the universe doubled in size these

clumps did not quite double, and the density differences slowly grew. Once the local gas

density was more than about twice the average value, then the local effect of gravity was

strong enough to overpower the universal expansion. Huge clouds of gas atoms thus

developed, their mutual attraction brought the local expansion to a halt, and they then

began to draw in on themselves - they started to collapse. Their slight density excesses

grew at the expense of their surroundings - "the rich get richer, the poor get poorer". After

about 1-2 billion years, the first super-clusters of galaxies were well established, within

which our own Milky Way galaxy emerged.

We're now going to follow one large gas cloud, with 8-50 times the mass of our sun – that

is, 8-50 solar masses - as it collapses. Large gas clouds create large stars, which undergo

all the processes of building up the heavy nuclei, so we can use this to learn how the all

the elements in the modern universe have been created.

the first stages of collapse

We now have a cloud of gas that has started collapsing; the atoms are falling in on

themeselves. You can perhaps think of the giddy feeling at the top of a fairground ride, as

you start to fall slowly at first, then faster and faster. Our gas cloud will experience millions

of years of such falling, as its atoms of matter come together. When this starts, the cloud's

average density is about 1 atom/10 cm3 (1 atom in your little fingertip), so the distance

between atoms is over 100 million times their individual size. The chances of collisions

between atoms is tiny, but the cloud is so huge, and there are are so many atoms, that

collisions will occur, and become more frequent, and more energetic. The gas cloud is

starting to compress under its own gravitational attraction, its own weight - this is the birth

of a star. Gravity and matter meet and compete, and this will be the constant theme

throughout the cycle of the star's life and death. We'll see that gravity will always win,

even to the extent of crushing some of the star out of its material existence, but the

universe will be immeasurably enriched in the process.

pumping up a bike tyre


McNeil, 23/04/11,

Krauss, Atom, p. 95. If we take a human "diameter" as 1 metre, then on a human scale the nearest person will be 100 million metres away… a quarter of the distance to the Moon.

McNeil, 23/04/11,

Delsemme, p.31, Hogan, p.21, and Silk, p.175.

..., 23/04/11,

Silk, p. 171.

McNeil, 23/04/11,

Silk, p. 171.

McNeil, 23/04/11,

Krauss, Atom, p. 93. et seq. See also Hogan, ch 7.


So, let's start with the process of compressing a gas, for example, in pumping up a bike

tyre. After only a few vigorous strokes, the end of the pump feels hot - where has that heat

energy come from? It's worth taking a moment to think about the processes going on

inside the bike pump - figure 3.47.

Figure 3.47: How compressing a gas heats it upA typical bike pump, with the piston drawn back will contain about 5 x 1021 particles of air -

five thousand billion billion. At a "normal" room temperature of about 20oC (close to 300K)

these air particles are moving at an average speed of close to 500 metres/second - half a

kilometre each second. The gas particles also collide with and rebound from each other.

In the air we breathe, the particles have a size of about 0.1 nm, and are on average about

3 nm apart, a distance they can cover in about 7 million millionths of a second (7x10 -12 s).

The average air particle will then be experiencing something like 150 billion collisions

each second, with other particles and with the walls of its container. It's the huge numbers

of these tiny collisions against the inside walls of the bike tyre that keep it inflated. This is

a world of unceasing and furious activity. And yet at normal temperatures the energies

involved are tiny. Remembering our rule of thumb, that a temperature of ~12,000K is

equivalent to an energy of 1 eV, we can see that at our normal air temperature of 300 K,


to tyre

gas particle rebounds slightly faster from the moving piston

piston moving in

collisions with the walls produce the gas pressure

gas particles collide with each other

average particle speed is about 500 m/s.

the end of the pump feels hot

the same number of gas particles are confined in a smaller volume - so the pressure increases…

…and the particles are moving faster…they are hotter - increasing the pressure still more.

when the pressure inside the pump is bigger than the tyre pressure, then air will flow into the tyre.

McNeil, 23/04/11,

Allday, p.257 and Weinberg, p.81.

McNeil, 23/04/11,

Is this section appropriate – here or anywhere? It’s a thought-provoking peek into the atomic scale of events in an act as familiar as pumping up a bike tyre. It also illustrates how collapsing gas clouds get hotter. I've used some simple physics here: taking air density as about 1.3 kg/m3, and the mass of an air "particle" as 5 x 10-26 kg (between oxygen and nitrogen), and the size of a diatomic oxygen molecule as about 0.1 nm (from… http://www.webelements.com/ ) So a 1m cube of air contains about 2.6 x 1025 particles, which will be about 3 nm apart. Knowing that the average particle ke = 1.5kT where Bolzmann's constant k = 1.4 x 10-23 J/K, then we can calculate this average ke as about 6 x 10-21 J or 0.03 eV. I've used the simple word "particle" here, to avoid burdening the reader with ideas of atoms or molecules. Most pre-university physics text books cover this in the section on the kinetic theory of gases: eg Nelkon and Parker (old but still sound), and Ken Dobson. This gives the total number of collisions in 1 second in 1cm3 of air as about… 150x109 x 2.6x1019 = 3.8x1030 So in the 4 billion years of Earth's history, there will have been about 5 x 1047, which is close to Lawrence Krauss's estimate, given earlier in this chapter.

McNeil, 23/04/11,

Marschall, Supernova, p. 126.


the average particle kinetic energy is a mere 0.03 eV - the slightest pat on the back for an

atom.

compressing leads to heating

In the bike pump the air particles gain speed when they rebound from the incoming piston.

Faster means hotter; compressing the gas heats it up. For a gas cloud in space, gravity is

the piston; the gas particles fall in on each other, converting potential to kinetic energy;

the gas cloud warms as it is compressed. The gas particles strike each other with more

and more energy as time passes and the cloud shrinks. At first the particles rebound from

each other with no loss of energy, but as the temperature rises, some of the energy of

collisions is emitted as light radiation. So in our shrinking gas cloud, we see gravitational

potential energy converted to kinetic energy and radiation, in roughly equal proportions.

The loss of energy as radiation means the gas cloud does not heat up so much, and this

speeds up the contraction.

the gas cloud starts to glow

The gas cloud thus starts to emit radiation, starting in the invisible infra red, and working

towards visible light. The intensity of radiation can be enough to blow away the outer

layers of the cloud, ejecting as much as half of its original mass. When the temperature

reaches several thousand degrees the hydrogen molecules and helium atoms have their

electrons stripped from them, leaving just naked nuclei. The last time the nuclei were free

of electrons was in the early universe, about 300,000 years old. The presence of charged

nuclei means that the radiation can't escape, but is trapped inside the gas cloud. The gas

cloud dims and its contraction slows as the internal temperature starts rising rapidly,

reaching several million degrees.

the hot gas cloud becomes a true star

Our huge diffuse gas cloud has compacted and heated up, and is emitting huge amounts

of energy as radiation - but it is not yet a star, for its energy output has come from

converting its gravitational energy to heat and light. Finally, with a core temperature of

about 10 million degrees, there occurs the first of a series of nuclear reactions that

generate energy - our gas cloud has become a true star. Figure 3.48 shows the

development of galaxies and stars, enormously dense concentrations of matter, from the

slight ripples in the CMB.


..., 23/04/11,

from… http://map.gsfc.nasa.gov/media/030651/030651_320.wmv These are screen shots of the QT format animation. This site provides a number of helpful animations of many aspects of the CMB data and what it tells us about the universe. Permission to use selected images has kindly been given by the NASA/WMAP Science Team (Paul Butterworth: email 18 March, 2009). Paul Butterworth links the nasa animation to… http://en.wikipedia.org/wiki/Timeline_of_the_Big_Bang

McNeil, 23/04/11,

I'm following Silk's distinction, though he has nuclear reactions starting at 1 million degrees, p.18. Kaufman and Freedman, ch. 20 cover the birth of stars.

McNeil, 23/04/11,

Recall the uiversal fog before recombination.

McNeil, 23/04/11,

The processes in this paragraph are graphically described by Krauss, chapters 7 and 8.

McNeil, 23/04/11,

See the virial theorem… http://hyperphysics.phy-astr.gsu.edu/hbase/astro/gravc.html#c2 and also Krauss, p.110. Gribbin, p.160, describes the collapse of a gas cloud, and the roles of molecules like carbon monoxide and water in cooling the cloud by radiating infra-red.

..., 23/04/11,

The recent England-Australia Test cricket series used “hot-spot technology” to determine if the ball had made any contact with the bat. Infra-red cameras detected the local rise in temperature where the ball had collided with the bat and set the molecules in the wood vibrating faster. See… http://en.wikipedia.org/wiki/Hot_Spot_%28cricket%29 (accessed 13 Jan 2011).


Figure 3.48: From ripples in the cosmic microwave background (CMB) to galaxies and stars: (a) the view of the CMB across the whole sky; (b) a cold spot in the microwave radiation; (c) within the forming clouds of gas the first stars light up (about 100 million years); (d) the early universe is lit up by a variety of stars and galaxies, and (e) the WMAP satellite looking through today's universe back in time to the CMB.We're ready now to look at the remarkable processes whereby our star assembles

protons into all the elements in the universe.

3.12.3 Stars and the elements

Our story of stars focusses solely on their rôle as nuclear furnaces that forge light

elements into heavier ones. We've seen how nucleosynthesis - the formation of the nuclei

of the elements - in the fireball was limited by the rapidly falling temperature and short

time available. Now we'll see stars recreate the conditions in the fireball and sustain them

for millions of years.

We'll continue the story of our gas cloud, now that it has become a large star, and this will

show us in one narrative, the main processes that have laid the foundations of our

material existence. ”Thus it is possible to say that you and your neighbor and I, each one

of us and all of us, are truly and literally a little bit of stardust”.

The basic nucleosynthesis reactions

the battle between gravity and matter


ab

c

de

McNeil, 23/04/11,

from Willy Fowler's Nobel prize address, at … http://nobelprize.org/nobel_prizes/physics/laureates/1983/fowler-lecture.pdf

McNeil, 23/04/11,

Kaufman and Freedman, p.550. I'm ignoring dark matter as well as dark energy, and sticking to well established knowledge.

McNeil, 23/04/11,

There’s a huge literature on stellar nucleosynthesis, a lot of it very well written and accessible to the non-technical reader, for example (in no particular order), Krauss, Sagan, Marschall, Croswell, Greenstein, Chown, Gribbin, Kaufman and Freedman, chapters 21 and 22. There is of course, a lot available on-line, which includes animations and stuff not available in books. Wikipedia has articles on just about everything, which give references and links to further material. A comprehensive resource is… http://rst.gsfc.nasa.gov/Sect20/A1.html A helpful summary of stellar eveolution in 2 formats at… http://chandra.harvard.edu/edu/formal/stellar_ev/story/story.pdf http://chandra.harvard.edu/edu/formal/stellar_ev/story/index.html (accessed 13 January 2011). A scholarly review is at… http://cococubed.asu.edu/papers/wallerstein97.pdf (accessed 13 January 2011).


A star reveals – in a sense, it embodies - the conflict between gravity and matter - (figure

3.49).

Figure 3.49: The conflict between gravity and matter: stars go round the gravity-matter circuit, undergoing a series of nuclear reactions, until all their fuel is exhausted.A gas cloud contracts under gravity, until its temperature and pressure are high enough to

ignite a nuclear fuel, when the cloud becomes an active star. The fuel “burns” and

releases the energy that supports the star’s weight, and halts contraction. When this fuel

is exhausted, further contraction under gravity increases the star’s core temperature until

the next nuclear fuel ignites. Each fuel represents one lap of the gravity-matter circuit.

Gravity always wins in the end, because it never gives up, whereas a star's nuclear fuel

supply is finite. The star spirals in towards its end, which is largely dependent on the mass

of the gas cloud from which it formed. Thus “stars live their lives on the brink of disaster”,

always on the inwards gravity-matter spiral.

Temperature and pressure are the dominant factors in a star’s gravity-driven life. They

decide when a star’s life begins, and we will see a star’s life end when the matter it is

made of can no longer sustain the temperature and pressure imposed on it.

the creation of the elements

Oliver Sacks has written how “I was pleased when I was told that we ourselves were

made of the very same elements as composed the sun and stars, that some of my atoms

might once have been in a different star. But it frightened me too, made me feel that my

atoms were only on loan and might fly apart at any time, fly away like the fine talcum

powder I saw in the bathroom”. For the stars have created the elements of our material

world, through a set of 3 basic nuclear reaction processes. These can be summarised as

follows…


gas cloud

?

Nuclear burning

Gravitational contraction

Exhaustion of fuel

Core heating

McNeil, 23/04/11,

Mason, p.50, and Wallerstein. I’ve reduced the classic 8 processes proposed by Burbidge et al. (commonly known as B2FH) to 3 types: 2 that build up bigger nuclei, and 1 that makes them smaller, omitting only the e-process. Wallerstein, section XV, discusses the development of the modern view of the e-process. The e-process was originally invoked to explain the great abundance of the iron-group nuclides in stellar nucleosynthesis. The original concept of nuclear equilibrium has since been extended to cover a hierarchy of statistical equilibrium conditions, and it is now recognised that usually only a limited quasi-equilibrium condition is possible. The original e-process is now regarded as inadequate to explain the origin of the iron isotopes in the solar system (p.1060). We can see conditions favouring the e-process in the pre-supernova cores of massive stars, and in the subsequent supernova explosion, both in the assembly of nucleons into Fe-group nuclides close to the neutron star, and in explosive burning a little further out. In all these cases, the very high temperatures allow nucleons to ‘re-shuffle’ so that the iron-group nuclides, with the largest binding energies, emerge. The fusion reactions have distinct mechanisms, but can be treated as conceptually the same, as the grouping of nucleons into progressively larger clusters, with the release of binding energy. I think this will help the general reader, without losing accuracy. I’ve put all the fusion processes together, from hydrogen to silicon, and treated the e-process as the natural end-point of the fusion sequence, where the nuclides have the largest binding energy possible. The fuels are treated as burning separately in a massive star, but in a supernova explosion we see separate nucleons assembled into He-4 and then smoothly to Fe-group nuclides. Similarly, explosive burning of any fuel will produce the iron-group, if the temperature exceeds ~5 billion degrees (WHW, p.1053). One imagines throwing a handful of protons and neutrons, or of any nuclei into a temperature above ~5 billion degrees, and seeing them assemble in seconds into a mix of iron-group nuclides.

..., 23/04/11,

Oliver Sacks, p. 5.

..., 23/04/11,

Marschall, p. 125.

McNeil, 23/04/11,

after a diagram at…http://www.astro.columbia.edu/~archung/labs/spring2002/lab03.html (accessed 13 January 2011).


enlarging nuclei

o fusion reactions - in which the main nuclear fuels “burnt” are: hydrogen, helium,

carbon, neon, oxygen and silicon, whereby nuclei combine into bigger clusters to

release binding energy, and generate heat

o single nucleon captures - including

neutron capture processes, where a nucleus picks up stray neutrons: the slow

"s"-process and the rapid "r"-process, and

proton capture – the p-process.

fragmenting nuclei - the "spallation" process, in which nuclei in space are broken into

smaller fragments by cosmic rays

In addition, any unstable nuclei will seek stability by a beta-decay process, whereby

protons and neutrons interchange by the weak interaction (described in the last chapter).

3.12.4 Describing nuclear reactions

How can we describe nuclear reactions? One of the important stellar reactions is the

"triple alpha" process, whereby three quartets of nucleons, each one (2p,2n), fuse

together into a cluster of a dozen, (6p,6n)…

3 (2p,2n) (6p,6n)

This is how the nuclear reactions in stars should, strictly, be described. This is the world of

protons and neutrons, that operates on its own terms. However, the nuclides created by

the stars are the foundations for our atomic world, made of elements we are familiar with.

Thus the nuclide (2p,2n) will become helium-4, with a total of 4 nucleons in its nucleus.

The nuclide (6p,6n) will become carbon when it is ejected from the star into space. When

this carbon atom aggregates with others it becomes the material carbon we know - the

"lead" in a pencil, the diamond in a ring, or your burnt toast under the grill. For now, it is

just a cluster of 6 protons and 6 neutrons - a total of 12 nucleons, jostling with other

nuclear clusters in the heat and pressure of a star’s core.

So the nuclear reactions will be described like this…

3 helium-4 (2p,2n) carbon-12 (6p,6n)



including the chemical element name as well as the nucleons. While this is cumbersome,

we can see the elements created, and also follow the arithmetic of the protons and

neutrons.

This reaction is also called helium burning. We normally think of burning as a chemical

reaction with oxygen like a bonfire or gas flame. But this is a nuclear reaction, in which

helium nuclei are consumed and transformed to something else. We can also think of the

reactions as "nuclear Lego”, where nuclei can be built up or broken down, just like

assemblies of standard Lego bricks. As with the first nuclear reactions in the cosmic

fireball, I'll show the nuclear reactions as moves in nucleon-space, and figure 3.50 shows

the major nuclear reactions in this way.

Figure 3.50: Some of the possible nuclear reactions in a star, shown as moves in nucleon-space.Humanity has grown up with the stars in the heavens: “In the great cities of the world, we

have detached ourselves from night. If you are a city dweller who doesn't believe this,

travel at least a hundred miles into the countryside, mount the highest hill, and stare at the

sky. It is not the same sky at all. … On a clear night in the mountains, you become part of

the sky. The stars reach out and touch you, and suddenly, you feel the embrace of a

galaxy”.

Our recent understanding of stars as nuclear forges is the latest and truest conception of

their nature. Perhaps as a heritage of the other ways we've seen them in the past, we

tend to anthropomorphise them; we speak of their life-cycles, of their birth and inevitable


-2 -1 +1 +2 neutrons, n

protons, p

+2

+1

-1

-2

gain a neutron

gain a proton

lose a proton

gain a helium-4 nucleus (2p,2n)- an alpha particlebeta-minus

decay: n p

beta-plus decay: p n

lose a neutron

..., 23/04/11,

“Stellar evolution” is a fairly common, but perhaps inappropriate phrase, for evolution occurs by reproduction with variation in a competitive environment. A star has a lifecycle: it forms, is “born” from a collapsing gas cloud, it consumes its nuclear fuel, metabolising (?) nuclides and emitting energy, and it comes to an end, peacefully fading as a small star, or in an violent explosion for a big one. There are parallels with human lifecycles.

..., 23/04/11,

Krauss, p.96.

..., 23/04/11,

I’ve followed the approach of Nicholas Short… http://rst.gsfc.nasa.gov/Sect20/A7.html …and also figure 7 in William Fowler’s 1983 Nobel Prize lecture… http://nobelprize.org/nobel_prizes/physics/laureates/1983/fowler-lecture.pdf

McNeil, 23/04/11,

check approval of the Lego people !! Lego lawyers' reply…Feb '09.

..., 23/04/11,

…and often the element name is all you get in a technical text.


death. We may find ourselves thinking of their purpose as being to create the heavy

elements, so we might talk of a star burning its nuclear fuel to generate the heat

necessary to halt the compression due to gravity. We will see that the sequence actually

is: the crush of gravity heats up the star's interior…which ignites a nuclear fuel…which

generates heat…which temporarily halts further compression…until that fuel runs out…

and the cycle repeats itself with the next nuclear fuel that “burns” at a higher

temperature…until all the available fuel is consumed.

Through gravity, the stars recreate the conditions in the fireball soon after the big bang.

But whereas the fireball cooled within minutes, the stars can maintain enormously high

temperatures and pressures for billions of years, and thereby accomplish what the entire

universe could not - the creation of all the viable clusters of nucleons, all the elements of

the periodic table - the basis for our atomic existence.

We're ready now for our newly-formed star's first nuclear reaction.

3.12.5 Hydrogen-burning - the proton-proton chain

no free neutrons

The cosmic fireball reactions easily started with protons combining with free neutrons, and

then numerous reaction steps to produce helium-4 (see section 3.11.6). But our new star

contains no free neutrons, so the nuclear reactions must start with only protons. We've

seen how neutrons are vital to stable clusters, so the star must somehow generate its

neutrons from the protons. Once a proton has changed to a neutron, it can join with

another proton to make hydrogen-2 (deuterium), and things can go on from there.

We'll focus on one proton, in the midst of the other protons dashing about, and repelling

them, due to their mutual electrostatic repulsion (figure 3.51).


..., 23/04/11,

Having described the quantum tunnelling in alpha-decay, I’m using simple representations of particle-waves here.

McNeil, 23/04/11,

Any deuterium present is rapidly consumed in the star’s first nuclear reactions – Krauss, p.126. I'm omitting the burning of hydrogen-2 (deuterium) and lithium-7, since these are tiny constituents of the star - see Krauss, p.126.


Figure 3.51: A single proton surrounded by a potential “hill”, repelling other protons; the vertical axis is not to scaleThis is the reverse of alpha-decay, where alpha particles escape from a large nucleus by

quantum tunnelling through the wall of the “energy well” that confines them. In the figure

above, our single proton is not so much in a well, as surrounded by a potential energy

“hill”, arising from the repulsion between protons. Other protons “climb” part-way up this

hill, come to a stop, and then “roll” back down again, as they are repelled. Here the

protons are shown as simple particle matter-waves.

a proton tunnels in

Even at 10 million degrees the average proton has around 1 keV of kinetic energy,

nowhere near enough to get over the energy hill; so it “climbs” part way up the slope, and

is then repelled. However, at any moment the protons have different speeds. Think of the

dodgem cars at the fair; at any one moment, a few cars will have stopped because of

head-on collisions, and a few will be moving extra fast due to collisions from cars behind.

In the same way, there is a distribution of speeds among the protons. About 1 in 10 million

protons has 10 times the average kinetic energy, and can climb further up the hill and get

closer. Even this rare proton has only about 1/100th the energy needed to get to the top of

the electric potential hill. But if it is on an exact collision course with the other proton, then

it can briefly get close enough that its matter-wave extends through the potential hill into

the inside; that is, there is a small but finite chance that the proton can tunnel through the

potential hill, and join the other proton inside. So, out of a multitude of proton collisions

one will result in a proton-proton pair.

the weak interaction transforms a proton to a neutron


3. a proton with an average kinetic energy of ~1 keV can only get a little way up the "potential hill" before it is repelled

4. a proton with ~10 KeV kinetic energy can get close enough for its matter-wave to penetrate through the hill, so it can tunnel through

1. a proton surrounded by a "potential hill", which repels other protons

p+

p+

p+

2. a proton needs ~1 MeV to reach the top of the "potential hill"

..., 23/04/11,

JohnGribbin describes the protons getting close enough so that “the extended edges of the wave packets can overlap” – p.82.

McNeil, 23/04/11,

Figures from Krauss, p.129. At T=10 x 106K, the energy, E = kT ~833 eV or nearly 1 keV. 1 proton in 10 million has 10 keV energy, so the energy barrier is 100 times this, that is ~1,000 keV, or ~1 MeV. The equation for electrical potential tells us that the potential energy of a proton at a distance of 1.5 fm from another proton is ~1 MeV. So a proton with 1 MeV of kinetic energy, on a direct collision course with another proton, will get no closer than ~1.5 fm, a separation about equal to the proton’s size. So the simple calculation matches Krauss – as it should! Gribbin points out that the collision has to be exactly head-on, and says it’s only 1 in 100 milliion protons manage to pentrate the potential energy barrier – p.87.

..., 23/04/11,

I’m following Krauss’s figures for the start of the hydrogen burniong process at 10 million degrees – p.129. Stars of about 1 solar mass burn their hydrogen steadily at ~15 million degrees, and more massive stars run even hotter – WHW, table I.


But we have seen that the nuclear force is not quite strong enough to bind a pair of

protons – there’s no such thing as a diproton. However, if one of the protons were to

transform to a neutron the pn combination would be stable. While this is energetically

favourable, it can only happen by the weak interaction, which is very slow, or to put it

another way, is very unlikely to happen. So we mostly see our two protons almost

immediately fly apart. However, there is just a tiny chance that, in the very short time they

are together, one of the protons transforms to a neutron, in which case we now have a

stable nucleus of hydrogen-2, or deuterium. We write this simple nuclear reaction as…

p + p hydrogen-2(p,n) + e+ (positron) + (neutrino)

neutrinos from the sun

All this is taking place now, deep in the heart of a star like our sun. How do we know it's

happening? The proof is in the neutrinos produced, which were first detected in 1988.

“The multitude of neutrinos generated by nuclear reactions in the solar core pass through

a half million miles of sun as if it were a thin sheet of glass, emerging in a matter of

seconds. They reach Earth in eight minutes, and pass through it with no resistance,

either. About 500 billion neutrinos from the sun fall on each square inch of ground in a

second. Our bodies are pierced by them unceasingly, day and night. They leave not a

trace.”

a very slow nuclear reaction

The star has taken the first step on the long road of building up bigger nuclei. It looks so

simple - there's no hint of its incredible difficulty. This reaction is so rare that it has never

been observed in laboratory experiments. Even in our own sun it takes an average proton

around 10 billion trillion collisions, lasting about 14 billion years, to to bind with another

proton and transform to a neutron at the same time. However, this unbelievable slowness

ensures that stars like our sun consume their fuel very gradually over billions of years,

thus ensuring that life can evolve slowly on Earth under steady conditions. But for now,

the universe and organic life just wait.

flipping a coin

There is an old story about a student who flips a coin to decide what to do that evening: if

heads, then go to the cinema; if tails, then go to the pub, and if the coin should stand on

its edge - then study. We've seen the three enormous obstacles that must be overcome:

(1) only very few protons have enough energy to get near enough to another proton; (2) of


McNeil, 04/23/11,

Chown, “The Magic Furnace”, p.122.

..., 23/04/11,

These figures from Gribbin, p.87. Chown, “The Magic Furnace”, p. 122 gives 10 billion years, as does Bertulani, p.350. Hey and Walters, p. 212 give 1 billion years.

McNeil, 04/23/11,

Hey and Walters, p. 210. http://en.wikipedia.org/wiki/Proton-proton_chain

McNeil, 23/04/11,

Marschall, p. 249.

McNeil, 04/23/11,

see Hey and Walters, p. 211-213. Marschall, p.249. Silk, p.17…"…absolute proof of an ongoing nuclear reactor…". How do we detect neutrinos? If the gallium-71 nuclide absorbs a neutrino with an energy >0.233 MeV it transforms to radioactive germanium-71, whose decay can be detected and counted - Clayton, p.272.

..., 23/04/11,

Krauss, p.127, Bertulani, p.347.

..., 23/04/11,

See back to section 3.2.6. “…an unstable nucleus containing two protons.” - Hey and Walters, p.212. Chown, Furnace, p.122…puts it like this: "the strong force is only very slightly too weak to bind helium-2" - that is 2 protons.


these, very few tunnel through the potential wall; and (3) it is extremely unlikely that either

of the pair of protons will transform into a neutron in the very short time before they repel

and separate. Thus, the p-p reaction is “just about the most inefficient nuclear reaction

imaginable”. Flipping a coin and getting three "edges" in a row is easy by comparison!

But, it's not impossible, only unlikely. If pairs of protons collide enough times, then the

nuclear reaction will happen. If you flip a coin enough times, then you will get not only an

edge, but at some point, three edges in a row will occur. And a star contains so many

hydrogen protons which are undergoing huge numbers of collisions each second, that the

proton-proton reaction happens in abundance. Our own sun "burns" some 600 billion kg

of protons each second - that's over 3 x 1035 protons - or nearly 400 trillion trillion trillion,

and it contains enough protons to do this for about 10 billion years.

We saw how a few of the lightest nuclei were assembled in the cosmic fireball, in the first

few minutes after the big bang, but this was very limited because of the rapid expansion

and cooling. “Stars, on the other hand, stay dense and hot for millions or even billions of

years”, long enough for the rarest, most unlikely of nuclear reactions to occur.

the proton-proton chain

So our star has fused two protons; this is the first stage in a series of reactions called the

proton-proton chain – figure 3.52.


McNeil, 04/23/11,

The full proton-proton reaction sequence is described in Williams, sections 14.3 and 14.4, and HyperPhysics, at… http://hyperphysics.phy-astr.gsu.edu/hbase/astro/procyc.html#c1 (accessed 24 Jan. 2011). and also… http://en.wikipedia.org/wiki/Proton-Proton_Cycle See also Mackintosh et al, ch.9. This diagram is similar to the one from Nicholas Short, at… http://rst.gsfc.nasa.gov/Sect20/A7.html …and shows the PPI chain, the main one (86%) of the three reaction branches - Bertulani, p.352, Mason, p.53, Williams, p.350. In the diagram I use the “bubble of ignorance” as described by Allday, p.211.

..., 23/04/11,

Chown, Furnace p. 169.

McNeil, 04/23/11,

600 x 109 kg / 1.67 x 10-27 kg gives us 3.6 x 1038 protons. Philip Ball, p. 107, Gribbin, p.87, and Marschall, p.127

McNeil, 04/23/11,

Chown, “The Magic Furnace”, p. 123.


Figure 3.52: The nuclear reactions in the proton-proton chainThe three steps in the proton-proton chain successively create deuterium (p,n), then

helium-3(2p,n), and finally helium-4(2p,2n). The two protons left over at the end are

recycled back into the mix. The reaction series creates two photons which will leave the

star as light radiation. The positrons will meet electrons and will annihilate to create more

radiation. The neutrinos scarcely interact with matter and escape from the star into space,

taking some of the reaction energy with them. Once two protons have been fused to make

hydrogen-2, then subsequent progress is much faster. Steps 2 and 3 only require the

strong nuclear force to bind the larger nuclei together, and can occur within minutes.


1. Two protons collide, and one of them transforms to a neutron - to make a nucleus of deuterium (p,n)…

2. …another proton collides…is gripped by the strong nuclear force…and a nucleus of He-3 (2p,n) is made…

3. …another He-3 joins with this to make stable He-4 (2p,2n)…releasing 2 protons in the process.

4

gamma ray photons slowly make their way though the star's outer layers and leave as sunlight

p

This can be summarised as…

He-4(2p,2n)

n np p

+ 2e+ + 2 + 2 + 16.7 MeV

positrons (anti-electrons) meet electrons and mutually annihilate to produce gamma radiation

neutrinosleave the star and fly through space at nearly the speed of light

p

neutrino

n

p

p

p

nn

n

pp

p

positrone+

gamma ray photon

pp

p

p

p

n

p

pp n

p

e+ recycled

the fomation of one nucleus of helium-4 releases 16.7 MeV of heat energy

..., 23/04/11,

Krauss, Atom, p. 127/8 and p. 133, and Mackintosh et al, say it’s only a matter of seconds – p.112.


where does the energy come from?

This is the summary of the hydrogen burning reaction…

4 protons + 2 electrons 1 helium-4(2p,2n) + 2 neutrinos + energy

(4x938.3) + (2x0.5) = 3754.2 MeV 3727.4 MeV mass loss = 26.8 MeV

The reaction sequence produces two positrons, which annihilate with two electrons back

to radiation energy, so these two electrons are included in the energy accounting. Four

unbound nucleons become bound into a nucleus, and the energy of their binding, which is

released as heat, is “paid for” by the particles’ loss of mass. Thus the mass loss of the

system of particles gives us the energy released.

Hydrogen burning, the rearrangement of four nucleons into one nucleus, releases 26.8

MeV of energy, about 6.7 MeV per nucleon. This reaction, when free nucleons are first

bound into a nucleus, releases an enormous amount of energy. We’ll see that subsequent

reactions, rearranging already bound nucleons into progressively bigger nuclei, are not

nearly so productive.

hydrogen burning sustains the star’s weight

This, then, is the proton-proton chain, otherwise known as “hydrogen burning". “The

energy liberated in these [nuclear] reactions yields a pressure … which opposes

compression due to gravitation. Thus an equilibrium is reached for the energy produced,

the energy liberated by radiation, temperature and pressure”. The reaction releases

energy, so the nuclei in the core get hotter, that is, move faster, thereby exerting a higher

pressure, which can support the weight of the star and halt the star’s contraction.

In the same way, the moving air particles inside a car tyre support the weight of the

vehicle. We deliberately increase the pressure in a tyre by pumping more air particles into

it, but the core pressure in a star depends on its temperature, which is governed by the

nuclear reactions taking place there.

A star of one solar mass burns its hydrogen steadily in its core at a temperature of ~15

million degrees and a density of ~150 g/cm3. It’s worthwhile pausing to think about

density, because this will be an important factor in the star’s life. Gold is one of the

densest metals, with a density of close to 20 g/cm3 – that is a gold lump the size of the tip

of your little finger would have a mass of ~20 grams. So, the hydrogen burning in the core

of a star is compressed to about 8 times the density of gold.


..., 23/04/11,

WHW, table I. Williams, p.350, and Bertulani, p.350, give densities of 100 g/cm3. Densities are given in different units – commonly kg/m3 and g/cm3, and authors use different ways to try to convey the significance of densities. I will give densities as the mass of a volume of 1 cm3, about the volume of the tip of the reader’s little finger. There is some variation in density values. I’ve used the data in WHW – this is a recent major review, and so I think values will be accurate and self-consistent. Where others say something markedly different I’ll note it. These figures are for a 1 solar mass star, in its H burning phase. Larger stars burn their hydrogen much faster, at a higher temperature and hence at a lower density.

..., 23/04/11,

The physical gas laws describe what happens when a mass of gas collapses under the weight of its own gravity – Delsemme, p.45.

..., 23/04/11,

Krauss, p.130.

..., 23/04/11,

Bertulani, p.341.

..., 23/04/11,

Williams sums the energies released (the Q-values) for each step in the pp-chain, ending up with a figure of 26.7 MeV, with each neutrino escaping with ~0.26 MeV – p. 350, as does the Wikipedia article… http://en.wikipedia.org/wiki/Pp_chain WHW say the neutrinos take 1.7 MeV for each He nucleus made, p.1026. The table mass.mas03 gives atomic masses, so subtracting the masses of electrons gives the masses of the bare nuclei, which are used here.he mass You can work out the energy released just from the mass loss of the system of particles. This approach agrees with values for hydrogen burning, and for subsequent stages – the burning of helium, carbon, oxygen and silicon – figures checked with Williams. You can equally well use mass excess values, available in the mass.mas03 database, and in Clayton. The maths is simple, but mass excess values are not straightforward, and they are for neutral atoms, not nuclei. So, I’ve used nuclear masses, and the general reader see a mass loss converted to energy, according to E=mc2. If the combined masses (A+B) are greater than the products (C+D) in the nuclear reaction A+B ( C+D, then the reaction will liberate energy – Clayton, p.294.

..., 23/04/11,

This is a slight simplification of the full reaction – Williams, p.348.


nuclear reactions are self-regulating

Stellar nuclear reactions are inherently self-regulating – as long as there is nuclear fuel

available to release energy. “If the core gets hotter, the pressure increases, causing the

gas to expand against gravity, thus cooling the core.” Conversely, if the reaction slows

down, the core cools, the pressure drops, and the star contracts. This then heats up the

core, and the nuclear reaction speeds up again. This is sometimes called a “natural

thermostat”, and this suggests that the star somehow “sets” the temperature, but of

course, it’s not like that. A star compacts, and its core heats up until a nuclear reaction

starts that can generate the heat energy to halt the compaction. We’ll see soon what

happens when a heat generating nuclear reaction is not available. Stars in the hydrogen

burning phase are very stable; a star the mass of our sun will burn hydrogen steadily for

about 10 billion years, though we will see later that this time depends on the star’s mass.

the star’s radiation takes a long time to escape

What is surprising is that energy is generated so slowly in the sun's core that a human

sized volume of our sun burns its nuclear fuel slower than a human converts food into

energy. Then this energy takes a very long time to escape. The proton-proton chain

reaction produces high energy gamma ray photons. Each photon's way is blocked by the

nuclei in the star's core, like a crowd of shoppers in a busy market. The photons bounce

back and forth in a zig-zag path, repeatedly scattered by the charged nuclei, and finally

emerge from the sun's surface with a range of smaller energies, spanning the

electromagnetic spectrum. After working their way out of the sun (a distance of about 700

million metres) in about 30,000 years - light would travel that distance in space in just 2

seconds - they then reach the Earth (200 times further) in about 8 minutes.

the proton-proton chain in nucleon space

Figure 3.53 shows the proton-proton chain as a set of moves in nucleon space.


McNeil, 23/04/11,

Chown, Magic F, p. 94. Christoph Schiller, estimates that the effective speed of light at the sun's core is about 10km/year, and that today's sunlight was generated 200,000 years ago. http://www.motionmountain.net/index.html

McNeil, 04/23/11,

Hey and Walters, p.210.

..., 23/04/11,

Delsemme, p.47.

..., 23/04/11,

For example, Delsemme, p.46, though he does explain how this negative feedback works.

..., 23/04/11,

Krauss, p.130.


Figure 3.53: The proton-proton chain as a set of three moves in nucleon spaceWith no help from free neutrons, protons have worked their way diagonally in nucleon

space to helium-4. We will see the further stellar nuclear reactions work their way steadily

up and to the right in nucleon space, creating ever larger nuclides.

entering the nuclear valley

Thus the hydrogen-burning star has entered the nuclear valley. The entire universe got

little further than this, though that was with the temperature and pressure rapidly falling.

We can look through the helium “pass”, and glimpse the nuclear terrain which the stars

will explore (figure 3.54).


H-1 (p) H-2(p,n)

He-3 (2p,n) He-4 (2p,2n)

(2)

+ p

0 1 2 3 neutrons, n

(3) 2He-3He-4+2p

NO freeneutrons!

free protons

(1) p + pH-2(p,n)

protons, p

lithium-Li,3

helium-He 2

Hydrogen-H 1

0

(4) the 2 protons are recycled back into the mix

H-3 (p,2n)

..., 23/04/11,

The nuclides that define, that are, this terrain have not yet been created, so can we say that the nuclear valley exists? Can we say that the rules already exist that govern nuclear structure, and hence the nuclear masses that determine the terrain? Does a falling tree…? Hmmm…I think it’s still a good picture, though.

..., 23/04/11,

Compare this with the production of He-4 in the cosmic fireball.


Figure 3.54: A view of the nuclear valley through the helium “pass”. The central black arc of stable nuclides separates the blue neutron-rich nuclides from the orange proton-rich nuclides. The largest nuclides form the peak in the far distance.

3.12.6 The carbon (CNO) cycle

a richer gas mix

The first generation stars, starting with only hydrogen and helium, have only the

enormously difficult proton-proton cycle as their first nuclear reaction. However, stars of

later generations form from a richer mix that includes heavier nuclei from previous

generations of stars. One of these, carbon, present even in small amounts, can greatly

speed up the production of helium - figure 3.55.


a proton – a hydrogen nucleus

the end of the arc of stable nuclides

..., 23/04/11,

There are various good diagrams available… Mackintosh et al, p.116, and http://en.wikipedia.org/wiki/CNO_cycle http://aether.lbl.gov//www/tour/elements/stellar/CNO.gif http://rst.gsfc.nasa.gov/Sect20/A7.html These show the beautiful repetitive cycle of reactions, but they don't show the nucleus-building "tide" advancing in nucleon space. I only show the “nuclear Lego” aspect here, so the emission of gamma ray photons, and beta decay products are not shown.

..., 23/04/11,

Normal hydrogen burning starts at about 10 million degrees, and the carbon cycle becomes significant above about 15 million degrees - Delsemme, p. 62, and Krauss, p. 154.


Figure 3.55: The CNO cycle produces helium: the C-12 nucleus follows a zig-zag path through nucleon space, capturing 4 protons and decaying twice, to end up as O-16, which emits a He-4 nucleus, and so returns to C-12. Proton capture is accompanied by gamma ray photon emission, and beta decay is accompanied by the emission of positrons and neutrinos – these are not shown here.A carbon-12 nucleus captures a series of 4 protons, 2 of which decay to neutrons. The net

result is the gain of 2 protons and 2 neutrons, and moving 2 squares diagonally, landing

on oxygen-16. This nucleus fragments to helium-4, and carbon-12, which undergoes the

cycle again. This cycle is incredibly fast compared to the proton-proton cycle, and can

produce a helium nucleus in about a day. We can see why this is called the CNO cycle,

since it creates in sequence the nuclei of carbon, nitrogen and finally oxygen. It's

described here because of its rôle in building hydrogen rapidly into helium. It's also the

major source of nitrogen in the universe - the same nitrogen we breathe on Earth. The

CNO cycle certainly powers many of the stars shining now, because they contain some

carbon and other nuclear species that had been produced by earlier generations of stars.

The first generation of stars would have been powered solely by the proton-proton chain.


protons, p

oxygen-O, 8

nitrogen-N, 7

carbon-C, 6

boron-B, 5

5 6 7 8 9 neutrons, n

C-12(6p,6n)

1. capture a proton

C-13(6p,7n)

N-13(7p,6n)

N-14(7p,7n)

N-15(7p,8n)

O-16(8p,8n)

2. first beta decay, pn

6. capture a fourth proton…

O-15(8p,7n)

5. second beta decay, pn

7. …and then the O-16 nucleus splits…into He-4 and C-12……and the cycle starts again…

3 and 4. capture two more protons

…back to He-4

..., 23/04/11,


McNeil, 23/04/11,

Croswell, p.172, and Clayton, p.76.

McNeil, 23/04/11,

Krauss, Atom, p. 154.


once around the gravity-matter circuit

We have now described the major energy generating process of the majority of stars, in

which hydrogen burns, building up helium in the star's core, like ashes in a fireplace. Our

star has consumed the first of its nuclear fuels, and has gone once around the gravity-

matter circuit (see back to figure 3.49). Let’s follow our star round the circuit again, and

see how the ashes from the first reaction become the fuel for the second.

3.12.7 Helium burning - the " triple alpha" process

when the hydrogen runs out

The accumulation of helium ash and the depletion of hydrogen slow down the hydrogen-

burning reaction. The weight of the outer layers can no longer be supported, and the star

contracts, further heating up the core. This is the profound difference between nuclear

“burning” in a star and our familiar chemical combustion. In a chemical combustion

reaction on Earth, once the fuel is consumed, the fire simply goes out. In a star,

everything is driven by gravity, like the inexorable demands of a blackmailer. Gravity

produces the enormous temperatures and pressures needed to start and sustain nuclear

reactions. The energy thus produced halts the star's contraction - as long as the reaction

has fuel to continue. When the fuel runs out, that reaction ceases, and gravity, now

unopposed, compresses the star further, driving up the temperature and starting the next

reaction in the series.

When the hydrogen burning ends, the temperature of the star’s core rises to ~120 million

degrees or more, its density increases to ~20 kg/cm3, the mass of a dozen house bricks in

your little fingertip, and the helium that has accumulated now ignites. The outer layers

also get hotter, so hydrogen-burning spreads to the layer outside the core.

improbability and coincidence

If we think of the star's reactions as "nuclear Lego", then we'll be comfortable with helium-

burning. If we can put together singles, pairs and trios of nucleons to make quartets, then

we can put together quartets to make bigger nuclei - all we need is higher temperatures

and energies - right? We've seen how a star accomplishes the fantastically improbable

first step of the proton-proton chain with ease. Like flipping a coin and getting an "edge"

this is merely improbable - flip enough coins enough times and you're certain to get

"edges". Helium burning, however, depends on a set of conditions that appear quite

coincidental. There is a universe of difference between improbability and coincidence.


McNeil, 04/23/11,

This links to the discussion of fine-tuning of the universe - the anthropic debate.

..., 23/04/11,

WHW, table I. Williams, gives 100 100 kg/cm3, p.352. Bertulani gives 0.1-100 kg/cm3, p.357. In a later paper, Woosley and Janka give a lower figure – 1.4 kg/cm3.

..., 23/04/11,

Marschall, p. 127. Hey and Walters, p.214, Adams and Laughlin, p. 48. et seq. The web-site of the George Smoot group… http://aether.lbl.gov//www/tour/elements/stellar/Hepoison.gif

McNeil, 23/04/11,

nasa http://rst.gsfc.nasa.gov/Sect20/A7.html Clayton, p.289, Bertulani, p.357, Williams, p.352, Silk, p. 133.


no stable clusters of 5 or 8 nucleons

The problem can be stated simply: there are no stable nuclei of 5 or 8 nucleons. A star

can't add a proton to helium-4, because a cluster of 5 is unstable. Similarly, two helium-4

nuclei can't fuse to make a cluster of 8. There appears to be no route to the heavy

elements, that comprise our physical world.

beryllium-8 is unstable but with a long lifetime

In the now very hot star's core two helium-4 nuclei have enough energy to collide and

stick. This gives a beryllium-8 nucleus, which is unstable, but with a mass only very

slightly greater than the original pair of helium nuclei, it has a unexpectedly long lifetime -

about 10-15 s. This seems vanishingly short to us, but it's quite long-lived in the nuclear

world, and so there is a small but sustained population of beryllium-8 nuclei in the star's

core - about one for every billion helium-4 nuclei. With trillions of collisions occurring each

second, this lifetime is long enough for about 10,000 encounters with a helium-4 nucleus,

in which the two might possibly fuse to make a carbon-12 nucleus. This is shown in figure

3.56.

Figure 3.56: The helium burning process: helium-4 nuclei successively fuse, creating an alpha-particle “staircase” in nucleon-space. (The cell outlined in red was the centre of the display)


1 2 3 4 5 6 7 8 neutrons, n

protons, p

oxygen-O, 8

nitrogen-N, 7

carbon-C, 6

boron-B, 5

beryllium-Be, 4

lithium-Li, 3

helium-He, 2

hydrogen-H, 1 1. Two He-4 nuclei join to make…

3. …to make stable C-12…

4. …some of which joins with another He-4 to make O-16.

2. …unstable Be-8, which lasts just long enough for a third He-4 nucleus to join…

There are no stable clusters of 5 or 8 nucleons

..., 23/04/11,

This shows the two principal He burning reactions: the triple-alpha process making C-12, and the further reaction that makes O-16 – WHW, p.1026, and Williams, p.352.

McNeil, 04/23/11,

Krauss, Atom, p.135, and Chown, Furnace, p. 169.

McNeil, 23/04/11,

A pair of helium-4 nuclei have a mass of 7454.8 MeV; beryllium-8, 7454.9 MeV. So the Be-8 is unstable, but with such a small mass difference, it has a long lifetime. Mackintosh et al give the lifetime as 10-16 secods – p.115. Clayton, chapter 6: Carbon.

McNeil, 04/23/11,

Chown, furnace, p. 145. Krauss, p. 134.


This shows the two main reactions in the helium-burning phase: the ‘triple-alpha’ process

in which three He-4 nuclei fuse to make a C-12 nucleus, and this C-12 then fusing with

another alpha particle to make O-16. Much of the oxygen in the universe comes from this

second nuclear reaction.

Beryllium-8 is like a stepping stone that starts crumbling under your weight, but holds up

just long enough for you to jump to the next stable position. The degree of instability of

beryllium-8 is critical. If it were more stable then on reaching the helium ignition

temperature the star's nuclear reactions would quickly transform all helium to carbon. The

sudden release of energy would blow the star apart, and this would prevent the

appearance of all elements heavier than carbon in our universe.

the excited carbon-12 nucleus

But with so few beryllium-8 nuclei available, it's vital that the collisions between beryllium-

8 and helium-4 have a high chance of leading to fusion. In the normal run of things even

10,000 collisions aren't enough. This is where coincidence first steps in. It turns out that

the carbon-12 nucleus can exist in an excited energy state which is just right for the fusion

of beryllium-8 and helium-4.

The combined mass-energy of the beryllium-8 and helium-4 nuclei is 7.37 MeV. They will

fuse most easily, if the product nucleus has a favoured energy of vibration that is close to,

but less than this value. It turns out that the carbon-12 nucleus has a favoured excited

state at an energy of 7.65 MeV - close but just too high. However, the temperature of 100

million degrees inside the star core is enough to raise the combined energy of beryllium-8

and helium-4 to just above the excited carbon-12 energy, so the fusion reaction can occur

with its greatest efficiency.

resonance in swings and radios

Think of pushing a child on a swing. The pendulum of the swing has a natural frequency,

say 1 complete swing every 2 seconds. We know that if we match the frequency of our

pushing to the swing’s natural frequency then we will transfer our energy to the swing

most quickly and build up a large amplitude. Our push and the swing have the same

frequency – we say they are in resonance. When we tune a non-digital radio, we alter the

natural frequency of the oscillating tuning circuit to match the frequency of the radio

signal. This “resonance allows a radio to be tuned so it is millions of times more sensitive

to radio waves of a certain frequency than signals of all other frequencies”.


..., 23/04/11,


..., 23/04/11,

Chown, Furnace, p. 177. For a technical description of the Be-8 He-4 reaction, and the role of C-12 resonance, see Bertulani, p. 358.

McNeil, 23/04/11,

John Barrow, “The origin of the Universe”…p.124. Fred Hoyle predicted this excited state, and his reasoning went something like this. The only way for carbon-based life forms to be here is if the helium-burning process can proceed efficiently, and that requires that the carbon nucleus has an excited state at just the right energy. So the experiment was done to look for it - and it was found at the predicted energy. Marcus Chown tells the story: "The Magic Furnace", chapter 13.

McNeil, 04/23/11,

Chown, Furnace, p. 177.

..., 23/04/11,

And also from neon burning, which we meet later – WHW, p.1029.


resonance and the formation of carbon-12

In a similar way, the carbon-12 nucleus has an excited resonant state at just the right

energy for it to be formed rapidly by the fusion of beryllium-8 and helium-4. We see nuclei

grow in steps of helium-4 nuclei, ascending an alpha-particle “staircase” in nucleon space.

Lawrence Krauss explains the cosmic significance of this: “Once carbon has been formed,

the gateway to creation of all the heavy elements that dominate our own existence on

Earth is opened”.

the energy yield of helium burning

The net result of the triple-alpha reaction is to combine three helium nuclei into one

carbon nucleus…

3 He-4(2p,2n) C-12(6p,6n)

(3x3727.4) = 11,182.2 11,174.9 mass loss = 7.3 MeV

So the rearrangement of 12 nucleons from three nuclei into one yields 7.3 MeV energy, or

about 0.6 MeV per nucleon. This is a lot less than the 6.7 MeV per nucleon produced by

hydrogen burning.

a trio of cosmic coincidences

Carbon-12's precise excited energy state is a remarkable coincidence, but there's more.

The newly made carbon-12 nuclei can combine with helium-4 to make oxygen-16. If this

reaction is too easy then all the carbon-12 will be transformed to oxygen-16, with

consequences for subsequent carbon-based life forms. If this reaction is too slow, then

there will be insufficient oxygen produced for these life forms to breathe. So, “A delicate

balancing trick is required if neither oxygen nor carbon is to be over-abundant in the

universe, at the expense of the other. ... Life is possible because nature has fine-tuned

the properties of three atomic nuclei. Beryllium-8 is unusually long-lived for an unstable

nucleus. Carbon-12 possesses the exact energy state needed to promote its production.

And oxygen-16 lacks an energy state that would promote its production at the expense of

carbon”.

We've thus seen two nuclear fusion processes: first, hydrogen burning, which is highly

improbable, and second, helium burning, which is contingent on a set of coincidences.

And now you and I, oxygen-breathing carbon-based life forms, are here to consider them.

after helium burning


..., 23/04/11,

Chown, Furnace, p. 177/8. Coincidences such as these convinced Fred Hoyle that “the universe was geared up for the emergence of living organisms. Life had arisen on earth not because of some zillion-to-one accident but because it was truly a cosmic phenomenon” – Barrow, p.178. Paul Davies explores this view in “The Goldilocks Enigma”. Mackintosh et al comment:“it has been estimated that if te strength of the nuclear force were to be changed by 1 part in 200, then the proportion of C-12 nuclei would be 30 times less. The universe is extraordinari;y finely tuned!” – p.115.

..., 23/04/11,

Williams gives the energy released as 7.3 MeV – p.352.

McNeil, 04/23/11,

Krauss, Atom, p, 136.

..., 23/04/11,

Mackintos et al, p. 115.


The star has now burned its first two nuclear fuels, and is developing a concentric layered

structure, with a core of carbon-12 and oxygen-16, surrounded by a layer of burning

helium, which itself is surrouded by burning hydrogen. The sequence of burning reactions

is spreading outwards through the star, so each layer comprises the ashes of the previous

reaction, that is now proceeding in the next layer out.

We’ve seen two of the nuclear fusion reactions; it’s now time to look at the other

processes that build up nuclei.

3.12.8 Filling in the gaps - capturing protons and neutrons

We have seen earlier the major role played by helium-4 nuclei in the alpha-decay of large

unstable nuclei, and we now see its rôle in the fusion reactions that assemble larger

nuclei in stars. The small size and great stability of the helium-4 nucleus, makes it a

favoured unit of currency in nuclear transactions, both fission and fusion. But how do the

stars create the intermediate nuclei, ones that are not multiples of (2p,2n)? And also, how

are the large nuclei, way beyond the stable maximum of lead, created? These are the

work of the nucleon capture processes.

The p-process – making proton-rich nuclides

We know that stars are rich in protons, so they will be constantly colliding with any nuclei

that are formed. If the energy of collision is enough to overcome their mutual repulsion,

then the two can fuse – the nucleus “captures” the proton. The first step in the CNO cycle

is an example of proton capture, which moves the nucleus one step upwards in nucleon

space (see back to figure xXx).

carbon-12 (6p,6n) + p nitrogen-13 (7p,6n)

However, this nucleus is unstable, and decays by the weak interaction…

nitrogen-13 (7p,6n) carbon-13 (6p,7n) + e+ +

and another form of carbon has been made.

In principle, proton capture can produce the proton-rich nuclei, that lie above the arc of

stable nuclei in nucleon space. Unstable nuclides will undergo beta-plus decay and move

back towards the stable arc, as we saw in the CNO cycle. Because nuclei and protons

repel each other, direct proton capture plays a minor rôle in building up nuclei, and is

associated with light nuclei (small charges) and high temperatures (high collision

energies).


..., 23/04/11,

For nuclei beyond the iron group, with >70 nucleons, the “abundances of the proton-rich isotopes, the r-process nuclei, and the s-products lie in the approximate ratios 1:25;50” – Mason, p.59. Bertulani shows the rp-process as producing nuclei up to ~35 protons – p.358. Mackintosh et al give a diagram showing nuclei with up to 58 protons, made by proton capture – p.117. Proton capture was proposed to explain the existence of the proton-rich nuclei. It is now recognised that neutron loss due to collisions with high energy photons is equally important. The “p-process” now includes this latter process of photodisintegration. http://en.wikipedia.org/wiki/P-process Actual proton capture is now known as the rapid proton, or rp-process… http://en.wikipedia.org/wiki/Rp-process See also Clayton, p.299, Bertulani, p.356-358.

..., 23/04/11,

The term isotope is introduced later.

McNeil, 23/04/11,

Clayton, p.299. Mason, p. 56-59, WHW, p.1055, and Wallerstein, p.1054.

..., 23/04/11,

Only the basic processes are covered here. There is a rich variety of capture processes, that are now quite well understood – WHW and Wallerstein.

..., 23/04/11,

Clayton, p.290. Helium burning also creates other nuclides, such as O-18, F-19 and Ne-22 – WHW, p. 1029.


Proton-rich nuclides can be produced by the removal of one or more neutrons from a

nucleus, following a collision with a high energy photon or neutrino, and this is the way

that large proton-rich nuclides are created.

Neutron capture processes

it all depends on the rate of neutron arrival

In contrast with protons, a stable nucleus can easily capture an uncharged neutron, and

this shifts its proton:neutron balance, moving it one step up the neutron-rich slope of the

nuclear valley. In time, this nucleus will undergo beta-minus decay (np) and move

diagonally back down the slope towards the arc of stability (see back to section zZz).

However, if the nucleus is hit by a second neutron before it has time to decay, then it is

driven another step away from the arc of stability, and further up the nuclear valley slope.

If the nucleus is subjected to a series of neutron collisions, then there is a competition

between the rate of neutron capture and the rate of nucleus decay. If the rate of neutron

arrival is slow, with a long time between captures, then the nucleus will be able to decay

back to the arc of stability before the next neutron arrives. If the rate of neutron arrival is

fast, then the nucleus will be driven far up the slope of the nuclear valley, and be unable

to return to the stable arc at the valley bottom. We see both neutron capture processes at

work in stars, the slow “s-process” and the rapid “r-process”, and they build up quite

different sequences of nuclei - figure 3.57.


..., 23/04/11,

Bertulani, p.367.

McNeil, 23/04/11,

Diagrams showing the full s and r nuclide series are in: Mackintosh et al, p.78 and 119, Bertulani, figure 12.9, Kris Heyde, fig. 7.2, and at… GSI Helmholtz Centre, at… http://www.gsi.de/forschung/kp/kp2/nuc-astro/HeavyElements_e.html and Marielle Chartier, University of Liverpool, at… http://ns.ph.liv.ac.uk/~mc/my_research/mass_measurements.html Short sequences of the s and r process nuclides are shown at… http://rst.gsfc.nasa.gov/Sect20/A7.html and also… http://www.astro.wesleyan.edu/~bill/courses/astr231/wes_only/element_abundances.pdf (accessed in 2010, but no longer available – 21 April 2011) A short sequence of s process nuclides is given in… http://en.wikipedia.org/wiki/S_process A short sequence of r process nuclides is in… http://www.riken.go.jp/engn/r-world/info/release/press/2010/100608/index.html (all accessed 22 April 2011) The r nuclide sequences, from a diversity of sources, are not identical, but fairly consistent. I’m content with this, especially as the exact r-process sequence, with its highly unstable nuclides, is a bit speculative anyway. References for neutron capture: Clayton, p.306, Mackintosh, p.118/9, Williams, p.354, Chown, p.185-193, Marschall, p.205, Bertulani, pp.356, 358, 367, Mason p. 56-9.


Figure 3.57: The s- and r-processes, starting at iron-56, in the nuclear valley. The path of the s-process meanders along the nuclear valley bottom, but the path of the r-process runs up the neutron-rich slope, and even goes outside the valley.The slow s-process

A star is a rich mix of nuclear reactions, some of which produce free neutrons. A major

source of neutrons is the carbon-13 mentioned above, which can step out of the endless

CNO cycle (figure 3.55) and undergo this reaction…

carbon-13(6p,7n) + helium-4(2p,2n) oxygen-16(8p,8n) + n

Neutrons from reactions like this hit and are captured by other nuclei. It is a very slow

process – there may be hundreds or even thousands of years between successive

captures. This means that any unstable nucleus formed will very likely undergo beta-

minus decay (n p) before the next neutron comes along. Notice that neutron capture

may take a nuclide into the beta-minus (np) decay zone, but it decays back to stability

before another neutron comes along. “The effect is that the nucleus works its way along

the floor of the energy valley towards heavier and heavier nuclei”. The s-process creates


slow s-process beta-plus (pn) decay rapid r-process

the slow s-process: the long delays between neutron captures allow the nucleus time to undergo beta-minus (np) decay back to the line of stability

the rapid r-process: very short delays between neutron captures, so the nuclei cannot undergo beta-minus (np) decay, and are driven up the nuclear valley slope

beta-plus (pn) decays are part of the s process

arc of stable nuclides

the s- and r-processes are shown here starting from the same iron-56 (26p,30n) nuclide

the unstable r-process nuclei decay back to the stable arc

..., 23/04/11,


McNeil, 04/23/11,

Clayton gives 10-100 years between captures -p.306, though Chown puts the rate at “1 neutron every 100,000 years or so” - p. 191. Clayton gives some illuminating figures: “In the helium-burning shell of an AGB star the flux of free neutrons is typically … 1016 neutrons per cm2 per second - and that condition may last 300 years ... during that time a huge 1026 neutrons strike each cm2. That is sufficient for 10 neutron captures ... the s process is patient, like a river." - p. 298.

..., 23/04/11,

The CNO cycle is widely accepted as the neutron source in red giant stars smaller than 8 solar masses – Chown, p.184. In massive stars it is mainly alpha capture by Ne-22 – WHW, p.1030. The s process nuclei are blown away into interstellar space, and this accounts for most of the s process nuclei in the galaxy – Clayton, p.307. Asymptotic Giant Branch or Hydrogen and helium burning alternate in AGB red giant stars, and core nuclei can be dredged up and mixed with the outer stellar layers, to undergo the s process - http://en.wikipedia.org/wiki/S-process The other main site for the s-process is in the cores of massive stars, with >12 solar masses, with the reaction… He-4(2p,2n) + Ne-22(10p,12n) ( Mg-25(12p,13n) + n – Clayton, p.307.


nuclei up to a size of about 90 nucleons, somewhat bigger than the most tightly bound

iron-group of nuclides, and well short than the largest effectively stable nucleus – bismuth

209.

the rapid r-process

If we imagine the s-process nuclides as produced by a dripping “neutron tap”, then in

contrast, the r-process nuclides are produced by a dam-burst of neutrons. Instead of a

wait of thousands of years between neutron captures, there may now be a thousand

captures per second. Each nucleus is hit by neutrons so frequently that it is driven high up

the slope of the nuclear valley, to become so unstable that its half life is comparable with

the time between successive neutron captures. The neutron deluge starts with an iron

‘seed’ nucleus, and produces a string of nuclei high up on the neutron-rich slope, moving

at great speed up the nuclear valley towards the largest nuclei at the end. The nuclides

high up the valley slopes have extremely short half lives, mere fractions of a second, and

so they can “move” through nucleon space very quickly.

Once the neutron deluge stops, each nuclide descends to the arc of stability at the valley

bottom, following a diagonal path set by beta-minus decay.

the three nucleon capture processes in nucleon-space

We can summarise the work of the three processes of nucleon capture on the nuclide

chart - figure 3.58.


..., 23/04/11,

The locus of the r-process in nucleon-space will fall short of the neutron drip-line. If you add a neuton to a nucleus on the drip-line, it “drips” out immediately – Mackintosh et al, p.78, Williams, p.354, Bertulani, p. 358. This is an allusion to the liquid drop model of the nucleus.

..., 23/04/11,

Clayton, p. 303. Mason says time intervals of 0.01 to 1 s - p.57.

..., 23/04/11,

Lead-208 is the largest truly stable nucleus, but Bi-209 has such a long half life it is often viewed as effectively stable. Mason says that the s-process terminates at bismuth-209, p.57.

..., 23/04/11,

WHW, p. 1031 – and the s-process in massive stars makes isotopes of sulphur, chlorine, argon, potassium, calcium and titanium. See also Chown, p. 186.


Figure 3.58: The three nucleon capture processes, shown in nucleon-spaceThe p-process creates many nuclides on the proton-rich side of the arc of stability. The s-

process creates many of the stable nuclei, as it meanders along the nuclear valley

bottom, taking perhaps thousands of years on each step, and going no further than the

largest stable nucleus. About half of the abundances of the elements between iron and

bismuth are produced by the s-process. The r-process creates, in a only few seconds, a

long arc of highly unstable, neutron-rich nuclides, all the way up to the largest known

nuclei. The arc is kinked where the nuclei have magic numbers of neutrons, conferring

additional stability. We will see later how the violent death of a large star provides the

explosive conditions that create the r-process nuclei.

a bigger context

We have looked at the how the nuclei at the two extremes of the nuclear valley are

created - the light nuclei by the burning of the first two fuels, and the heavy nuclei by the

explosive r-process. These processes mark the beginning and end of a star’s lifetime. We

are ready now to see these reactions in a bigger context, and look at the life cycles of

stars.

3.12.9 The life cycles of stars

stars of different masses


the rapid proton capture rp-process creates light proton-rich nuclei

the rapid r-process creates neutron-rich nuclei up to the heaviest known nuclei

the slow s-process creates nuclei along the arc of stability, but only as far as the biggest stable nucleus

50 82 126 magic neutron numbers

iron ‘seed’ nucleus, Fe-26p,30n), the start of the r-process

..., 23/04/11,

http://www.gsi.de/forschung/kp/kp2/nuc-astro/HeavyElements_e.html

..., 23/04/11,

There’s some variation between sources regarding the p-process. Mackintosh et al show the rp-process arc lying between roughly (20p,20n) and (50p,50n), as do GSI; Bertulani puts it between (5p,5n) and (35p,35n); Marielle Chartier has the rp-process running nearly up to bismuth. Mason says proton capture produces nuclides up to nearly 200 nucleons – fig. 5.3, and states that all three processes between them have created the 10 stable isotopes of tin, from Sn-112 to Sn-124 – p.59.


The Hertzsprung-Russell (HR) diagram plots a star's total energy output (magnitude or

luminosity - with our sun set at 1.0) against its surface temperature (calculated from its

spectrum - hence the "spectral class" plotted on the x-axis) – figure 3.59.

Figure 3.59: The Hertzsprung-Russell (HR) diagram plots luminosity on the right hand y-axis against surface temperature on the x-axis. Stars burning hydrogen lie on the diagonal main sequence; massive stars are bright, hot and short lived, and small stars are dim and cool and have long lives.All the stars burning hydrogen lie on the main sequence line, running diagonally across

the diagram. At the lower left are the white dwarf stars - very hot, but small, so they don't

emit much radiation energy. In contrast, the red giant stars in the upper right are cooler,

but so big that they emit large amounts of energy.

Larger stars run hotter and have shorter lives

Small objects, of less than ~0.1 solar masses, do not ignite their hydrogen, and only glow

in the infra-red as they slowly relase their gravitational energy, thus being somewhere

between stars and giant planets. A star needs to start out with at least ~0.1 solar masses

to achieve a core temperature of 10 miilion degrees and start hydrogen burning, and such

a star would then slowly burn hydrogen for an estimated 800 billion years, longer than the

current age of the universe.

The greater a star’s mass, the faster it must burn its nuclear fuel in the core, to support

the greater weight of the surrounding layers, and so we see that more massive stars on

the main hydrogen-burning sequence have shorter lifetimes. Thus, a star with 30 solar


mass: 30 Ms

Lifetime: 1 MyT ~25 MK

mass: 1 Ms

Lifetime: 10 BYT ~12 MK

mass: 0.3 Ms

Lifetime: 800 BYT ~10 MK

mass: 15 Ms

Lifetime: 10 MY

Key"30Ms" = 30 solar masses "60My" = 10 million years;"800By" = 800 billion years "25MK" = 25 million degrees Kelvin

..., 23/04/11,

Delsemme, p.56.

..., 23/04/11,

Williams, p.347. Delsemme puts the threshold at 0.3 solar masses – p.56.

..., 23/04/11,

Delsemme, p.58.

..., 23/04/11,

diagram from HyperPhysics… http://hyperphysics.phy-astr.gsu.edu/hbase/starlog/starc.html#c1 Added data from Delsemme, p.47. Silk, p.135, gives these times on the main sequence… 30 solar mass; 1 miilion years, 15 solar mass; 10 million years. Woosley and Janka give 11 My for a 15 solar mass star.

McNeil, 23/04/11,

A thorough description from… http://hyperphysics.phy-astr.gsu.edu/hbase/starlog/starc.html#c1 See also Kaufman and Freedman on stars: chapters 19-22. Wallerstein describes stellar life cycles in detail.


masses consumes its hydrogen (protons) in only about 60 million years, and has 10,000

times the energy output of our own sun. So, “the more fuel a star starts off with, the

sooner it runs out. This is because the more massive the star is, the hotter it needs to be

to balance its gravitational attraction”. Meanwhile, “the lower-mass stars consume their

hydrogen fuel in a very frugal manner … [and] ... scrape along from year to year while

spending virtually nothing. In contrast, the most massive stars bear an uncomfortable

resemblance to rich and profligate heirs, who run shamelessly through a multimillion dollar

estate over the course of a single weekend”.

Our sun - the life cycle of a star of 1 solar mass

Figure 3.60 shows the life cycle of our sun - a modest star of 1 solar mass.

Figure 3.60: The stages in our sun's life cycleOur sun is about half way through its 10 billion year phase of hydrogen burning, with a

core temperature of about 15 million degrees. We've seen that when the hydrogen-

burning reaction slows, the core heats up further, so hydrogen burning spreads outwards,

and heats up the outer layers of the star, which then swell. The surface area increases

faster than the energy output, so the star surface cools, and becomes redder, even


The end of the sun's life5: The hydrogen in the core is consumed, and now starts burning in the outer layers…the sun leaves the main sequence…

6: …swells to a red giant…helium is now burning in the core…the outer layers are being blown away…7: …so there is a

helium/carbon core inside big planetary nebula…

8: …which has now blown away, exposing the core as a white dwarf, with about half the sun's original mass…

1: A large cool gas cloud…

2: …collapses under gravity and heats up, so that….

3: …H-burning starts, and the new sun joins the Main Sequence…

9: …which now slowly dims and cools.

4: …where it steadily burns hydrogen in its core for about 10 billion years.

McNeil, 04/23/11,

Hey and Walters, p.214. George Smoot… http://aether.lbl.gov/www/tour/elements/stellar/stellar_a.html (accessed 18 Jan 2011)

McNeil, 04/23/11,

Our sun is a second or third generation star, so it contains other elements besides hydrogen and helium, but this does not alter the main features of its life cycle… Delsemme, p. 56. Sources on the sun's lifecycle… HyperPhysics http://hyperphysics.phy-astr.gsu.edu/hbase/astro/herrus.html#c4 Silk, p.133-5, is a helpful summary Nicholas Short is very helpful, as always… http://rst.gsfc.nasa.gov/Sect20/A5.html (accessed 18 Jan. 2011) American Association of Variable Star Observers http://www.aavso.org/ and for example, Mike Simonsen “Variable stars and the stories they tell”, in the Presentation Library National Maritime Museum… http://www.nmm.ac.uk/upload/img_400/1_20020624100906.gif (accessed 18 Jan. 2011) http://www.astro.columbia.edu/~archung/labs/spring2002/lab03.html (accessed 18 Jan. 2011) Adams and Laughlin, p. 48. Mason, ch. 7. Stuart Taylor describes the solar system, and the role of chance in its formation, and in the evolution of life. Our solar system formed ~4.5 billion years ago out of interstellar gas that was enriched by ~2% by mass with heavier nuclei from previous generations of stars – Hogan, p. 128.

..., 23/04/11,

HR diagram from HyperPhysics. Carl Sagan, chapter 9, shows how the Earth will change as the sun goes through its red giant phase. A number of sources for this sequence: Silk, p.133, Delsemme, ch.3, Taylor, p. 34 http://www.astro.columbia.edu/~archung/labs/spring2002/lab03.html (accessed 18 Jan. 2011) Woosley, lecture 13-14 http://ucolick.org/~jjfang/ay12/0301_section.pdf American Association of Variable Star Observers… http://www.aavso.org/ National maritime Museum.. http://www.nmm.ac.uk/upload/img_400/1_20020624100906.gif (accessed 18 Jan. 2011) Wallerstein shows the life cycles of a number of starsof different masses – fig. 1.

..., 23/04/11,

Adams and Laughlin, p.45.

..., 23/04/11,

Stephen Hawking , "A Brief History of Time", p.83.

..., 23/04/11,

Data on star lifetimes from Delsemme, table 3.2.


though it is emitting more luminous energy. The star is moving through its red giant

phase. Our sun is expected to increase its luminosity 10,000 times and expand 200 times,

thus engulfing the Earth, though its surface will cool slightly to 4,000K. The core

temperature, however, continues to rise under gravitational contraction, until at ~120

million degrees helium ignites, burns for about 100 million years, and builds up carbon in

the sun’s core. With hydrogen continuing to burn steadily outwards through the star, the

outer layers are inflated and the star separates into a highly dense core of helium/carbon

surrounded by a swollen nebulous cloud of gas, rich in carbon and nitrogen, most of

which is blown away.

a white dwarf star

The core is not massive enough to raise its temperature high enough to ignite carbon, the

next fuel after helium. So, with no further nuclear reactions accessible, it shrinks and

becomes a white dwarf star, that just cools slowly for billions of years. A white dwarf star

is very hot, but, being small is not very luminous, so it is at the bottom left in the HR

diagram.

matter under pressure

Even a modest sized star, like our sun, has a mass of about 2 x 1030 kg – two thousand

billion billion billion kilograms. The matter at the star’s centre must support this enormous

weight. How does matter behave under such extreme pressures?

We have seen earlier that in the recombination event, when the universe cooled to a

temperature of ~3,000 degrees, each nucleus gathered enough electrons in “orbit” around

itself, to make a neutral atom. But for atoms in the hot core of a star, the electrons have

too much energy to remain in settled orbits, so the “nuclei are immersed in a sea of free

electrons that tend to cluster near the nucleus”.

degenerate electrons and the Chandrasekhar mass

The matter in the core of a modest sized star is supported by this sea of free electrons,

which exert an ’electron degeneracy pressure’, which is itself a consequence of the

uncertainty principle. The uncertainty in the positions of the electrons is equivalent to tiny

motions, and these exert a pressure. In a white dwarf star, “this pressure is exerted by

electrons that are crowded together as the collapse of the star squeezes atoms together

until they overlap”. We can think of the electrons' matter-waves starting to overlap each

other, and because the electrons are fermions, they can't occupy the same volume, they


..., 23/04/11,

Silk, p.134.

..., 23/04/11,

Silk, p.134. “The Pauli exclusion principle prevents two electrons with the same quantum numbers occupying the same volume. The size of this minimum volume is determined by the de Broglie wavelength of the electron. Since this wavelength becomes smaller as the momentum of the electron increases, the electrons move faster and faster as the pressure increases…. when the electrons are moving with velocities close to the speed of light, the Pauli principle applied to electrons will prevent any furthe rcollapse of the core” - Hey and Walters, p.216. So it is the uncertainty principle again – as the electrons are packed tighter together, with less uncertainty in their position, the uncertainty in their motion increases, but their speeds cannot exceed light speed, so this limits the star’s compaction. Also Delsemme, p.51, Marschall, p.129, Williams, p.347, and… http://hyperphysics.phy-astr.gsu.edu/hbase/astro/whdwar.html#c3 Degenerate matter is not in the same atomic state as we know on Earth, “we must assume the electron shells are partly crushed” and the “ electron quantum rules cannot be upheld” – Bertulani, p.336. An object such as a white dwarf star, comprising degenerate electrons is unusual in that (1) adding more matter to it makes it shrink rather than get bigger, and (2) heating it does not make it expand – Adams and Laughlin, p.49.

..., 23/04/11,

Bertulani, p.347.

McNeil, 23/04/11,

Hey and walters, p.216.

McNeil, 23/04/11,

Clayton, p.77, and Croswell, p.172

..., 23/04/11,

WHW, table I and Silk, p.133. This is a very simplified version of events – see, for example, Adams and Laughlin, p.49.

McNeil, 04/23/11,

HyperPhysics. Delsemme, p.56 gives 20x larger and 100 times brighter.


must always be separate. So “the material of the star stiffens and, even without the

benefit of additional energy, supports its own weight“.

This degenerate electron pressure brings the gravitational contraction to an end in a

moderate sized star, such as a white dwarf. However, the greatest mass that degenerate

electrons can support is about 1.4 solar masses, known as the Chandrasekhar mass, in

which matter is compressed to a density of about 1000 kg/cm3, that is, 1 tonne/cm3, or

about the mass of a small car in your fingertip.

degenerate neutrons and neutron stars

In stars up to ~1.4 solar masses the electrons are viable independent particles, and can

sustain the pressure at the core. However, in a more massive star, the pressure forces

the particles so close together that the protons in the nuclei are able to capture the free

electrons, and form neutrons by the reaction…

p+ + e- n0 +

with the excess energy being carried away by the newly created neutrinos.

Thus, the star’s structure of separate nucleons and electrons is destroyed, and “all nuclei

decay and only a ‘puree’ of neutrons is left”. The star’s core becomes, in effect, a single

gigantic nucleus, composed only of degenerate neutrons - a neutron star. A typical

neutron star has the mass of two suns crammed into a sphere 20 km in diameter. The

density is now hard to imagine – approaching 1 billion tonnes of matter in each cubic

centimetre. We can think perhaps of a moderate sized mountain compressed into your

fingertip, seemingly an impossible condition, but this is the density of the nucleus of every

atom our world is made of. Nucleons are independent entities with a viable physical

presence at these extreme pressures. Yet there is a limit to what even nuclear matter can

withstand, and the greatest mass that degenerate neutrons can support is ~2-3 solar

masses.

black holes

Beyond this mass, physical matter undergoes total collapse into a black hole, from which

no light can escape. The theory of General relativity predicts that space-time is distorted

by gravity. Thus in a gravity field clocks tick and crystals vibrate more slowly. This is a tiny

effect in the Earth’s gravity field, but big enough that the highly accurate clocks on the

satellites used in the GPS navigation system need regular correction. The enormously

strong gravity field near a black hole slows down time so much that radiation from it is so


..., 23/04/11,

PHYSICS IoP magazine…

..., 23/04/11,

Silk, p.142.

..., 23/04/11,

Black Holes Kaufman and Freedman, ch, 24. Stephen Hawking, "A brief History of Time", ch. 6, and "The universe in a nutshell", p.110-117. A helpful review and links… http://en.wikipedia.org/wiki/Black_Hole http://en.wikipedia.org/wiki/Stellar_mass_black_hole An interactive site at… http://hubblesite.org/explore_astronomy/black_holes/home.html If an object is compressed to less than its Schwarzschild radius then its own gravity will collapse it into a black hole. Every object has its own Schwarzschild radius, which is proportional to the mass: for our sun it's about 3 km, for Earth it's 9 mm. A neutron star with more than about 3 solar masses will be smaller than its Schwarzschild radius. http://en.wikipedia.org/wiki/Schwarzschild_radius Hey and Walters, p. 223. Gribbin "Stardust", p. 136, p.144. These are predeicted by Einstein’s General Theory of Relativity. Their properties are theoretically well described, and their identification is becoming increasingly confident. Interesting as black holes are, they are not part of our story. See Silk, section 8.5, Williams, section 14.11.

..., 23/04/11,

Rod Nave… http://hyperphysics.phy-astr.gsu.edu/hbase/astro/pulsar.html#c3 Williams says neutron stars cannot exceed ~2 solar masses – p.368, while Silk says 3 solar masses – p.145. Gribbin name it as the Oppenheimer-Volkoff limit, ~3 solar masses – p.136.

..., 23/04/11,

I’ve gone for the round number of Silk, p.136. Pogosian gives the same value of average density. Bertulani, p.373, gives 108 , Hyperphysics gives 2x108 tons/cm3. Wikipedia’s article on neutron stars gives the core density as ~6x108 tons/cm3… http://en.wikipedia.org/wiki/Neutron_star

..., 23/04/11,

Dmitri Pogosian, at http://www.ualberta.ca/~pogosyan/teaching/ASTRO_122/lect19/lecture19.html Bertulani gives 1.4 solar masses in 20km radius, and an average density of 100x106 tonnes/cm3 – p.369.

..., 23/04/11,

Delsemme, ch.3, Silk, p.135, Williams, section 14.3.

..., 23/04/11,

Delsemme, p.52. The star’s weight is now supported by neutron degeneracy pressure – Silk,p.136.

..., 23/04/11,

Silk, p.136, Hey and Walters, p.220, Woosley and Janka, Marschall, p.132.

..., 23/04/11,

The consensus seems to be ~1000 kg/cm3. Hey and Walters say I teaspoonful would weigh several tons - p.216, and HyperPhysics also gives 5 tons/teaspoonful – and I take 1 teaspoon as 5 cm3. Delsemme states that degenerate electrons can sustain a core density of 100 tons/cm3, that is, 100,000 kg/cm3 – p.51.

..., 23/04/11,

Silk, p.134, and … http://hyperphysics.phy-astr.gsu.edu/hbase/astro/whdwar.html#c3 http://en.wikipedia.org/wiki/Chandrasekar_limit This mass depends on the ratio of electrons to nucleons, and is ~1.4 solar masses – Bertulani, p.371, Silk, p.135, though Williams puts it at 1.2-2 Solar masses – p.362.

..., 23/04/11,

Marschall, p. 129.


redshifted that it disappears from view.Thus a stellar core that collapses to a blck hole has

reached the "end of time", and is "cut off forever from the rest of the universe", for not

even light can escape.

nuclear fuel thresholds

we can thus see a series of thresholds for nuclear fuel burning and stellar fates – figure

3.61.

Figure 3.61: A summary of stellar nuclear fuels and fates (Ms stands for solar mass)This graph outlines the story so far, and also shows the way ahead. The more mass a star

starts out with the more nuclear fuels it can burn. A star with more than ~0.1 solar masses

can ignite hydrogen, and with more than ~0.25 solar masses it can burn helium to carbon.

The graph shows that stars with more than ~4 solar masses can then ignite carbon, with

more than ~8 solar masses they can ignite oxygen, and with more than ~15 solar masses

they can make it all the way to the last fuel, silicon.

A star’s outer layers are blown away in the process of burning its nuclear fuel, especially

helium, and so it ends up as a much smaller remnant. A star starting out with ~8 solar


1

10

100

1000

10000

0 5 10 15 20 25initial star mass (solar masses)

core

tem

pera

ture

(mill

ions

of d

egre

es K

)

ends up <1.4 Ms

white dwarfends up 1.4 to-~3 Ms

neutron starends up >~3 Ms

black hole

hydrogen burns: 15MK

helium burns: 120MKcarbon burns: 700MK

oxygen burns: 2,000 MK

silicon burns: 3,300 MKoxygen burns: 2,000 MKneon burns: 1,500 MK

carbon burns: 700MK

helium burns: 120MK

hydrogen burns: 15MK

core pressure ~1 ton/cm3 the maximum for

degenerate electrons

core pressure ~1billion tons/cm3 the maximum for degenerate

neutrons

..., 23/04/11,

I’ve used data from Delsemme, tables 3.1 and 3.2. However, this does not really square with WHW, who state that <~8 solar masses stars do not ignite carbon, and end up as white dwarfs (p.1035), and that C and Ne ignite in stars >~11 solar masses – though this figure is uncertain, and that stars of >~11 solar masses can burn their nuclear fuels all the way to silicon - p.1037.

..., 23/04/11,

I felt the reader – and I - would benefit from a diagram laying out the overall pattern, but it was a bit more effort than I expected. There are no diagrams, and data is here and there and not all that consistent. Nuclear burning temperatures from Woosley, Heger and Weaver (WHW), table I and figure 8. The temperatures of burning nuclear fuels depend on the star’s mass, and are higher than their ignition temperatures. For simplicity, I give each fuel one burning temperature, using data for the advanced stages of burning in a massive star – WHW, fig.8. These temperatures are fairly consistent with others, such as Williams and Bertulani. Outcomes for starting masses consistent with Dmitri Pogosian, at… http://www.ualberta.ca/~pogosyan/teaching/ASTRO_122/lect19/lecture19.html (accessed 2 Feb 2011). Stellar masses to ignite burning reactions from Delsemme (table 3.1) and Williams (section 14.3) – and fairly consistent. Figures for H and He burning are for a 1 solar mass star – we go on to look at a 15 solar mass star, which runs hotter.

McNeil, 23/04/11,

Delsemme, p.52.

..., 23/04/11,

Hawking, "A brief History of Time", p.88.

..., 23/04/11,

Silk, p.143.


masses will lose so much material that a core of ~1.4 solar masses is all that remains.

This is the most that can be supported by degenerate electrons, so stars up to ~8 solar

masses end their lives as white dwarf stars, composed mainly of carbon and oxygen.

Stars with initially between ~8 and 20 solar masses end up as neutron stars, and beyond

this they end up as black holes.

going all the way

It's time now to look at a massive star, in which the nuclear reactions will consume all the

available fuels, and end up with a neutron star core. We will see how the enormous

temperatures and pressures developed by a massive star bring about its inevitable

destruction, but also enrich the universe.

3.12.10 The alpha-process - burning all the way to iron

We will follow the life of a large star, starting out with about 15 solar masses, which has

burned helium-4 largely to carbon-12, with some oxygen-16, by the triple-alpha process.

Once helium-4 is depleted in the star's core, gravitational compaction pushes up the

pressure and temperature, and now heavier nuclei are built up by the alpha process. This

is a mix of nuclear reactions, broadly based on transactions involving helium-4 nuclei, at

steadily increasing temperatures - up to more than 3 billion degrees. These enormous

temperatures introduce two new features to nuclear fuel burning. The first is the entrance

of neutrinos into the stellar economy. We have seen that a star in the hydrogen burning

phase loses a small portion of its energy by the emission of neutrinos, in addition to light

radiation. With rising temperatures this becomes more significant, and with a core

temperature more than ~500 million degrees, neutrino losses come to dominate the star’s

‘energy budget’. Second, at the very high temperatures of the alpha-process the thermal

photons have enough energy to break up nuclei, and this ‘photo-dissociation’ becomes an

important feature of nuclear burning reactions. So in addition to the normal fusion

process, whereby two smaller nuclei merge into a larger one, we also see a progressive

rearrangement of nucleons, involving the capture of protons, neutrons and alpha particles.

The alpha process has four distinct stages, distinguished by their principal fuels - carbon,

neon, oxygen, and finally silicon. The first and third burn by simple fusion reactions; the

second and last involve the break up of nuclei by thermal photons.

carbon burning


..., 23/04/11,

WHW, p. 1032, Williams, p.353, Clayton, glossary, Bertulani adds neon-burning to the sequence, p.364.

..., 23/04/11,

Generally, the alpha process seems to be depicted as mainly alpha particle captures, building up the alpha nuclei. Wallerstein writes of protons, neutrons and alpha particles being present in the star core during the alpha process reactions – p.1034-1036. Bertulani refers to the large number of free neutrons – p. 366. WHW describe the many nuclear reactions involving single nucleon capture.

..., 23/04/11,

WHW, p.1032.

McNeil, 23/04/11,


..., 23/04/11,

References for this section: WHW, Williams, section 14.4, Bertulani, sections 12.8-9, Delsemme, ch.3, Clayton, glossary, Wallerstein, Kaufmann and Freedman, Gribbin, Krauss, and NASA, at… http://imagine.gsfc.nasa.gov/docs/teachers/elements/imagine/05.html This is the original name; it’s now treated as carbon, neon and oxygen burning – WHW, p.1016. There is a short Wikipedia article, lacking references… http://en.wikipedia.org/wiki/Alpha_process

..., 23/04/11,

This is for a star of about 15 solar masses. Consequently the burning processes run hotter than for a 1-solar mass star – hydrogen at ~35 million, and helium at ~180 million degrees – WHW, fig. 8. A useful illustrated summary at… http://aether.lbl.gov/www/tour/elements/stellar/stellar_a.html Temperatures, fuel lifetimes and densities are from WHW, table I and figs, 1 and 8, for a 15 solar mass star, with fairly comparable figures from Bertulani, Williams, Delsemme- table 3.1, and Kaufmann and Freedman.

..., 23/04/11,

Bertulani states that a star of ~25-35 solar masses is big enough to leave a remnant core of more than a few solar masses – p. 375.

..., 23/04/11,

Approximate figures here – Silk says stars >~30 Ms (p.143) and Williams says >~10 Ms (p.362). I’ve just taken an average – we only want to get the overall shape of things. WHW fig. 12 suggests that <~8 solar mass stars become white dwarfs, and then become neutron stars up to ~25 solar masses.

..., 23/04/11,

Adams and Laughlin, p.53, and Silk, p.135.


At a temperature of ~700 million degrees, the carbon-12 nuclei have enough energy to

overcome the mutual repulsion of their 6 protons, and they collide and undergo a set of

competing reactions, producing a mix of different nuclides. We’ll look at this stage in some

detail, because it illustrates the important features of the alpha process – figure 3.62.

Figure 3.62: The mix of nuclear reactions that constitute carbon burning. The main reactions produce neon (Ne-20) and sodium (Na-23), but magnesium (Mg-24) and oxygen (O-16) are also directly produced. Further reactions produce oxygen and also neutrinos. The most energetic photons have enough energy to create electron-positron pairs, a few of which then create neutrino-antineutrino pairs.The diagram shows a tumble of nuclear reactions, producing a variety of nuclei in the size

range 16-24. Some of the reactions involve single nucleons. Thus, a loose proton reacts

with a carbon-12 nucleus to give nitrogen-13, which is unstable and undergoes beta-plus

(pn) decay to carbon-13, which then reacts with a helium-4 nucleus to give oxygen-16.

The main nuclear products of carbon burning are: oxygen-16, neon-20, sodium-23, and

magnesium-24.

neutrinos enter the scene


Ne-20 + He-4(10p,10n) (2p,2n)

C-12(6p,6n)

C-12(6p,6n)

Na-23 + p(11p,12n)

C-12(6p,6n)

N-13(7p,6n)

O-16 + He-4 + He-4(8p,8n) (2p,2n) (2p,2n)

Mg-24 (12p,12n)

C-13 + e+ + (6p,7n)

O-16 + n(8p,8n)

carbon-burning core700 million degrees

helium-burning shell180 million degrees

hydrogen-burning shell35 million degrees

energy-bearing neutrinos escape from the stellar core

beta-plus (pn) decay

e+ + e-

electron-positron pair

+ neutrino

antineutrino pair

..., 23/04/11,

The principal products are isotopes of oxygen neon, sodium, magnesium, aluminium, with a little silicon and phosphorus – WHW, p. 1032. Also see Bertulani, p.363, Williams, p.353

..., 23/04/11,

There is a standard notation for nuclear reactions – Williams, p.101, but while it is concise, it would not help the general reader. Besides, I want to give a pictorial representation. Because there are a lot of labels, I have used the chemical symbols for the nuclei. There aren’t many and I think the reader will cope with this. The C-burning reactions are fromWHW, p.1032, Williams, p.353, Bertulani, p.362, and Clayton, p.280. The Wikipedia article is fully consistent with these writers, and gives further sources… http://en.wikipedia.org/wiki/Carbon_burning_process The temperatures are from WHW, fig,8 for a 15 solar mass star in the advanced


The star’s core has now regained the temperature of the universe when it was about 3

minutes old (see back to section 3.11.6). At 700 million degrees, the average photon

energy is about 60 keV, that is, about one eighth of the mass-energy of an electron or

positron. The fastest thermal photons have enough energy to sustain a substantial

population of electron-positron pairs, either directly, or by colliding with a nucleus in the

core. Furthermore, “when the electrons meet and annihilate with positrons, a neutrino-

antineutrino pair is occasionally produced. These neutrinos escape the star with ease and

force the burning to go faster to replenish the loss”. Figure 3.62 shows that some of the

nuclear reactions themselves produce additional neutrinos. From the carbon-burning

stage onwards “the dominant energy loss from the star is due to neutrinos streaming out

directly from the stellar nuclear furnace, rather than by [light] photons from the surface”.

Even a main sequence star, burning hydrogen, loses a small fraction of its energy – about

6% - as neutrinos (see back to section 3.12.5). The carbon-burning star core loses ~84%

of its energy as neutrinos, leaving very little to oppose gravitational collapse, and forcing

the nuclear burning reactions to go even faster. Moreover, the nuclear reactions are

yielding less energy. The rearrangement of 24 nucleons from two carbon nuclei into one

neon and one helium nucleus…

C-12 (6p,6n) + C-12 (6p,6n) Ne-20 (10p,20n) + He-4 (2p,2n) mass loss 4.7 MeV

releases only 4.7 MeV, about 0.2 MeV per nucleon. So the star is losing energy at a faster

rate, and the nuclear fuel is yielding less energy. Temperatures and densities are starting

to stretch the imagination. The temperature is approaching one million degrees, and the

core density is ~240 kg/cm3 , around a quarter ton of mass in your little fingertip).

summary of carbon burning

Helium burning produced nuclei ~12-16 nucleons in size, and carbon burning has raised

this to a size range of ~16-24 nucleons. We are seeing a mix of fusion reactions, and also

the nuclei reshuffling themselves, mainly by transferring alpha particles, but also single

nucleons. The star is building up a multi-layer structure, with the latest reaction in the

core, surrounded by the previous reactions in successive outer layers.

neon burning and photo-dissociation

When carbon is depleted, the star compacts further, driving the core temperature to about

1,500 million (1.5 billion) degrees, and now the most energetic photons start breaking the

up the neon-20 nuclei, that were assembled in the carbon burning phase.


..., 23/04/11,

WHW. Fewell (p. 656) states that photodissociation starts at ~1 billion degrees, with neon burning, as does Bertulani, p.364.

..., 23/04/11,

Williams, p.353.

..., 23/04/11,

Bertulani gives a much larger figure of 3,000 kg/cm3 - p.363.

..., 23/04/11,

Carbon burning to magnesium-24 yields 14 MeV, but this is a fairly minor reaction – Bertulani, p.364. Bertulani quotes one estimate that when all the secondary reactions of carbon with oxygen and with neon, are included, each pair of C-12 nuclei releases ~ 13 MeV of energy – p.364. This is still ~0.5 MeV/nucleon, less than the yield of helium burning. WHW state that the yield of Ne burning is ~1/4 that of C burning, in terms of energy/gram – p.1033.

..., 23/04/11,

Bertulani, p.364.

..., 23/04/11,

Woosley and Janka - they write that temperatures of 1 billion degrees or more, are adequate to maintain a large thermal population of electron-positron pairs. See also Adams and Laughlin, p.55, and Wikipedia… http://en.wikipedia.org/wiki/Carbon_burning_process (accessed 15 Jan 2011). Above 100 million degrees, an electron and positron can annihilate to a neutrino-antineutrino pair with the release of 1.02 meV, either by W exchange or through a Z0 intermediate state. There are numerous reactions producing neutrinos – Williams, p.360/1. Woosley and Janka give some figures (in solar units) for luminosity (photon emission) and for neutrino losses for a 15-solar-mass star: H burning: 28,000 and 1,800 respectively; He burning: 44,000 and 1900 ; C burning: 72,000 and 370,000; Si burning: 75,000 and 1.3 x 1011. From these I infer that in H burning the neutrino fraction of total energy losses is 1800/(1800+28,000) ~6%; in He burning it’s ~4%, in C burning it’s ~84%, and in Si burning it’s virtually 100%. Williams states that ~2% of our sun’s total energy is removed by neutrinos – p. 357.

..., 23/04/11,

Following Weinberg, and using E=~kT, p.81.


neon-20 (10p,10n) + photon oxygen-16 (8p,8n) + helium-4 (2p,2n)

The loose helium then reacts with surviving neon…

helium-4 (2p,2n) + neon-20 (10p,10n) magnesium-24 (12p,12n)

This photo-dissociation process - breaking up by light - has reshuffled the two neon

nuclei. We will see this process become increasingly significant as the star continues to

heat up. At the end of the neon-burning stage the star’s core contains mainly oxygen-16,

magnesium-24 and silicon-28.

oxygen burning

The core temperature now rises to ~2,000 million (2 billion) degrees, and compresses to a

density of ~7 tonnes/cm3, which is sufficient for oxygen nuclei to overcome their mutual

repulsion and react directly, with two main outcomes:

1) oxygen-16 (8p,8n) + oxygen-16 (8p,8n) silicon-28 (14p,14n) + helium-4 (2p,2n)

2) or… sulphur-32 (16p,16n)

Oxygen burning also creates a large range of nuclides up to ~40 nucleons in size, with the

alpha-nuclei being favoured, due to their higher binding energies. Thus a number of

familiar elements appear: phosphorus, sulphur, chlorine, argon, potassium and calcium.

silicon burning

The core temperature now exceeds 3 billion degrees, and the density is about 40

tonnes/cm3, the mass of around 40 cars in your little fingertip. Now the thermal photons

have so much energy that the processes of photo-dissociation and fusion are finely

balanced. A simple but helpful view is of highly energetic photons breaking up Si-28

nuclei, and the fragments - protons, neutrons and alpha particles - then being

incorporated into larger nuclei. Nuclei in the size range 28-65 nucleons thus compete for

survival under the barrage of photons, and the larger nuclei are favoured because they

are more tightly bound. So, “nuclei with smaller binding energies are destroyed by photo-

dissociation in favor of their more tightly bound neighbors, and many nuclear reactions

involving alpha-particles, protons, and neutrons interacting with all the nuclei … take

place”. This has been called "nuclear melting" to distinguish it from nuclear burning.

antarctic penguins

At the South Pole, the male Emperor penguins huddle together for warmth, as they guard

their eggs through the antarctic winter. The bigger the huddle, the more penguins there


McNeil, 23/04/11,

Clayton, p.309.

..., 23/04/11,

Bertulani, p.366.

..., 23/04/11,

The basic e-process - Wallerstein, section XV B.

..., 23/04/11,

Wallerstein, p. 1036. Si burning is not the simple fusion 2 Si-28 ( Ni-56 WHW, p.1034.

..., 23/04/11,

WHW, fig. 1 and table 1.

..., 23/04/11,

Wallerstein, p.1034. Also Bertulani, p.365, Gribbin, p.142, NASA., and WHW, p.1033.

..., 23/04/11,

Bertulani, p.365, and Wallerstein, p.1036, and WHW, p.1033. The heavy s-process nuclides beyond the iron-group built up during He, C and Ne burning are destroyed by photo-disintegration – WHW, p.1033.

..., 23/04/11,

Woosley and Janka.

..., 23/04/11,

Secondary reactions produce a host of isotopes in the size range of 24 nucleons up to iron-group nuclei with ~60 nucleons. Also, this is when a significant neutron excess can be established – WHW, p. 1033.

..., 23/04/11,

Bertulani, p.365, Williams, p.353. Also called photo-disintegration. Photodisintegration A good article, with many external links at… http://en.wikipedia.org/wiki/Supernova Marschall, p.132, Adams and Laughlin, p.56, Woosley and Janka. Delsemme, p. 51. The total reaction.. 2 Ne-20( O-16 + Mg-24 yields ~4.6 MeV, or ~0.1 MeV/nucleon.


are in the warm interior. In a similar way, the nucleons in a stellar core gather in

progressively bigger and more tightly bound huddles, against the rising fury of the

photons.

growth v. fragmentation

The nuclear melting process thus broadly favours the tightly bound iron-group of nuclides ,

at the peak of the binding energy curve, with 56-60 or so nucleons. Silicon burning

produces a clutch of nuclides in this range: isotopes of chromium, manganese, iron,

cobalt and nickel. However, in the enormously hot stellar core the processes of nuclear

fragmentation and growth are finely balanced, and the viability of a nuclear huddle

depends as much on its resistance to photo-dissociation as on its binding energy. Under

these circumstances, the two main products of silicon burning are Ni-56, the most tightly

bound of the series of alpha-nuclei, and “the natural end of the alpha process”, and Fe-56,

with the third highest binding energy of all the nuclides. The balance of products is

sensitive to the neutron population in the stellar core, so that a small neutron excess

favours the formation of the more Ni-56(28p,28n), and a large excess favours the neutron-

rich Fe-56(26p,30n), which has an excess of 4 neutrons.

a poor energy yield

The transformation of two silicon-28(14p,14n) nuclei into one nickel-56(28p,28n) nucleus

yields little energy…

Si-28 (14p,14n) + Si-28 (14p,14n) Ni-56 (28p,28n) mass loss = 10.9 MeV

This rearrangement of 56 nucleons yields 10.9 MeV, that is, only 0.2 MeV per nucleon.

Ni-56 is unstable, and undergoes beta-plus decay twice to Fe-56...

nickel-56 (28p,28n) n + cobalt-56 (27p,29n) n + iron-56 (26p,30n)

Thus Fe-56 is the favoured nuclide in the star’s core, either produced by silicon burning or

by the decay of Ni-56. Iron-56 has an excess of 4 neutrons, and we'll see these neutrons

play an important rôle later.

silicon is the last nuclear fuel

The iron-group of nuclides have the highest binding energies of all the nuclides, and are

at the peak of the binding energy curve. We’ve seen how the nuclear burning reactions

fuse smaller nuclei into bigger reduce mass and hence release binding energy to support

the star’s weight. However, this only works as far as the iron-group nuclides. Beyond


..., 23/04/11,

Wallerstein, p.1034, and WHW, p. 1034 describe how the main Si-burning products are the most tightly bound nuclides at the value of neutron excess pertaining in the core.

..., 23/04/11,

Statement and quote from Fewell, p.656; figures in the mass.mas03 table. This is tricky, but here’s how I think things go. The nuclei in the star’s core are mainly multiples of He-4, because they have built up by the successive addition of alpha particles. Ni-56 has the largest binding energy of the alpha-nuclei (Fewell) but is some way off the arc of stable nuclides, which now have more neutrons than protons. Ni-56 is about 60th in the ranking of binding energies per nucleon, so there are many nuclides with higher binding energies. Fe-56 has the third largest value of all the nuclides, behind Fe-58 and Ni-62. My understanding of Fewell is that the finely balanced competition between growth and photo-dissociation favours Fe-56, even though it hasn’t got the largest binding energy/nucleon. Fewell suggests that Ni-62 has a small abundance is because it is less resistant to photo-dissociation - see Rod Nave on this... http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin2.html#c1 Fewell points out that many established physics texts state Fe-56 as having the largest binding energy per nucleon, which is incorrect; Both Bertulani (p.366) and Williams (p.356) cite Fe-56 as being the most strongly bound nucleus. Fewell suggests that the prominence of Fe-56 in the literature is due to the the model of supernova nucleosynthesis, which explains its great abundance. Ni-56 can undergo two beta-plus decays, first to Co-56, and then Fe-56, and this seems another source of Fe-56. These two decays yield heat energy, but the half lives of 6 days and 77 days respectively, are too slow to meet the star’s urgent need for energy. We’ll see Ni-56 appear in supernove SN1987A later. The silicon burning reaction is sometimes described as a process of progressive capture of fragments that “slowly reorganises two Si nuclei into one Ni nucleus” (Clayton, p.309), but it seems to be more elaborate than this. The core of a large silicon-burning star is usually described as iron, with some writers staing it as "Fe-Ni" (Bertulani, fig.12.15), or iron-group elements (Williams, fig.14.1 and Gribbin, fig.8.3). What’s important for the reader is that the iron-group of nuclei, with ~56-60 nucleons, and the largest binding energies, are no good as nuclear fuels for the star. Fusion reactions beyond the iron group don’t yield heat energy, instead they take it in – they cool down rather than heat up. The consumption of silicon is the end of the line for the star.

..., 23/04/11,

Fewell, p.656.

..., 23/04/11,

Marschall, p.128, Gribbin, p.142, Woosley and Janka, table 1.


these, further enlargement increases mass, and takes energy from the star, rather than

donate it. So, with the “burning” of silicon, the star has used its last nuclear fuel.

The main steps in the alpha process are shown in nucleon space, in figure 3.63.

Figure 3.63: The alpha process starts with carbon-12 and builds ever larger nuclei, up to the iron group of nuclidesthe alpha-nucleus staircase

The dotted line in the diagram marks the series of alpha-nuclei, that have equal numbers

of protons and neutrons. We can see the alpha process as a kind of stair-case, working in

regular steps diagonally upwards in nucleon space. The tightly bound helium-4 nucleus is

the preferred currency of "trade", so the alpha-nuclei are favoured, and we'll see later that

they are noticeably more abundant in the universe than the ones in between. The major

fuels in the alpha process - helium-4 (2p,2n), carbon-12 (6p,6n), oxygen-16 (8p,8n), neon-

20 (10p,10n) and silicon-28 (14p,14n) – are all even-even nuclei, and two of them are also


5 10 15 20 25 neutrons, n

iron-56 has the third largest binding energy of all the nuclides, and is favoured by a large neutron excess

1. C-12 burning creates O-16 and nuclides up to ~24 nucleons…

2. …oxygen burning creates Si-28 and other nuclides, up to ~40 nucleons…

3. …finally, silicon burning creates nuclides in the tightly bound iron group, mainly Ni-56 and Fe-56…

protons, p

zinc-Zn,30copper-Cu,29

nickel-Ni,28cobalt-Co,27

iron-Fe,26manganese-Mn,25

chromium-Cr,24vanadium-V,23

titanium-Ti,22scandium-Sc,21

calcium-Ca,20potassium-K,19

argon-A,18chlorine-Cl,17sulphur-S,16

phosphorus-P,15silicon-Si,14

aluminium-Al,13magnesium-Mg,12

sodium-Na,11neon-Ne,10fluorine-F,9oxygen-O,8

nitrogen-N,7carbon-C,6

boron-B,5beryllium-Be,4

lithium-Li,3helium-2,He

hydrogen-H,1

the line of alpha-nuclei, where proton and neutron numbers are equal

Ni-56 has the largest binding energy of the alpha-nuclei, and is favoured by a small neutron excess

C-12 (6p,6n)

clusters containing…56 60nucleons nucleons

..., 23/04/11,

Williams, pages 58 and 132, and see back to section zZz.

McNeil, 04/23/11,

Chown, Furnace, p. 179, Mason, p. 42/3, and… http://www.tufts.edu/as/wright_center/cosmic_evolution/docs/text/text_stel_6.html (accessed 18 Jan. 2011) Fewell, p.656.

..., 23/04/11,

Screenshot from the 2-D NUCLEUS program. This is a “busy” diagram, but I wanted to show the start and end and the main steps of the alpha process, all on the same diagram. I’ve omitted Ne burning – the diagram is busy enough as it is. The proton axis is a roll-call of the emerging elements. The last stage, silicon melting, produces predominantly Ni-56 and Fe-56, so I have shown this in the small circle in the diagram.


doubly magic. This means they all have larger binding energies than nuclei in between,

and so are the major fuels in the alpha-process.

a roll call of familiar elements

The proton axis in figure 3.63 is a roll call of elements, many of them familiar - the fluoride

in toothpaste, the sodium in salt, the aluminium in a saucepan, the silicon in beach sand

and computer "chips", the chlorine in bleach and swimming pools, the calcium in our teeth

and bones, the chromium in shiny electroplate, the iron in a car body and in our red blood

cells, the cobalt in the magnets in our earphones and in vitamin B12. The foundations of

the substances of our material world and of ourselves are emerging from the stellar

inferno.

an onion-like structure

The star has now developed a structure of concentric layers, like an onion, with a different

nuclear fuel burning in each layer (figure 3.64).

Figure 3.64: The interior of a star of 15 solar masses at the end of silicon burning. The blue scale shows how much of the star’s mass is inside that point – for example, there are about 3 solar masses of matter inside the helium burning shell; the yellow scale gives the approximate distance from the star’s centre.


15

12

9

6

3

0Includedmass(solarmasses)

stellar surface

hydrogen burning to helium-4, 35 million K(also nitrogen)

helium-4 burning to carbon-12, 200 million K(also oxygen, neon)carbon-12 burning to oxygen-16, 800 million K(also neon, sodium, magnesium)oxygen-16 burning through to silicon-28, 2 billion K(also phosphorus, sulphur, chlorine, argon, potassium, calcium)silicon-28 burning through to nickel-56, 3.3 billion K(also titanium, chromium, manganese, iron, cobalt)iron-56 and nickel-56 core, 7 billion K

centre of the star

300million

5,000

1,000

0distancefromcentre(km)

..., 23/04/11,

Diagram based on Marschall, p. 132, and Williams, fig.14.1, and extra elements from Gribbin (p.142), Wallerstein, and from… http://imagine.gsfc.nasa.gov/docs/teachers/elements/imagine/05.html Temperatures from Woosley and Janka, and compatible with Delsemme, table 3.1, Wallerstein, and Kaufmann and Freedman, ch.22. WHW, figure 9, is fairly consistent with this, but shows the basic structure is more complicated than this, due to convective mixing. Diagrams giving included mass are fairly common, but it’s not so easy to get an idea of size. The very approximate size scale from Williams, table 14.10, and Dmitri Pogosian gives the He burning shell as ~10,000 m diameter, lecture 18… http://www.ualberta.ca/~pogosyan/teaching/ASTRO_122/lect18/lecture18.html WHW, p.1054, give the radii of the Si and O burning shells as 1200 and 6400 km for a 15 MS pre-supernova star of solar metallicity. These sets of figures are thus fairly consistent. However, Woosley gives the radii of a 25 solar mass star and of the shells burning He, O and Si as 900, 1, 0.1 and 0.01 RS respectively, where RS is the solar radius, 700,000 km - Woosley’s Review of lectures 15-16… http://ucolick.org/~jjfang/ay12/0308_section.pdf (accessed 2 Mar 2011). These figures make the shells much larger, by a factor of ~10, and would seem to be too big to participate in the explosive fuel burning that comes in the supernova explosion.

McNeil, 23/04/11,

A common simile - Adams and Laughlin, p.55 Marschall, p. 132, Bertulani, fig. 12.15, Williams, fig. 14.1.


This cross-section gives us the star’s nuclear burning history. The main fuels are laid out

in the sequence of their ignition, with subsidiary reactions building up the nuclei that are

not in the alpha sequence. At the base of each layer a burning front is moving outwards,

igniting the nuclear ash produced by the layer above. The silicon burning reaction has

accumulated around 1-2 solar masses of nickel and iron at the star’s core, and continues

to add to this. The nuclear fuels are burning in a dense small core about the size of the

earth, at the centre of a huge, bloated star, that may be as big as the orbit of Jupiter.

3.12.11 running out of nuclear fuel

climbing the binding energy curve

The star has in its lifetime gone six times around the gravity-matter circuit, and burned the

major nuclear fuels: hydrogen, helium, carbon, neon, oxygen, and finally silicon. Each

burning rearranges the nucleons into bigger nuclear configurations, which are more tightly

bound. The binding energy released in each stage of burning has sustained the star, and

temporarily halted its gravitational collapse. We can follow the sequence on the nuclear

binding energy curve – figure 3.65.

Figure 3.65: The sequence of fusion reactions generating heat in a star of 15 solar masses.As long as the fusion reaction moves the arrangement of nucleons further up the binding

energy curve, mass will be lost, energy will be released, and the star’s collapse is


0

1

2

3

4

5

6

7

8

9

0 10 20 30 40 50 60nucleons

Bin

ding

Ene

rgy/

nucl

eon

(MeV

)

burning hydrogen (4p) helium (He-4) + 27 MeV11 million years at 35 million degrees (MK)Energy: ~7 MeV/nucleon

burning helium carbon2 million years at 180 MKEnergy: ~0.6 MeV/nucleon

burning carbon neon2,000 years at 700 MKEnergy: ~0.2 MeV/nucleon

burning oxgen silicon30 months at 2,000 MKEnergy: ~0.3 MeV/nucleon

burning silicon nickel/iron18 days at 3,300 MKEnergy: ~0.2 MeV/nucleon

..., 23/04/11,

The binding energy curve is a smoothed plot for selected nuclei in the mass.mas03 database. Rather than try to get some kind of “average” set of figures, I’ve taken times and temperatures taken from one source - WHW for a star of 15 solar masses. These figures are from a recent authoritative source They fom a self-consistent set, that also agrees fairly well with other sources: Kaufman and Freedman, ch 22, and Krauss, ch.8 for one of 30 solar masses, and Lodders… solarsystem.wustl.edu/our%20reprints/2008/2008Lodders-SLAS-elementalabundances2.ppt, http://en.wikipedia.org/wiki/Silicon_burning_process …also… http://imagine.gsfc.nasa.gov/docs/teachers/elements/imagine/05.html

..., 23/04/11,

Williams, p.349 and Bertulani, p.373.


deferred for a little longer. Hydrogen burning, the first reaction, that binds the free protons,

yields a huge amount of energy, ~27 MeV for binding 4 nucleons, about 7 MeV per

nucleon. After helium, the curve is much less steep, and the burning reactions have to

rearrange more and more nucleons to get even a small energy yield.

Moreover, the star is losing energy at an ever faster rate, through neutrino emission, and

so each successive nuclear fuel lasts a fraction of the time of the previous one. The

figures in the graph show that from hydrogen to oxygen, each fuel lasts only about one

tenth as long as the preceding fuel.

In the star's onion-like structure we can see the retreat of matter under the inexorable

pressure of the star’s gravity. Successive reactions throw new fuels on the nuclear fire,

but these yield less and less energy. Finally, the nuclear rearrangements of the alpha

process produce nuclei in the iron group, with around 56-60 nucleons, and having the

largest binding energies of all the nuclides. Enlarging these will take energy in rather than

give it out. There is no viable way that a nuclide like iron-56 can be made to yield further

binding energy; the star’s core has finally exhausted its nuclear fuel.

the “death” of a star

The gas cloud that started maybe 10 million years before as the merest breath of slightly

cooler hydrogen and helium in almost empty space, has become a turmoil of heavy nuclei

and high energy photons of radiation. The nuclear reactions that for so long have powered

this star and kept its enormous gravity at bay, have now run their course. Millions of years

have homed in on a single day, and on the last seconds at the end of the last day.

Lawrence Krauss captures this moment well for a star originally of 30 solar masses: “For

10 million years all of the nuclear reactions holding the star up against gravitational

collapse have been leading to this single last gasp. Almost 10 million years of hydrogen

burning, followed by 1 million years of helium burning, 100,000 years of carbon, 10,000

years of oxygen, and then a single day for the rest of the trip. Once it is over there is no

hope. In fact, the dense inner core of the star, now surrounded like an onion by shells of

oxygen, carbon, helium and hydrogen, is about to undergo one of the most traumatic

events in all the visible universe”.

a technical note - why not nickel-62?

I don’t feel I‘ve understood why the main products of silicon burning are Ni-56 (~60 th in the

binding energy rankings) and Fe-56 (3rd highest binding energy). If binding energy is so


..., 23/04/11,

This is just for this first draft. This has been taken as far as it can for now; it’s time to move on – Jan 16 2011.

McNeil, 23/04/11,

Krauss, Atom, p. 139.

..., 23/04/11,

Krauss points out this pattern for a star that starts out as a gas cloud of 30 solar masses –p.138. The times in the graph are from Woosley and Janka,


important in the competitive melée of the super-hot core, then why not Ni-62, Fe-58, and

then Fe-56, which are ranked one, two and three?

Fewell suggests that the Ni-62 nuclide is preferentially photo-dissociated. And certainly,

resistance to break-up by photons would be as important as binding energy, in deciding

which nuclides are produced by silicon burning.

Wallerstein and WHW suggest that the proton/neutron ratio plays a rôle in deciding which

nuclides are made. WHW describe how the main Si-burning products are those nuclides

that are most tightly bound at the value of neutron excess pertaining in the core. Both

sources say that a small neutron excess in the core leads to the preferential formation of

Ni-56, with its proton/neutron ratio (Z/N) of 1.0. If the neutron excess increases, then the

preferred products will be the neutron-rich isotopes of iron, Fe-54, Fe-56 or Fe-58. WHW

state that for still greater neutron/proton ratios “the equilibrium shifts to heavier isotopes”,

and note that the most tightly bound nucleus is Ni-62.

Wallerstein (fig. 24) relates binding energies to the proton/neutron ratio: “Fe-56 could be

made as itself in equilibrium if the ratio of neutrons to protons in the nucleosynthetic

environment were around 0.87. … In fact, nature seems to have chosen to assemble

most of the solar system’s iron-group nuclei in matter that had equal numbers of neutrons

and protons. In this case Ni-56 was made and later decayed to Fe-56.” (fig. 24 caption)

The figure below plots binding energies against cluster size for three sets of nuclides, with

different proton/neutron (Z/N) ratios.

The bottom graph shows that Ni-56 is the most strongly bound of the alpha-nuclei, that

have equal number of protons and neutrons (Z=N, so Z/N=1.0). The second (blue) plot

shows how a slight neutron enrichment has increased binding energies, and that Fe-56 is

the most tightly bound nuclide of this group. Further neutron enrichment (the green plot)

has not really increased binding energies, and now it’s Fe-58 and Ni-62 that are the

tightest bound.


..., 23/04/11,

I don’t understand what this means - how or why nature should “choose” the way things are.

..., 23/04/11,

Clayton says that the almost equal numbers of protons and neutrons means that Fe-56 is synthesised as the radioactive isotope Ni-56 – p. 237.

..., 23/04/11,

Wallerstein, p.1034 and 1060, fig.24. WHW, p.1035.


Perhaps the more neutron-rich nuclides, even with their large binding energies, simply

cannot be made unless the stellar environment is sufficiently neutron-rich to start with. So

silicon burning in a core with very few excess neutrons produces mainly Ni-56, with equal

numbers of protons and neutrons. Increasing the neutron excess enables the production

of Fe-56 and the even more neutron-rich Fe-58. Ni-62 is then competing with these two

nuclides that have almost the same binding energies, but are smaller, and hence easier to

form by aggregation.

So, maybe this is why Ni-62, the most tightly bound nuclide of all, is not produced in

quantity in the star’s core: it is more susceptible to photo-dissociation, there needs to be a

bigger neutron excess in the core, and if there are the excess neutrons available to make

it, then it is in competition with two other nuclides, Fe-56 and Fe-58, that are equally

tightly bound but significantly smaller.

3.12.12 Supernova

A star’s continued existence as a self-supporting entity depends on a continuing nuclear

burning reaction, and the ability of the matter in its core to withstand the conditions there.

A star that is massive enough to make it to the silicon burning stage imposes

temperatures and pressures that atomic matter, comprising nucleons and electrons, can


8500

8550

8600

8650

8700

8750

8800

52 54 56 58 60 62 64 66

nucleons

BE p

er n

ucle

on (k

eV)

Z/N = 1.0 Z/N = 0.85 to 0.90 Z/N = 0.80 to 0.85

Fe-56(26p,30n)Z/N = 0.87

Ni-62(28p,34n)Z/N = 0.82

Ni-56(28p,28n)Z/N = 1.0

Fe-58(26p,32n)Z/N = 0.81

McNeil, 23/04/11,

Type Ia supernovae are thought to originate in white dwarf stars in binary systems, which gather material from their paired star, exceed the Chandrasekhar mass, and collapse. Type I supernovae produce no neutrinos, and result in total destruction of the white dwarf star, and the production of elements up to, but not beyond, iron – Marschall, p.139, Gribbin, p.141. A typical type Ia supernova produces about 0.6 solar masses of iron (Croswell, p.171, Gribbin, p.147), and these have have produced most of the iron in the universe – Silk, p. 227. Type II supernovae occur when the core of a massive fuel-burning star collapses – Woosley and Janka, and Marschall, ch.6. Type II supernovae produce little iron, but most of the universe’s other heavy elements – Silk, p.227. I focus just on the type II supernova, because it shows all the processes of nucleosynthesis. There are many excellent descriptions of the processes in a type II supernova by authoritative writers - Adams and Laughlin, p.56; Bertulani, section 12.12; Chown, p.205; Delsemme, p.51; Gribbin, p.141; Krauss, p.144; Marschall, p. 131; Mason, p. 60; Silk, p.136; Williams, section 14.9; Woosley and Janka, for example. Most describe the core collapse being triggered by the exhaustion of the last fuel, Silicon, though a few say that collapse starts with the core exceeding its Chandrasekhar mass, and can no longer be supported by degenerate electrons. All say that photo-dissociation of Fe-56 and neutronisation occur. The processes inside a supernova are complicated (see for example the review paper by Woosley et al), and the sequence depends on the balance of conditions within the star. Here I attempt to provide a simple representative account of nucleosynthesis when a massive star ends as a type II supernova. I’ve used mainly the recent authoritative reviews of Woosley, and academic texts by Bertulani and Williams, and tried for consistency with other writers. There's an account of stellar evolution, with links and images at… http://chandra.harvard.edu/edu/formal/stellar_ev/story/index.html (accessed 26 Feb 2011) Nicholas Short gives a helpful review, with lots of images at… http://rst.gsfc.nasa.gov/Sect20/A6.html (accessed 10 Feb 2011).


only just withstand. If the core gives way, then the entire stellar edifice collapses. The core

and the outer layers of the star now go very different ways. We'll first tell the core's tale.

The Core's Tale

The life cycle of the core of a massive star “can be thought of as just one long contraction,

beginning with the star’s birth, burning hydrogen on the main sequence, and ending with

the formation of a neutron star or black hole. Along the way, the contraction ‘pauses’,

sometimes for millions of years, as nuclear fusion provides the energy necessary to

replenish what the star is losing to radiation and neutrinos.” Eventually a nickel-iron core

of about 1.5 solar masses builds up, in effect, a white dwarf inside the massive star . The

temperature is so high that even the tightly bound iron-group nuclei are on the edge of

being fragmented by the high energy photons, and the pressure is so high that the

degenerate electrons can only just support it.

The iron-group nuclides are at the top of the binding energy curve, and so no further

energy can be released by nuclear fusion reactions. The star has exhausted its nuclear

fuel at the core, yet continues to lose energy at an enormous rate through the emission of

neutrinos. Consequently, the core contracts, and its temperature and pressure increase

even further, and the matter in the core finally gives way. Two processes then occur that

remove energy from the core, speeding its collapse.

the unravelling of iron-56

The core temperature rises as high as 7 billion degrees, and the thermal photons now

break up the iron-56 nuclei back to alpha particles from which they were formed…

Fe-56 (26p,30n) 13 He-4 (2p,2n) + 4 n mass gain 124 MeV

This rearranges the nucleons into less tightly bound nuclei, with a mass gain, which is

‘paid for’ by taking energy from the star’s core, so it collapses faster.

electron capture and ‘neutronisation’

As the increasing core pressure pushes the density past 10,000 tonnes/cm3, the electrons

are “squeezed into iron-group nuclei”, initiating the ‘neutronisation’ reaction, in which an

electron and a proton combine to make a neutron, and emit a neutrino.…

e- + p+ n0 + - ~0.8 MeV


..., 23/04/11,

Williams, p. 361.

..., 23/04/11,

Woosley and Janka.

..., 23/04/11,

Bertulani, p.372. The mass gain calculated from table mass.mas03 matches Bertulani’s figure.

..., 23/04/11,

Woosley and Janka. Krauss puts it at 5 billion degrees - p.147. Gribbin has the iron nuclei breaking up when they fall into each other - "Stardust", p. 143.

McNeil, 23/04/11,

Marschall, p.131

..., 23/04/11,

Woosley and Janka.

..., 23/04/11,

I don’t think that a sequential account is possible in a general description of a supernova. In the last stage of a supernova’s life, matter is in extremis, and the core is sustained by three things (1) continued silicon burning, (2) the iron-group nuclei resisting photo-dissociation, and (3) the degenerate electrons supporting the core’s weight. As with a three-legged stool, if any one of these gives way, then the whole collapses. WHW state that “so long as there is an active burning shell within the core, it will not collapse; contraction leads to accelerated nuclear burning and expansion. However, the iron core does not grow by radiative diffusion, but by a series of convective shell-burning episodes, the last of which overshoots the (generalised) Chandrasekhar mass.” – p.1046. I infer that the core may well exceed the Chandrasekhar mass, but can’t collapse as long as there is an ongoing nuclear burning reaction. Collapse then occurs as soon as the reaction ceases. So, things come together, there is a confluence of events that make the star’s continued existence impossible.

..., 23/04/11,

Following Williams’s approach, p.362.


Electrons and protons effectively disappear, leaving only neutrons. This removes all the

electrons that have so far supported the star's weight, and also consumes energy to

create a burst of neutrinos, which leave the core.

doomed either way

Both these processes are inevitable, for failure in one respect triggers failure in the other.

These two events, the unravelling of iron-56 nuclei and the neutronisation reaction,

remove a huge amount of energy from the star, maybe as much energy as it has radiated

in its entire lifetime. The star’s core collapses so fast that it falls inward at about one

quarter of the speed of light, shrinking from about the size of the Earth (~12,000 km

diameter) to a sphere about 60 km in diameter, in around 1 second.

a neutron star

What happens to the core now depends on its mass. If it is more than ~3 solar masses it

collapses completely to a black hole, the "ultimate triumph of gravity over matter”. But for

cores less than ~3 solar masses, the outcome is a sphere of neutrons, a neutron star;

with a density of around 200 million tonnes/cm3 – “a mass in excess of a million Earths

confined to a region the size of a small city, and with a mass of Manhattan contained in

each cubic centimetre of material”.

Neutron stars are “at the limit of density that matter can have, the subsequent step being

a black hole”. This state of matter would seem beyond comprehension, yet a neutron star

resembles a giant atomic nucleus, and it has about the same density as the tiny nuclear

nuggets that are at the heart of every atom, from hydrogen to lead. But, whereas an

atomic nucleus, with its specific balance of protons and neutrons, has an identity, a

neutron star, has none. During the star’s lifetime, the protons and neutrons in its core

were continually being rearranged in ever-bigger nuclei as star burned its series of fuels.

Now that rich diversity of nuclei has been unravelled by the temperature, and crushed to

neutrons by the pressure. “Matter this compact loses its atomic identity. Electrons blend

with protons; intervening space effectively disappears.” A neutron star represents matter

in its most lumpen condition, faceless and nameless, imprisoned by its own gravity.

As far as the stellar core is concerned the star's patient "journey-work", building large

nuclei over millions of years, has been undone in an instant. It is “as if the real business of

the star, the conjuring of nuclei, was now monumentalised as a giant nuclear tombstone”.

But the core is only a fraction of the entire star. Its collapse has produced a huge amount


..., 23/04/11,

I’m omitting consideration of the manner of the core’s collapse, and the possible shock wave that ensues. This is still being discussed – see Bertulani, p.374 and Woosley and Janka.

..., 23/04/11,

Timothy Ferris recounts the evolution of stars and their role in creating nuclear matter in ch.14 of "Coming of Age in the Milky Way".

McNeil, 23/04/11,

Walt Whitman's often quoted line… "I believe a leaf of grass is no less than the journey-work of the stars". Walt Whitman , Song of Myself, Leaves of Grass (1881-82) http://www.whitmanarchive.org/published/LG/1881/poems/27 Quoted, for example, in Clayton, p.6, and Marschall, p.197.

..., 23/04/11,

I’m thinking of the neutron star as a whole, and in terms of nuclear identity. "A neutron star is not just a large nucleus of neutrons: it will contain electrons and protons, and the surface layers will probably contain nuclei” - Williams, p.364, and also http://en.wikipedia.org/wiki/Neutron_star Hydrogen is defined by the single proton that is at the heart of each atom, but there is no atom with a nucleus only of neutrons. It is ironic that an isolated neutron is unstable, and decays to a proton in a matter of minutes. Neutrons are only stable in vast numbers in a neutron star, stabilised by their enormous gravity.

..., 23/04/11,

Marschall, p.187.

..., 23/04/11,

The origins of many nuclides are summarised in WHW, table III, and in detail by Clayton.

..., 23/04/11,

Bertulani, p.369, who gives their density as 100 million tonnes/cm3.

..., 23/04/11,

Krauss, p.147.

McNeil, 23/04/11,

It should perhaps be called a proto-neutron star, and not wholly neutrons, but “a gigantic neutron-rich nucleus” – Woosley and Janka. Neutron stars Kaufman and Freedman, ch. 23, Hey and Walters, p.221, Silk, p.135, Marschall, p.186. http://hyperphysics.phy-astr.gsu.edu/hbase/astro/pulsar.html#c1 Delsemme, p.53. As the star collapses, it spins faster and faster, in the same way ice skaters spin faster when they pull in their arms, to bring more of their mass near to their axis of rotation. Pulsars (PULsating stARs) are neutron stars that spin many times each second. The pulsar at the heart of the crab nebula spins 30 times a second. In comparison, out own sun spins once in about 27 days. For pulsars… http://hyperphysics.phy-astr.gsu.edu/hbase/astro/pulsar.html#c4 http://science.nasa.gov/NEWHOME/help/tutorials/pulsar.htm

McNeil, 23/04/11,

Gribbin, "Stardust", p.144.

..., 23/04/11,

Silk, p.145, and Gribbin, p. 144.

..., 23/04/11,

Woosley and Janka, text and table I, and Krauss, p.144.

McNeil, 23/04/11,

Gribbin, p.142


of energy, creating enormous temperatures and numbers of neutrinos and neutrons, and

these now erupt into the star’s outer layers, and the creation of nuclei continues there.

The outer layers’ tale

It is not easy to convey the scale of a supernova explosion, for it is not just the sheer

amount of energy, but also the speed of its release.

a gravity-powered neutrino explosion

A nickel/iron core of ~1.4 solar masses, collapsing to a neutron star, converts about 10%

of its mass to energy, releasing about 1046 J in a few seconds. The supernova’s light

output can increase by a factor of 100 billion, so that for a few days it can outshine the

100 billion or so stars in its galaxy.

But this visible light is only 1% or so of the supernova’s total energy output. The other

99% of the energy is in the form of neutrinos, one for each proton in the star’s core, about

1058 in total, streaming out from the core at almost the speed of light.

We have already encountered neutrinos, and seen that they “are the closest thing to

nothing that one can imagine. …they have … virtually no effect whatsoever on ordinary

matter… under ordinary conditions, a neutrino could penetrate millions of miles of lead as

if it were window glass.” About 500 billion neutrinos from the sun hit every square inch of

ground every second, and pass through the Earth with no resistance. "Our bodies are

pierced by them unceasingly, day and night. They leave not a trace."

But now deluge of neutrinos produced by the supernova is enormous, about 1058 in a few

seconds, and the inner layers of the star are so compressed that the densely packed

nuclei deflect the streams of neutrinos. As far as the neutrinos are concerned, “the star

offers ony a slight resistance – just a bit more than if nothing was there at all”. They

expend only about 1% of their total energy in penetrating the star’s outer layers, but even

this amounts to about 1044 J, more energy than the Sun emits in its entire lifetime. This

rips protons and neutrons from the star’s neutronising core, and creates a shock wave

that blows away the star’s outer layers in the supernova explosion.

The numbers on their own perhaps convey little, but pause to consider that only 1% of the

total energy released in the supernova explosion is enough to blow away the star’s outer

layers, at least several solar masses, at speeds of thousands of kilometres per second. A

supernova has been summarised as a “gravity-powered neutrino explosion”. In fact,

without neutrinos, the core mass would just keep increasing from layers of star falling on it


..., 23/04/11,

Woosley and Janka.

..., 23/04/11,

Up to 40,000 km/s in the case of SN1987A – Marschall, p.245.

..., 23/04/11,

Krauss, p.147, Bertulani, p.342, Gribbin, p.145.

..., 23/04/11,

I’m ignoring the ‘thermal’ neutrinos produced by the enormous temperature in the collapsing core – Marschall, p.134, Bertulani, p.374 and Gribbin, p.145. It’s now accepted that the cause of the supernova explosion is not core collapse and bounce, but the emission of a huge flux of neutrinos – Woosley and Janka., WHW, p.1047.

..., 23/04/11,

Marschall, p.136. And note that 1044 is 1/100th of 1046.

..., 23/04/11,

Marschall, p.136.

..., 23/04/11,

Some neutrinos may only travel a few centimetres before they are deflected by a nucleus – Marschall, p.136 – a far cry from millions of miles of lead.

McNeil, 23/04/11,

Marschall, p.249.

..., 23/04/11,

Marschall, p.132 and 133.

..., 23/04/11,

This draws on the detailed observations of supernova SN1987A, which exploded February 23, described by Marschall, ch. 11, Chown, "The Magic Furnace", p.204, Gribbin, p147, Woosley and Weaver, and in a useful article in Wikipedia, which has a lot of links to pictures, animations and ongoing observations… http://en.wikipedia.org/wiki/SN_1987A (accessed 23 March 2011) Woosley and Weaver, and also Marschall tell the story of the scientific community's collaborative work to observe and understand this unique event. One of the remarkable features of the story is how the sudden brightening of one star was observed and reported by several people all round the world, within hours of its happening.

..., 23/04/11,

Croswell, p.178 and Marschall, p.249 say 1058, and Gribbin, p.143, gives 1057.

..., 23/04/11,

Williams, p.362 and p.348.

..., 23/04/11,

Woosley and Janka, Williams, p.362, Bertulani, p.374. Williams says 3-10s, Woosley and Janka say ~10 seconds (fig.5), as does Clayton – eg. the section on F-19, p.103. Williams says the neutronisation reaction produces its huge numbers of neutrinos in about 1 second, which then take 3-10 seconds to burst through the star’s outer layers – p.362-3.


until it collapsed as a black hole, and no supernovae would occur. It’s a nice irony that

gravity is by far the weakest of all the physical interactions, and neutrinos are by far the

least substantial of all matter particles.

two sets of reactions

The nickel/iron core with a radius of ~1,000 km, has collapsed to a neutron star with a

radius of ~10 km or so, and a temperature of ~100 billion degrees. The photo-dissociation

of iron has produced free neutrons, and the neutronisation reaction has released a flood

of neutrinos, which rip protons and neutrons free from the neutron star’s surface. These

neutrinos drive a hot neutron-rich ‘wind’ lasting about 10 seconds, through the inner layers

of the star, inducing two main types of reactions there. The reactions in the innermost

layers rapidly assemble nuclei from free protons and neutrons, while a little further out, the

nuclear fuels that have remain unconsumed undergo explosive nuclear burning. Figure

3.66 shows the main nuclear reactions in these layers.


..., 23/04/11,

This is based on Woosley and Janka, fig.5, with additional data from WHW. If I’ve understood things correctly, the protons and neutrons in the inner layers of the star are assembled into iron-group nuclides within a few seconds. This does not seem to square with the common statement that the lives of stars are so long because their nuclear reactions, especially hydrogen-burning, are so slow. It does seem that the fusion reactions can be very quick indeed. Perhaps the hydrogen burning phase of a main sequence star is so long because the star is not losing energy through neutrino emission, and the fusion reaction yields so much energy. If you have only £100 pound notes in your wallet, then you only need to spend them very slowly.

..., 23/04/11,

The reactions of the neutrinos and anti-neutrinos with nucleons produce an excess of neutrons – Woosley and Janka, WHW, p.1057.

..., 23/04/11,

Gribbin, p.145.

..., 23/04/11,

Strictly speaking, this should be called a proto-neutron star (PNS) – Woosley and Janka. This raises the question: what takes the place of the collapsed core? JohnGribbin (p.145) writes that the core collapses in a few seconds, leaving the proximate inner layers “essentially hanging unsupported above the void … like … a cartoon character who runs off a cliff and hangs motionless in the air until he notices what has happened”. The protons and neutrons are ripped from the neutron star surface by the neutrino wind, and erupt into this empty space. Stars of different generations, with different starting compositions, produce the same mix of r-process elements. This suggests that these r-process elements form from the neutron star, which has the same composition, rather than from the star's inner layers - Woosley and Janka.

McNeil, 23/04/11,

Woosley and Janka.


Figure 3.66: The nuclear reactions in the outer layers of a star in a supernova explosion; bK stands for billion degrees.rapid nucleus assembly – the r-process

The neutrino wind leaves the neutron star with a temperature of ~50 billion degrees, then

it cools, and at ~10 billion degrees, and about half a second after the start of the neutrino

wind, the free protons and neutrons it carries have started assembling into alpha particles.

Very shortly after this, as the temperature drops further, these alpha particles themselves

combine into larger nuclei, especially those in the iron-group, that have the greatest

binding energies.

The wind now carries iron-group nuclei, alpha particles and neutrons. In the next couple of

seconds, as the temperature falls below 1 billion degrees, the iron-group nuclei rapidly

capture neutrons, to build up the r-process nuclei. There can be as many as 100 neutrons

for each iron-group nucleus, and the neutron flux is so great that “the time to capture a


2

5

10

100

100,000

13,00010,000

5,0003,700

1,000

100

10

1

Temperature, in billions of degrees

Distance from centre, km

1. A ‘wind’ of neutrinos blows for ~10 s, carrying free protons and neutrons out into the star.

5. …Fe-group nuclei capture n r-process nuclides (T~0.1bK, t~2 s).

4. …alphas combine Fe-group nuclei + free n (T~6bK, t~0.55 s)…

3. …p + n combine alpha particles + free neutrons (T~10bK, t~0.5 s)…

2. Neutrinos rip free protons and neutrons from the neutron star (t=0)…

neutron star

~1.4 ~3 included mass (solar masses)

Explosive burning, in a few seconds, of super-heated Si, O and C.For T>~5 billion degrees, complete burning to iron-group nuclides.For T between ~2 and 5 billion degrees, incomplete burning to a diverse mix of nuclides.

the turbulent neutrino wind stirs up the material in the star

..., 23/04/11,

WHW, p.1057.

..., 23/04/11,

Clayton, p.303.

..., 23/04/11,

Any alpha particles that remain uncombined by the time the temperature drops below ~3.5 billion degrees, are ‘frozen out’ of the reaction sequence – Clayton, p.273. Animations of the alpha-rich freeze-out at… http://nucleo.ces.clemson.edu/home/movies/alpha_rich/ (accessed 5 March 2011).

..., 23/04/11,

This is the basic e-process in action, at the enormous temperatures close to the neutron star. The relative abundances of the iron-group nuclei are decided by the quasiequilibrium that applies at this high temperature – see Clayton, Glossary and section on Fe-56, Wallerstein sections II and XV,

..., 23/04/11,

These are thought to come from the newly-formed neutron star, rather than from the inner layers of the original massive star – Woosley and Janka.

..., 23/04/11,

T>~4 MeV, that is a temperature of E/k = 4 x 106 /8.6 x10-5 = ~50 x 109 K – WHW, p. 1057.

..., 23/04/11,

This largely from Woosley and Janka, WHW (section VIII B), and Clayton, p.273 and 303. This basic account is generally accepted, though the details are not worked out – Woosley and Janka. The r-process is thought to occur most commonly in the collapsing core of a type II supernova – Clayton, p.306. The Nuclear Astrophysics group at Clemson University provide animations of some nucleosynthesis processes… http://nucleo.ces.clemson.edu/home/movies/ (accessed 4 March 2011) Temperatures and times are from their animation of the r-process, which starts with material at a radius of 5.6 km and a temperature of 40 billion degrees.


neutron is generally less than one-thousandth of a second”. This drives the nuclei high up

the slope on the neutron-rich side of the nuclear valley, until they can hold no more

neutrons. When the neutron flux ends, these unstable nuclei undergo a series of beta-

minus decays, maybe as many as twenty, taking them down to the nuclear valley floor

(see back to the section on the r-process). It is this r-process that produces, in a couple of

seconds, an array of nuclides heavier than zinc, containing from about 70-250 nucleons.

The amount produced is small, only about one millionth of a solar mass, but this is about

one third of the mass of the Earth, and includes some rare nuclides.

explosive burning

All nuclear fuels will burn very fast if they are heated well above their normal temperature

of burning in a stellar core. A fuel that will burn steadily for millions of years, will undergo

explosive burning in seconds if super-heated. This explosive burning can occur at

temperatures above 2 billion degrees – about 13,000 km out from the neutron star - and

will go to completion if the temperature exceeds 5 billion degrees – about 4,000 km out

(see figure xXx). Much of the shells that have been steadily burning carbon, oxygen and

silicon lie within 13,000 km, and so these fuels will undergo explosive burning - partial in

some regions, complete in others. Any nuclei within ~4,000 km of the neutron star will in a

few seconds be processed all the way to iron-group nuclei.

Explosive burning yields the same nuclides as stable core burning, but the presence of

free alpha particles and protons and neutrons results in a rich diversity of nuclides, from

~20 up to ~90 nucleons, well beyond the iron-group, and rich in protons and neutrons.

a few nuclides are formed by fragmentation

A tiny fraction of the torrent of neutrinos knock single protons or neutrons out of nuclei on

their way through the star’s outer layers. In this way boron-11 is created from carbon-12…

+ C-12 (6p,6n) B-11 (5p,6n) + p

and similarly, neon-20 (10p,10n) loses a proton to make fluorine-19 (9p,10n).

High energy photons can eject neutrons, to produce proton-rich isotopes of many

elements, such as mercury, tungsten, barium and xenon.

spallation in deep space - the last nucleosynthesis mechanism

We now look at the last stellar mechanism for creating nuclides, and this occurs, not

inside stars, but outside them, and operates, not by fusion, but by fragmentation. The


..., 23/04/11,

To name the more common. Generally, these are known as p-process elements, proton-rich nuclides that can be produced by protn capture as well as by neutron loss - WHW, p.1055, and Wallerstein, table VI and p.1056.

..., 23/04/11,

Croswell, p.179.

..., 23/04/11,

WHW, p.1056, and Clayton, sections on B-11 and F-19, and some of the universe’s N-15 is produced by neutrinos.

..., 23/04/11,

WHW, section VIII B, detail the main products of explosive burning of carbon, oxygen and silicon, under various conditions, and the situation is complicated. So I think it’s enough to state that a diversity of reaction products forms. Delsemme states that free neutrons are captured by carbon, oxygen and silicon to create a mix of heavy nuclei – p.52.

..., 23/04/11,

The basic e-process, producing the iron-group nuclides.

..., 23/04/11,

Figures from WHW, section VIII, and Clayton, Glossary on C, O, and Si burning, and the r-process. WHW put the O- and Si-burning shells are between ~1200 and 6400 km out, in a pre-supernova 15 solar mass star, of solar metallicity – p.1054. They give this temperature profile (in billions of degrees, bK)for a 25 solar mass star: 5 bK,~3700 km; 4 bK, 5,000 km; 3 bK, 7,000 km; and 2 bK, 13,000 km. These values are in the diagram.

..., 23/04/11,

Woosley and Janka, Clemson University Astrophysics group: http://nucleo.ces.clemson.edu/home/movies/r_process/ (accessed 10 March 2011). and personal communication from Stan Woosley, 16 April 2009. Also Krauss, p.148, Mason, p.60.

..., 23/04/11,

Clayton, p.303.


supernova explosion blows the nuclei in the star’s outer layers into space at enormous

speeds, up to near the speed of light. These high speed nuclei, known as cosmic rays ,

are predominantly single hydrogen protons and helium nuclei, and travel through ‘empty

space’ for very long times, typically 10 million years.

However, the space between the stars, the interstellar medium, is by no means empty,

but contains a scattering of particles of different kinds, typically around a few tens to a few

hundreds in a cubic centimetre. This is better than any vacuum that can be made on

Earth; compare it with ‘thin air’, which has nearly 30 billion billion gas particles in a cubic

centimetre. The particles are primarily atoms of hydrogen and helium, from the brief burst

of nucleus formation after the big bang. There is also a small proportion of larger nuclei,

produced by earlier stars and supernovae. Finally, there are a large number of molecules

of compounds, such as water, formaldehyde and ethanol, in great clouds between the

stars.

The high speed cosmic ray particles, on their long journey through space, collide with

larger nuclei, and break them into smaller nuclear fragments; this is the spallation

process. The fragmentation of carbon and oxygen nuclei is the source of small nuclides,

such as Li-6 (3p,3n), Be-9 (4p,5n) and B-10 (5p,5n). None of these nuclides can be

formed in stars; they are all destroyed before hydrogen burning starts.

Thus the massive star’s last act, as it explodes as a supernova, is the creation of rare light

nuclides by spallation collisions, that will occur far away and long after the star has

ceased to exist.

Supernova SN1987A

The first outward sign of a supernova is the burst of neutrinos, followed by the blast wave,

and the rapidly expanding shell of gas - within a day this is an incandescent ball a billion

miles across.

The neutrinos from the explosion of supernova SN1987A, a star of ~15 solar masses in

the large Magellanic cloud about 160,000 light years away, were detected on Earth at

07.35 GMT on 23 February 1987, followed some hours later by the visible light. This was

the first supernova observed in that year, hence its designation, and also the first to be

visible to the naked eye for nearly 400 years - the previous one being in 1667.

Two neutrino detectors picked up a total of 19 neutrinos passing through Earth in about

12 seconds. The probabilities of detecting these extremely elusive particles is minute, so


..., 23/04/11,

One in Japan (Kamiokande II), and the other in America (IMB). These had been set up to look for proton decays, and determine if it truly is a stable particle. Williams (sections 13.7 and 14.10) and Marschall (p.249) describe how they function, and their observations of SN1987A. Gribbin (p.148) says 22 neutrinos, Silk (p.137) says 11, Chown (p. 206) says 19. Williams (p.366) states that Kamiokande detected 11 neutrinos, and IMB another 8. The Wikipedia article agrees and says another detector, Baksan, picked up another 5. Marschall has used the figures from the first two detectors, which are water-based, and left out data from the third detector, which uses a different method. http://en.wikipedia.org/wiki/SN_1987A (accessed 23 March 2011) The different writers give a diversity of figures, and I’ve tried to compare these, to check consistency, and also get my own “feel” for this cataclysmic event. Figures in this paragraph from Marschall, p.251-6. I’ve taken a trillion as 1012. Marschall states that ~1 in a thousand trillion neutrinos is detected ( ~1 in 1015), and that the 19 neutrino captures indicate a hundred thousand trillion (1017) passed through the detectors – but the capture ratio suggests a total of only 20 thousand trillion – there seems to be an inconsistency. Also Marschall states that the solar neutrino flux is ~80 billion/cm2/second, and that normal neutrino events were rare, well under 1 per day. Williams states that the neutrino flux from SN1987A was ~5 billion/cm2, while Gribbin puts it at 100 billion/cm2. If we consider 1058 neutrinos emitted at the centre of a sphere of radius 160,000 light years (~1.5 x 1021 metres), with an area of ~2.9 x 1043 m2, then there will be ~35 billion neutrinos passing through each square cm of the sphere. There seems to be some broad consistency, though I don't see how a simple geometrical calculation gives a value one third that of Gribbins, and how Williams's figure is one fifth of Gribbins's, and also much less than the solar flux, though I may be missing something. Marschall and Williams agree on the amount of energy released by SN1987A - ~1046 J (Marschall gives 1053 ergs, where 107 ergs = 1 J). Williams gives the source temperature as 3-5 MeV, that is ~40-60 billion degrees, while Marschall puts it at "over 10 billion degrees". Gribbin gives 1058 neutrinos - p.148. The neutronisation process takes about 1 second (Williams, p.362), so why does the neutrino burst on Earth last around 10 seconds? Marschall says that because the neutrinos "take some time dodging atoms on their way out of the exploding star, it takes about 10 seconds for them all to get away" - p.255. The 12 seconds is for the Kamiokande events, which span ~12.5 seconds, and the IMB a bit less than 6. What is significant is the powerful confirmation of the models of neutron star formation, for the 12 second span is very close to the calculated time for core collapse - Gribbin, p. 149.

..., 23/04/11,

Williams, p.364.

..., 23/04/11,

The first neutrinos were detected at Kamiokande at 7.35.35 am, and at IMB 6 seconds later. However, their clocks were not synchronised, and it’s assumed that the neutrinos arrived at both sites within a fraction of a second. Detectors now take their times from GPS satellites. See… http://nu.phys.laurentian.ca/~fleurot/supernova/ which also gives the actual neutrino detection record. Woosley and Weaver give 7.36 am.

..., 23/04/11,

Gribbin, p. 147. Delsemme puts it at 180.000 light years, Silk, p.137, says 50,000 light years. The Chandra X-ray telescope web-site gives 160,000ly… http://chandra.harvard.edu/photo/2007/sn87a/ (accessed 23 March 2011).

..., 23/04/11,

Gribbin, p.149, and Marschall, p. 258.

..., 23/04/11,

Marschall, ch.11, Williams, section 14.10, Woosley and Weaver, Gribbin, p.147, Marcus Chown, p.207, Nicholas Short… http://rst.gsfc.nasa.gov/Sect20/A6.html

..., 23/04/11,

Strictly, anti-neutrinos – Williams, p.365, Woosley and Weaver, and… http://en.wikipedia.org/wiki/SN_1987A (accessed 23 March 2011)

McNeil, 23/04/11,

Marschall. p.137.

..., 23/04/11,

Clayton, sections on C-12 and O-16, and sections on lithium, beryllium and boron.

..., 23/04/11,

Clayton, p.283, Mason, p.62, Croswell, p.179.

..., 23/04/11,

See back to the section on collapsing gas clouds.

..., 23/04/11,

Mason, ch.6, Clayton, p.290, and http://en.wikipedia.org/wiki/Interstellar_medium (accessed 23 March 2011)

..., 23/04/11,

Clayton, p.283 and 290. http://en.wikipedia.org/wiki/Cosmic_rays (accessed 5 March 2011)


these few captures tell us that a total of about one hundred thousand trillion neutrinos

(~1017) passsed through the two detectors. From the captured numbers and energies we

can calculate that the supernova emitted about 1058 neutrinos, at a temperature of over 10

billion degrees, carrying a total energy of around 1046 Joules, “more energy than an entire

spiral galaxy gives off in a year’s time. … The neutrinos give us a glimpse into the interior

of a hellish fireball. It verges on the miraculous - 19 flashes of light in subterranean

darkness reflect the blazing heart of a supernova.”

The neutrinos were emitted when the star's core collapsed to a neutron star; the light was

emitted when the blast wave broke through the star's surface, maybe a day later. For the

next 160,000 years the neutrinos raced through space towards their meeting point with

Earth, with the light from the blast wave, only a few hours behind. By the time neutrinos

and Earth converged at the same location, we had evolved from the stone age, and had

learned how to make telescopes and neutrino detectors, and derive mathematical models

of supernovae.

The blast wave from SN1987A blew threw outer layers of the star at speeds up to about

40,000 km/second, more than one tenth the speed of light. These layers contained a huge

amount of radioactive nickel-56, and the decay processes, first to cobalt-56 and then to

iron-56, “produced gamma rays (energetic photons) that heated the surrounding gas to

incandescence”. Six months later, when the debris had thinned, a rich mix of heavy

elements were detected, including, iron, calcium, strontium, nickel, cobalt, argon, carbon,

oxygen, neon, sodium, magnesium, silicon, sulphur, chlorine, potassium and calcium. It's

estimated that the the amount of nickel alone was about 7% of the mass of our sun, or

about 23,000 Earth masses. Thus we see supernovae as not only creators, but also

distributors of the elements.

Figure 3.67 shows SN1987A about 20 years after the explosion. The shock wave, about 1

light year across, is clearly visible, but the distribution of the elements is not yet apparent.


McNeil, 23/04/11,

Marschall, p. 258, and Gribbin, p.149…taking the mass of the sun as 2x1030 kg, and of the Earth as 6x1024 kg. Also… http://hubblesite.org/newscenter/archive/releases/2007/2007/10/full/ estimates 20,000 earth masses of iron.

McNeil, 23/04/11,

Marschall, p.258-260, and Woosley and Weaver.

..., 23/04/11,

Marschall, p.258.

..., 23/04/11,

Marschall, p. 245, and Woosley and Weaver.

McNeil, 23/04/11,

Marschall (ch.11) describes the sequence of observations around the world of the rapidly brightening supernova. What is remarkable is how it appears that the entire universe is constantly being observed by someone, somewhere. An insignificant star in one of millions of distant galaxies suddenly brightens, and someone spots it. About 3 hours after the neutrino burst had been automatically recorded (though not yet analysed), the brightening supernova was photographed by R. H. McNaught in Australia (though he didn't develop and examine the film till later) and about 20 hours later it was seen by Ian Shelton in Chile - Marschall, p. 253, and Woosley and Weaver. Silk (p.137) states that 11 neutrinos were captured about 4 hours before the first visual observation in Chile - not consistent with other writers. A burst of neutrinos was recorded 4 hours before the main burst - an unexplained event - Woosley and Weaver, and Marschall, p. 254.

..., 23/04/11,

Both quotes from Marschall, p.254.


Figure 3.67: SN1987A 20 years after the explosion.Anoher supernova, Cassiopeia A, around 300 years after its explosion, clearly shows the

elements in the expanding star remnants, now about 10 light years across – figure 3.68.


Debris from the supernova blast. Hidden inside this is a neutron star or black hole

The supernova shock wave is colliding with a sphere of material ejected from the star maybe 20,000 years before it exploded.The ring is here about 1 light year across, so debris has been hurled into space at speeds of up to 20 million miles/hour - about 9,000 km/s, "seeding" space with heavy elements

A pair of bright stars in SN19878A's galaxy, the Large Magellanic Cloud

a b

c d

McNeil, 23/04/11,

Description and image… http://hubblesite.org/newscenter/archive/releases/star/supernova/2007/10/full/ (accessed 23 March 2011) and… http://en.wikipedia.org/wiki/Cassiopeia_A (accessed 23 March 2011)


Figure 3.68: X-ray images of the supernova remnant Cassiopeia A, viewed about 300 years after the light from its explosion reached Earth. The expanding cloud of debris is about 10 light years across.(a) broadband image, using all X-ray wavelengths, showing the gas at about 50 million degrees(b) Image using selected X-rays to show the silicon nuclei in the hot gas - this shows a bright jet of silicon-rich gas breaking out of the left side(c) the distribution of calcium nuclei(d) the distribution of iron nucleiThe colours show the intensity of the X-rays: from yellow as the most intense, then through red and purple to green as the least intense.And finally, “where did the energy come from to produce the sound and fury which is a

supernova explosion? Energy is conserved: who paid the debts at the end? Answer:

Gravity! … The ultimate energy source in the stars which produce the greatest amount of

energy is gravity power.”

3.13 Review3.13.1 Seeding inter-stellar space - the "cosmic stock-pot"

enriching the universe

About 100 million supernovae have erupted in the Milky Way galaxy since it was formed

about 10 billion years ago. They and other stars have steadily enriched our galaxy in all

the elements beyond helium. These have been added and stirred into the mix like the

ingredients of a family stock-pot. Just as the meals taken from the stock-pot reflect the

history of what has been put in, so the compositions of new stars and solar systems

reflect the growing enrichment of the galaxy “Dust to dust, atom to atom, the cycle of

stellar life and death progresses - with a larger fraction of heavy elements in each

successive generation.”

The earliest stars, starting with only hydrogen and helium-4 could make only a limited

range of primary nuclei: for example, carbon-12, oxygen-16, silicon-28, calcium-40 and

iron-56. It's only later generations of stars, that start with these primary nuclei already in

the mix, that can go on to make secondary nuclei, neutron-rich isotopes of the primary

nuclei, for example, carbon-13(6p,7n) and oxygen-18(8p,10n).

"Supernovas are the engines of creation. Not only do they give birth to new elements, but

they scatter those elements to the currents of space." The amounts of heavy elements

that one supernova can produce are awesome; for example, about 1,500 Earth masses of

sodium, 5,000 of aluminium and 30,000 of magnesium. Supernovae also disturb the local


McNeil, 23/04/11,

I use Donald Clayton's word; see the appropriate sections in his book.

McNeil, 04/23/11,


McNeil, 23/04/11,

Clayton, p.280.

McNeil, 23/04/11,

Marschall, p. 207.

McNeil, 23/04/11,

There are old stars on the outskirts of our galaxy that contain 100 times less iron than does our sun. It takes many supernovae to build up the abundances of heavy elements build up to the values we see today in our region of the galaxy - Krauss, Atom, p. 171.

McNeil, 23/04/11,

Marschall, p. 211. Gribbin, p.157, uses the idea, too, and gives the estimate that in the Milky Way galaxy about 10 solar masses of interstellar dust are re-worked into new stars each year. This is a very small amount, but it means that over 10 billion years it's 100 billion solar masses - about a third of the galaxy's mass - a lot.

McNeil, 23/04/11,

Marschall, p. 212, though Chown, "The Magic Furnace", p. 210, puts it 10 times higher, at 1 billion. Delsemme, p.60, states that within only 1 billion years of the first nuclear reactions all of the half billiion stars of more than 5 solar masses in the Milky Way galaxy had scattered their new elements into space. The figures differ somewhat, but we get the general idea.

..., 23/04/11,

Frank Shu, quoted in Timothy Ferris, "Coming of age in the Milky Way", p.280.

McNeil, 23/04/11,

The Chandra X-ray observatory… http://chandra.harvard.edu/photo/2000/cas_a062700/casa_comp_420.jpg (accessed 23 March 2011)


gas clouds which can trigger the formation of new stars and solar systems from the now

enriched galactic mix.

our solar system

Our solar system, formed around a third generation star, contains about 74% hydrogen

and 24% helium, nearly the same mix as in the very earliest universe. But now the heavy

elements add a small but crucial 2% to this mix. The large planet Jupiter, with about 25

times Earth's gravity, is close to the composition of the overall solar system. The Earth,

with its weak gravity, was not strong enough to hold on to these light gases, so they blew

away to leave our rocky planet, containing just about every nuclide created in the big bang

and the stars.

3.13.2 Reviewing nucleosynthesis processes

a stellar time-line

Figure 3.69 shows the life of a large star as it accelerates towards its destruction as a

supernova.

Figure 3.69: Scales of temperature and time for the stellar nuclear processes up to supernova/tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

s-processSlow neutron capture

H-burning

alpha processburning C, Ne, O, Si - up to Fe

He-burning

Supernova explosionexplosive burning

r-process - rapid n capture

Tim

esca

le:

-bill

ions

of

year

s

-mill

ions

of

year

s

-thou

sand

s of

yea

rs

-yea

rs

-day

s

-hou

rs-m

inut

es-s

econ

ds

1 billion 1 million 1 thousand 0 Time, years

Temperature, K

10 billion

1 billion

100 million

10 million

McNeil, 23/04/11,

Based on fig. II.4 in B2FH, and a version is available at… http://rst.gsfc.nasa.gov/Sect20/A7.html (accessed 6 March 2011) As before, this is limited to stars that end up as type II supernovae.

McNeil, 23/04/11,

Chown, , "The Magic Furnace", p. 211.

McNeil, 23/04/11,

Marschall, p. 200.

McNeil, 23/04/11,

Bill Bryson briefly recounts the cosmic events leading up to the formation of the Earth in part I of his book.

McNeil, 23/04/11,

http://en.wikipedia.org/wiki/Jupiter

McNeil, 23/04/11,

Taylor, p.40. Gribbin, p.162, gives figures close to these, and says that 1-2% of the pre-solar system cloud was in the form of solid grains. A small %, but since the cloud's held > 1 milliion solar masses, this is more than 3 billion Earth masses - plenty enough for new rocky planets.

McNeil, 23/04/11,

Chown, "The Magic Furnace", p. 211.

..., 23/04/11,

Delsemme, ch.4, Mason, ch.7-9, and Stuart Taylor provide good accounts of the formation of our solar system. The story of the formation of 'our' solar system, our neighbourhood in the universe, is very interesting, and I'm virtually ignoring it. This might seem surprising, but we have seen how the cosmos and stars have created all the nuclides possible, and now they are brought together, as atoms, in planet Earth. So the next step is to see how these atoms relate to each other - this will be in the next chapter. A Solar System simulator, at…http://space.jpl.nasa.gov/

McNeil, 23/04/11,

Gribbin, p.163, describes meteorite evidence that suggests the formation of our solar system was triggered by a nearby supernova.


We can see the temperature climb a thousand-fold, and the time scale shrink from billions

of years to mere seconds.

3.13.3 Flooding the nuclear valley

We have viewed the creation of the nuclides as a series of moves across the nuclear

chess board, starting from just a single proton in the bottom left corner – recall the view

into the nuclear valley through the helium “pass” (figure 3.54). Figure 3.70 gives an

overview of the individual processes.

Figure 3.70: The activity of nuclide building surges up the nuclear valleyWe can see the sequence of fusion reactions building up single protons to nuclides in the

iron-group. The slow s-process works its way along the arc of stability as far as the last

stable nuclide. The p- and r-processes make the proton- and neutron-rich nuclides, on

either side of the arc of stability. Only the spallation process is not shown here.

We can imagine the energy level rising with the temperature in the star, and like a tide,

flooding the "nuclear valley", with waves of nucleosynthesis surging ever higher. Neutron

and proton capture processes wash up the valley sides to produce unstable nuclei, which

then decay back towards stability. When the nuclei are ejected into the coolness of space


Start here, with single protons!

Fusion processes – making bigger nuclideshydrogen burning to helium…

…helium burning to carbon…

…carbon burning through neon, oxygen, silicon, to iron.

p-process - makes unstable proton-rich nuclei, which decay back towards stability

(s-process) – slow neutron capture makes stable nuclides from iron to lead-208

(r-process) - rapid neutron capture makes unstable neutron-rich nuclei, which decay back towards stability

the r-process makes nuclei beyond the last stable nucleus

iron-56


the stable clusters of protons and neutrons remain. In a similar way, when the tide

recedes, it leaves a series of rock pools, small stable bodies of water in isolated hollows.

the array of nuclide species

Every single nuclide is an individual, descended from a series of ancestors via a

sequence of nuclear reactions. Some are ancient, only a few minutes less old than the

universe itself; some are only a few thousand years old, like the nuclei made in supernova

1987A; some have taken millions of years to assemble by the infinite patience of stars;

some were made in a fraction of a second in the convulsions of a supernova; and a few

are here as the fragments of something larger, that was broken up by collisions in deep

space. All have their different stories to tell; each is as unique as a biological species -

“the personalities of [tantalum-181] and of [iron-56] differ as dramatically as those of the

seagull and the tiger”.

3.13.4 Nuclides and the stellar eco-system

We think of the the populations of living species in terms of a dynamic ecological balance.

Each species feeds on other species lower down the food chain, and is itself devoured by

"predator" species higher up. Species populations then reflect the numbers that can co-

exist within this ecological balance.

Nuclei within stars also have populations (usually called abundances). They are created

by the fusion of smaller nuclei, and are "consumed" by being fragmented, or capturing

protons or neutrons, or by being incorporated into yet bigger nuclei. The population of a

nuclide species is decided by the balance of such nuclear reactions. Nuclear populations

vary hugely; for example, carbon is nearly 100 million times more abundant than gold in

the solar system. The different conditions that can exist within stars create environments –

that we can perhaps think of as habitats - that favour one group of nuclei rather than

another, with consequences for their populations. We have seen how a big star develops

an onion-like structure, shortly before it explodes as a supernova. We can see this

perhaps as a range of habitats, within which certain nuclei are favoured, with nuclei in one

habitat "feeding" on the product of the next layer up, thus the carbon-burning layer

consumes the ashes of the helium burning in the layer above.

3.14 The emergent atom3.14.1 From nuclides to atoms


McNeil, 23/04/11,

Solar system abundances, data from http://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements (accessed 9 March 2011).

..., 23/04/11,

There seem to be some limited correspondences between nuclei in stars and living species in an ecosystem. Clearly nuclei don't evolve as living species do. But each nuclear species "feeds" on some species, and is "consumed" by others. The conditions inside the star determine the "fitness" of this nuclear species, and hence its population abundance. We've seen how a star is a dynamic structure, that changes through a life-cycle, and the nature and populations of nuclear species created within the star also change accordingly.

McNeil, 23/04/11,

Clayton, p. 2 et seq.

..., 23/04/11,

Clayton, p.7.

..., 23/04/11,

Its radiation and neutrinos only reached us in 1987.

..., 23/04/11,

Donald Clayton's book "Handbook of Isotopes in the Cosmos", is a superb description of the nuclear clusters from hydrogen-1 up to gallium-71. Even if you can't follow all the technical detail, you get a feel for each nuclide as a unique individual, with its origins, history and character.


With the life cycle of a massive star, and its end as a supernova, we have seen all the

reactions that create the nuclides. We now imagine them blown out into cool interstellar

space, and gathered into a newly forming solar system, no longer subject to the enormous

temperatures and pressures in a star. The unstable nuclides will undergo decay in their

own time, but the stable nuclides, ejected from the stellar ecosystem, are isolated by their

mutual repulsion, and will remain fixed and unchanging.

Since every proton carries one unit of positive charge, each nuclide carries as many units

of positive charge as it has protons. As the nuclides ejected from the star cool in space

each attracts a number of negative electrons, equal to the number of protons, to achieve

overall neutrality. Each nuclide acquires and tethers a number of electrons, that would

otherwise be roaming freely, thus creating a new physical entity, an atom - figure 3.71.

Figure 3.71: Quarks are bound by the strong colour force into nucleons, which are bound by the nuclear force into nuclei, which bind electrons to themselves.

parallels and contrasts


-

-

-

the nuclear force binds a group of protons and neutrons into a nuclear cluster, overcoming the protons’ mutual repulsion

p+

p+p+n0

n0

n0n0

the electrons are electrically attracted to the nucleus, overcoming their mutual repulsion

u

ud

the strong force binds a colour-neutral trio of uud quarks into a proton

the numbers of protons and electrons are equal, so the atom is electrically neutral

the mutual attraction of all nucleons means a nuclide has no centre...

... whereas an atom has a central nucleus

removing a neutron requires an energy input of ~5 MeV...

...whereas an electron can be removed with only ~5 eV

~5 eV

~5 MeV

-

p+

n0

McNeil, 23/04/11,

I have used an isotope of lithium as a simple example. The quark-gluon diagram is a simple representation of a proton. Likewise, the Li-7 atom is highly simplified. The electrons, of course, don’t sit in a circular orbit centred on the Li-7 nuclide – we’ll get into electron orbitals in the next chapter. The neutron separation energies for Li-6 and Li-7 are 5.7 and 7.3 MeV respectively - Clayton, section on lithium. The first electron separation energy for Li is about 5 eV - Williams, fig. 8.1. I have “rounded” the neutron separation energy down to 5 MeV, to bring out the ratio of 1 million. The electron separation energy depends on the number of electrons in the outer shell, and 5 eV is typical for elements like Li with only one electron in the outer shell. Similarly, neutron separation energies vary. Williams states that it is ~6 MeV for heavy nuclei (p.83), though they range from ~8-14 MeV for the iron isotopes - Clayton, iron section.

..., 23/04/11,

The first nuclides, of up to 7 nucleons, were made within the univere’s first quarter of an hour, and the first atoms based on these nuclei were made by about 300,000 years. This is a hierarchical, not a historical, account of the universe. So, now that we have all the elements present, we can properly be introduced to atoms, and to their emergent chemistry.


This diagram shows the parallels and contrasts between the worlds of quarks, nuclides

and atoms. First, the colour charges of quarks are balanced in a single proton or neutron,

so it has overall colour neutrality. Similarly, the electrical charges are balanced in an atom

so it has overall electrical neutrality. Second, in a stable nuclide, the nuclear force binds

protons and neutrons together against the mutual repulsion of the protons. In a similar

way, the electrons are bound to the central nuclide against their mutual repulsion.

However, because the nuclide holds all the positive charge, the atom has a centre,

whereas a nuclide cluster has none.

disparities in mass and energy

The atom displays big disparities in mass and in energy. The nuclide shown here is a

quite small inhabitant of the nuclear valley, only 7 nucleons, about 6,500 MeV, but this

hugely outweighs the combined mass of the electrons, which is a mere 1.5 MeV. The

electrons are like party balloons tied to a bag of bricks. There are also big differences in

energy; it takes about a million times less energy to remove an electron from a neutral

atom than to remove a neutron from the nuclide. This is an indication of how much weaker

the electrical force is compared to the nuclear force. If we think of nuclides as deaf to

anything less than a shout or a bellow, then an atom will respond to the merest whisper.

At the moderate temperatures on Earth, where the interaction energies are small, the

electrons will readily respond where the nuclide will not. The low energy interactions

between atoms are now mediated wholly by their attendant electrons. We can perhaps

think of the electrons tethered to the nuclide in figure 3.71 as resembling a bunch of party

balloons tied to a bag of bricks. A breeze blows, the balloons sway and gently bump into

each other, but the bag of bricks is indifferent to these gentle interactions.

the nuclide is now an atomic nucleus

The nuclide is no longer in its normal habitat, interacting with other nuclides, and with an

identity defined by its mix of protons and neutrons. It is now the inert centre of a cluster of

tethered electrons, whose number is fixed by the proton number, with the neutron number

now irrelevant - it is now an atomic nucleus. We have seen how very soon after the big

bang the quarks were bound up into protons and neutrons, and have never been seen

free since that time. The same has now happened to the nuclides, for under Earth’s gentle

conditions they are inaccessible, each buried deep inside its atom.

the electrons are now confined


..., 23/04/11,

The relative energies of nucleons and atomic electrons is ~400,000 – Barrow and Tipler, p.320. Thus 1 tonne of a nuclear explosive has the same energy yield as several hundred kilotonnes of conventional chemical explosive, usually tnt. The theoretical limit for nuclear fusion weapons is 6 megatonnes of tnt per tonne of nuclear explosive , a ratio of 6 million – see http://en.wikipedia.org/wiki/Nuclear_weapon_yield


The electrons were created by the time the universe was a second old. In themselves,

they appear negligible - a swarm of light particles, mutually repelling each other, and

apparently incapable of any corporate activity. However, each nuclide, by its positive

charge, gathers a specific number of electrons around itself. In this unnatural arrangement

the electrons are forced to co-exist in a confined space, contrary to their urge to be

separate. This is the atom, a conflict of inclinations, a totally novel construct in the

emergent universe.

We have seen how a cluster of nucleons in a nuclide form pairs and closed shells of

magic numbers, making a structured community. We will see in the next chapter how, with

nuclei acting as stable foundations, electrons also form structured, hierarchical

communities, and create wonderful patterns of intricacy and elegance, opening up a

whole new level of complex possibilities.

the chemical elements

So, instead of the 2-dimensional array of nuclides as we have seen them in the nuclear

valley, we now have a series of about 100 neutral atoms, that is, tethered clusters of 1 to

about 100 electrons. These are the chemical elements. Thus we shift our focus from the

nuclides, as they have been made in the big bang and the stars, to the elements, as they

interact chemically on Earth, mediated by their electrons. It is sometimes said that the

elements were created in the stars, and this is approximately true. It is only the nuclides

that are made in the stars, but each then captures the electrons to create an atom that

behaves as a chemical element.

from the 2-D nuclear valley to a 1-D series of elements

We’ll use some specific elements to illustrate the shift from a world of nuclides to a world

of atoms and chemical elements - figure 3.72.


..., 23/04/11,

I want to sidestep any dispute about how many elements there are. John Emsley describes element 118, ununoctium (with 118 protons in the nucleus, so Z=118). The last truly stable element is lead (Z=82). Bismuth (Z=83) is often classified as stable, because its half life is so long. There are less than 82 stable elements, because some, for example, technetium (Z=43) and promethium (Z=61), have no stable isotopes. There are practical uses for the elements americium (Z=95), curium (Z=96) and californium (Z=98). So I will put the number of elements as “about 100”. The origins of the elements’ names is covered by John Emsley, and also… http://www.nndc.bnl.gov/publications/preprints/origindc.pdf (accessed 29 March 2011). “So how many elements are there? I do not know, and neither does anyone else … We have no idea what the limit mught be.” – Philip Ball, p.91. There are around 92 naturally occurring elements, but even that is approximate.


Figure 3.72: How the chemical elements are related to the nuclides, and how nuclear decay transforms one element to another.The diagram shows the 2-dimensional array of the smallest nuclides in the nuclear valley,

where diagonal lines connect members of a nuclear cluster “family”. However, in Earth’s

low energy environment, we only discriminate the number of electrons that are tethered

around each nuclide. Thus all nuclides containing 3 protons, shown in the horizontal box,

will each gather three electrons, to make atoms that are chemically identical, and distinct

from atoms with different numbers of electrons. In this way we distinguish the atomic

elements, depending on their number of tethered electrons, and the first few are listed up

the side of the diagram. So we move from a 2-dimensional array of nuclides in the nuclear

valley, to a 1-dimensional series of atomic elements.

isotopes - same proton number, different neutron number

Thus atoms with the same number of electrons, whilst chemically identical, can have

different nuclides at their centres. An element can have a number of isotopes - atoms with

the same number of protons, but a different number of neutrons. These different nuclides

may have very different ancestries.


1 2 3 4 5 6 7 8 9 neutrons

Li-6(3p,3n) and Li-7(3p,4n) are stable isotopes

Li-5 (3p,2n) ejects a proton to become He-4 (2p,2n)

Li-8 (3p,5n) undergoes beta-decay to (4p,4n), which then alpha-decays to He-4 (2p,2n)

the isotopes of lithium, Li, all have 3 protons

Nuclides acquire electrons to match the proton number. So...

...8 electronsoxygen

...7 electronsnitrogen

...6 electronscarbon

...5 electonsboron

...4 electronsberyllium

...3 electronslithium

...2 electronshelium

...1 electronhydrogen

N-13 (7p,6n) decays to C-13 (6p,7n)

C-14 (6p,8n) decays to N-14 (7p,7n)

Cluster size...

9

8

7

6

5

protons

8

7

6

5

4

3

2

1

McNeil, 23/04/11,

For example, the noble gas xenon, has 9 stable isotopes, whose nuclei are made by different stellar processes - Clayton, p.304. Mason describes the origins of the 10 stable isotopes of tin – p. 59.

McNeil, 23/04/11,

"Isotope" - from the Greek: "iso" = same, "topos" = place, position. So all isotopes of an element, having the same number of protons, are at the same position in the periodic table of the elements. We'll meet the periodic table in chapter 4. The isotopes are chemically identical, but the differing masses of their nuclei can affect physical processes like evaporation and diffusion. We separate uranium-235 from uranium-238 by exploiting its slightly faster rate of diffusion – Emsley on uranium. Another example: oxygen has two stable isotopes – O-16 (99.8%) and O-18 (0.2 %), and the lighter of these evaporates more easily, “just as a sparrow takes flight more easily than an albatross”. The ratio of O-16 to O-18 in ice and the shells of marine animals is used to find global temperatures in past ages – Philip Ball, p. 130, and Emsley, p.304. The number of stable isotopes is often related to nucleon magic numbers. For example, the metal tin, with the magic number 50 protons, has 10 stable isotopes - Williams, p. 132.


lithium old and new

The element lithium, defined by its 3 electrons in each atom, has a number of isotopes,

shown in the horizontal box in figure 3.72, of which only 2 are stable - Li-6 (3p,3n) and Li-

7 (3p,4n). The lithium-6 isotope is only made by cosmic rays from dying supernovae

undergoing spallation reactions in deep space. In contrast, much of lithium-7 was made in

the last nuclear reaction in the cooling fireball after the big bang. Thus, in a lithium battery

in a calculator or a mobile phone, some of the nuclei will be quite recent, maybe just pre-

dating the solar system, while others will will be only a few minutes younger than the

universe itself.

lithium can transform to helium

The nuclides in the other lithium isotopes are highly unstable, for example, (3p,2n) and

(3p,5n) each decay to the nuclide (2p,2n), which creates atoms with only 2 electrons,

making the element helium. Speaking in chemical terms, we say that these unstable

isotopes of lithium decay to helium. This would appear to be magic, or nonsense - how

can a soft grey metal become a light gas? But we can see that the protons and neutrons

in the atom’s central nuclide rearrange themselves, and electrons are then acquired or

discarded to match the new proton number. With their enormously bigger energies, the

nuclides take precedence, and the electrons arrange themselves accordingly.

carbon and nitrogen

Figure 3.72 shows some other element transmutations. The elements carbon and

nitrogen are defined by the number of electrons in their atoms, 6 and 7, respectively. The

loss or gain of just one electron will transform one element into the other. Thus the

neutron-rich nuclide in the isotope C-14 (6p,8n) undergoes beta-minus decay to become

(7p,7n), one of the two stable isotopes of nitrogen. Conversely, the proton-rich isotope N-

13 (7p,6n) decays to become the stable isotope C-13 (6p,7n). The atomic transformations

involve gaining or losing one electron, and because electrons can be removed from atoms

so easily, there are always temporarily free electrons available, if needed.

the transmutation of the elements

These specific examples show how the decay of an unstable nuclide decays and adjusts

its electronic “clothing”, removing or adding “garments” to maintain neutrality. The atomic

elements are not immutable, but can be transformed from one to another, driven by the

enormous energies contained within the atomic nucleus.


McNeil, 23/04/11,

Ray Mackintosh et al comment how at the turn of the twentieth century the firmly held belief was that atoms never change – p.26. This resembles the early 19th century view of biological species as immutable –“…God had created species for a particular rôle in Nature. Since this rôle was fixed from the beginning, there was no need for species to vary; evolution made no sense of God’s purpose” – Cadbury, p.138.

McNeil, 23/04/11,

Clayton, on lithium, and http://en.wikipedia.org/wiki/Lithium (accessed 5 March 2011)


Oliver Sacks writes: “The feeling of the elements’ stability and invariance was crucial to

me psychologically, for I felt them as fixed points, as anchors, in an unstable world. But

now, with radioactivity, came transformations of the most incredible sort. What chemist

would have conceived that out of uranium, a hard tungsteny metal, there could come an

alkaline earth metal like radium; an inert gas like radon; a tellurium-like element,

polonium; radioactive forms of bismuth and thallium; and, finally, lead - exemplars of

almost every group in the periodic table? … Radioactivity did not alter the realities of

chemistry, or the notion of elements; it did not shake the idea of their stability and identity.

What it did do was hint at two realms in the atom - a realtively superficial and accessible

realm governing chemical reactivity and combination, and a deeper realm, inaccessible to

all the usual chemical and physical agents and their relatively small energies, where any

change produced a fundamental alteration of the element’s identity.”

the volcano

We can imagine standing on the edge of a volcano, looking down to the rich plant life

growing on the volcanic slopes, and also down into the crater full of red hot magma. The

plant life is suited to the cool lower slopes, and would be destroyed if exposed to the heat

of the volcanic core. The magma churns deep in the crater, and takes no account of the

plant life above. Standing on the lower slopes, you would have no hint of the huge

energies in the red hot rock underneath. If the volcano erupts, people die and the plant life

is destroyed, but after a while the people return and plant life re-establishes itself, and

things go on as before.

An atom is similar to a volcano, in that there is a huge turbulent nuclear core, deep

beneath the superficial electrons. If the nucleus is stable, there is no hint of this enormous

energy. But when an unstable nucleus erupts, this huge energy reveals itself, and the

electrons are disrupted. But then, like the volcano, electrons gather round the new

nucleus and neutrality is achieved once more. With the volcano, things go on as before,

but with another layer of fertile soil. But with nuclear decay, the balance of the nucleons,

and specifically the number of protons has changed. Consequently, the number of

electrons in the atom changes, and a new atom, with its own chemical identity appears.

The protons and neutrons in their nuclear world, take no account of what the electrons are

doing "outside". A cluster of nucleons is only concerned with its own stability, so it will

decay by any process that is energetically suitable - interchanging protons and neutrons,

or down-sizing by ejecting an alpha particle (2p,2n). The electrons regroup around the


..., 23/04/11,

Oliver Sacks, p.286.


new nucleus, and a “new” atom, with its own physical and chemical identity appears. The

nucleons pay no more attention to the electrons, than a volcano heeds the humans living

on its slopes.

Element abundances and ancestries

Figure 3.73 shows how the different nucleosynthesis processes have created the

abundances of the elements in our solar system.

Figure 3.73: The abundances of the elements in the solar system.The atomic number, on the x-axis, gives the number of protons in the nuclei, which determines the element.The vertical scale gives relative abundances, referenced to 1million silicon nuclei. This is a logarithmic scale, so the silicon abundance is 6 since log 1,000,000 = 6. Thus oxygen (O) at about 7 on the abundance scale is about 10 times more abundant than silicon (Si). Fluorine at 3 has about For every million silicon there are ~1,000 atoms of fluorine (log 1,000 = 3), and only ~1 atom of beryllium (log 1 = 0). It may help to view the positive numbers on the y-scale as the number of zeros after 1.Many of the elements are named with their standard chemical symbol.Hydrogen and helium, the big bang nuclei, vastly outnumber all others. After them, there

is a steady downward trend to larger and rarer nuclei, that is punctuated by the dip for the

light elements (containing 3-5 protons, that is, Z=3-5) and the peak for the highly stable

iron-group nuclei (Z=~26). Abundances decrease steadily after the iron-group peak, since

“element formation beyond this point costs energy”. The graph covers all the elements up

to uranium (Z=92), with most being identified with their chemical symbol. Only elements

43, technetium, and 61, promethium are missing, since they have no stable isotopes./tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023

Big bang - the first 3 elements (hydrogen to lithium) in the first three minutes

Nuclear fusionBurning H to He to C to O to Si to Fe/Ni- elements 1 - 26 (hydrogen to iron)

Iron peak: the most tightly bound nuclei

Neutron capture: by the slow s-process, and the rapid r-process in supernovae - elements beyond iron

Spallation: cosmic rays in deep space - elements 3 - 5 (lithium to boron)

..., 23/04/11,

Emsley.

..., 23/04/11,

Williams, p.356.

McNeil, 23/04/11,

Abundance data data from Katharina Lodders. This is consistent with Williams, fig.14.3, Delsemme, fig.4.2, and Mason, p.58, which itself is from… Fowler, Rev. Mod. Phys. 56, p.149, 1984, and online at… http://nobelprize.org/nobel_prizes/physics/laureates/1983/fowler-lecture.pdf (accessed 7 March 2011). A helpful presentation on elemental abundances at http://solarsystem.wustl.edu/2008/08/21/stl-astro-society-talk/ (accessed 7 March 2011). The plots of Woosley and Fowler seem to show more pronounced peaks at mass numbers around 75, 130 and 200. It’s standard practice to plot element abundances relative to 1 million Si atoms.

..., 23/04/11,

Hmmm....maybe getting poetic, but it may give a feel for the enormous energy difference between the nuclear and electronic realms, the unpredictability of nuclear decay, and the way the electrons will regroup around a new nucleus created from the decay of the old one.


“Only the first five lightest elements owe their abundances to origins outside stars - the

first three in the big bang and the fourth and fifth (beryllium and boron) by cosmic ray

interactions with interstellar atoms. From the stars came all the rest. From atomic number

6 (carbon) to atomic number 94 (plutonium) we look to the stars This range of atomic

numbers includes all the common elements of human experience on Earth, save for the

hydrogen within water that blesses the earth’s surface.”

The abundances are clues to the natures of the nuclides in the elements, and the

processes that made them. The prominent “iron-peak” is due to the extreme stability of

these nuclei – “it’s very easy to make iron, for example; but it turns out to be very hard to

make fluorine or boron”. The elements lithium to boron (Z=3-5) do not feature in stellar

fusion processes, and were made by spallation processes, as fragments of carbon and

oxygen nuclei in deep space. Fluorine (Z=9) is comparably rare, and if abundances were

represented as dwellings, then “fluorine would be a shack between two mansions” –

between oxygen and neon. Like the elements lithium to boron, fluorine is largely made by

a spallation process, the ejection of a nucleon from neon (Z=10) in a supernova

explosion.

Elements with more stable nuclei tend, understandably, to be more abundant. We have

seen that nucleons pair up, so nuclides with even numbers of protons or neutrons tend to

be more stable. Thus the abundances follow a zig-zag path across the graph, with the

elements with even numbers of protons (such as C, O, Ne, Si, S…) being consistently

more abundant than their odd-number neighbours. We have seen that nuclides are more

stable if they contain “magic numbers” of nucleons - 2, 8, 20, 28, 50, 82 or 126. The graph

shows a slight abundance peak for elements 50-55, for these have around 50 protons and

82 neutrons. Similarly, lead, with 82 protons and 126 neutrons, has a large abundance.

Every different isotope of every element has its own genealogy - its nuclear ancestry.

Donald Clayton writes, “If I could write an epic poem, I would lyricize over the history of

the universe writ small by their natural abundances. I would rhapsodize over the puzzling

arrangements at different times and places of the thousand or so different isotopes of

some ninety chemical elements. These different arrangements speak of distant past

events”.

Lawrence Krauss describes a very plausible history of a nucleus of oxygen: “Each atom of

oxygen on Earth, by its very existence, suggests a veritable treasure trove of detailed

history: the life and death of millions of stars, the slow dynamic evolution of our galaxy,


McNeil, 23/04/11,

Krauss, p.172.

McNeil, 23/04/11,

Clayton, p. ix. Who says all scientists are unfeeling and lacking imagination?

..., 23/04/11,

I have kept this simple. Other authors discriminate more detail – eg Williams, fig. 14.3.

..., 23/04/11,

Clayton on fluorine, and Croswell, p.179.

..., 23/04/11,

Croswell, p.179.

..., 23/04/11,

Stan Woosley, quoted in Croswell, p. 178.

..., 23/04/11,

Clayton, p.6.


and indeed the history of matter from well before the galaxies existed. Our oxygen atom

began life as 16 particles”, which fused to create 4 He-4 nuclei, 3 of which then made a C-

12 nucleus. “Finally, 2 particles, the nucleus of carbon and the nucleus of helium, are

brought together from originally disparate parts of the cosmos, with completely different

individual histories, to make a single nucleus, the nucleus of oxygen.”

3.14 The emergent atomWe are now in a position to review the series of emergent physical structures that

culminate in an atom – figure 3.74.


..., 23/04/11,

This recapitulates diagrams for the nucleon and nuclide shown earlier ion this section. This will be extended to include the binding of atoms into molecules and other giant structures. We have seen how the stong colour force in the nucleon extends to bind nucleons together. This is analagous to the way electron orbitals in single atoms extend to bind atoms together into molecules – Close p.105.


Figure 3.74: Successive stages in the emergence of the atom. From bottom to top…(a) the emergent nucleon, (b) the emergent nuclide, and (c) the emergent atom.


Z+

e-

e-

e-

Z+e-

Continuous interactions between nucleus and electrons, mediated by photons…(or…“photons bind electrons to the nucleus”)

… an atom - a stable cluster of electrons, around a massive nucleus, containing up to ~100 electrons.

A stable atom has emerged, a tethered cluster of a number of electrons equal to the number of protons in the nucleus, and sensitive to very small energies

free electrons are stable, but are isolated by their mutual repulsion.We don’t (usually) consider the protons and neutrons inside the nucleus.

u

u d

u

u

d

Continual interactions between quarks, mediated by gluons carrying colour charge…(or…“gluons bind quarks”)

… a colour-neutral quark trio (baryon); only 1 baryon is stable – the proton, the neutron is nearly stable.

A stable cluster of quarks has emerged, as a proton, with a large mass, and carrying a unit of positive electric charge

quarks and gluons cannot exist in isolation

… a nuclide – a cluster of protons and neutrons, with the mutual repulsion of the protons overcome by the stronger nuclear attraction between all nucleons. About 300 varieties of nuclide cluster are stable.

p+

p+p+n0

n0

n0n0

p+ n0

Continuous interactions between protons and neutrons, mediated by pions…(or…“pions bind protons and neutrons”)

A stable nuclide has emerged, with a large mass, and carrying as many units of positive charge as it contains tethered protons

free protons are stable, but are isolated by their mutual repulsions, and unstable neutrons cannot exist in isolation.We don’t consider the quarks inside protons and neutrons.

a

b

c





emergent nucleons

We have seen how the colour-neutral quark trios we know as protons and neutrons,

emerge from the continual interactions between quarks, mediated by the gluons that carry

colour-charge.

emergent nuclides

Protons and neutrons are each surrounded by a “cloud” of virtual pions, that “flutter”

between them if they are close enough, creating “an invisible, evanescent web … binding

them together”. It is perhaps ironic that neutrons, in themselves unstable, are essential to

the stability of all nuclides. Thus a nuclide emerges from the continual interactions

between all the nucleons, mediated by pions. The nuclear force, an extension of the

colour force operating inside each nucleon, binds protons against their mutual repulsion,

into compact massive centres of electric charge, with up to ~100 units of positive charge.

emergent atoms

The atom emerges from the electrical attraction between a positively charged nuclide and

electrons, mediated by the exchange of photons of electromagnetic radiation. The nuclide

thus tethers a cluster of electrons, against their mutual repulsion, and creates the range of

chemical elements, based on atoms with up to ~100 electrons.

a hierarchy of emergent structures

Thus we have a hierarchy of emergent structures, each one building on the one below,

and introducing a totally novel feature into the universe. So the strong colour force creates

a stable cluster of quarks, a nucleon, that can have 1 unit of positive electric charge; the

residual nuclear force binds these into clusters, carrying up to ~100 units of positive

charge; and these provide the foundation for the electromagnetic force to gather clusters

of up to ~100 electrons.

Each level in the hierarchy is created and sustained by continual activity in the level

below. At each level in the hierarchy, the activity “underneath” is subsumed in a new

emergent behaviour: quarks “disappear” inside nucleons, and nuclides “disappear” inside

atoms.

atoms can interact and yet keep their identity

A nuclide has no centre, so small nuclides can fuse to make one larger cluster, and lose

their identities in the process. But the electrons in an atom respond to very small energies,


..., 23/04/11,

And we’ll see this go further…a biological cell is a cluster of specific biochemical reactions “bound” by its DNA, and a human being is a cluster of thoughts, emotions attitudes and ideas. We’ll see how atoms combine through low energy electronic interactions. Thus the massive nuclei are moved about, directed by the transient, low energy electronic interactions. In a similar way, each of us is driven and moved about by our ideas, thoughts and feelings, transient and persistent, subtle and simple.

..., 23/04/11,

See back to the previous section for the sources of these quotes. I’m keeping this simple; at very close range, other particles are transferred in addition to single pions.


about one millionth of those involved in nuclide interactions. So atoms can interact and

join together through their electrons, and yet retain their unique identities within a larger

structure. A nucleus binds as many electrons as it has protons, to make a neutral atom.

Thus the nucleus of an atom decides its identity as an element, yet the nucleus plays no

part in the chemical interactions of the electrons. Electrons can be removed with very

small energies, and the nucleus remains unchanged; the elemental identity is fixed

regardless of any electronic rearrangements. In a similar way, someone remains the

same individual regardless of what clothes they wear.

3.15 The next chapterIn the next chapter we will leave behind quarks, gluons and nucleons, and look at the

behaviour of communities of tethered electrons – we will enter the realm of the chemical

elements. The 3,000 or so known nuclides can create only about 300 stable

combinations, which then become the nuclei of only ~100 different elements. Yet we will

see these few elements create millions of different stable combinations – the different

compound substances of our material world. We will see the amazing structures that

communities of electrons can create. “We could get rid of every electron in our bodies,

and we would never notice the difference if we stood on a scale. Yet, in spite of their puny

heft, electrons may be the most important particle in nature, at least to us, because they

determine almost every observable aspect of our existence.”


..., 23/04/11,

Krauss, p.80. Natalie Angier – the entire complement of electrons in an adult body weigh only ~20 grams. How do we work this out? Say an average body mass is 80 kg, and being composed of the light elements, with a 1:1 prton to neutron ratio in their nuclei, then this body will contain 40 kg of protons. Since each atom has as many electrons as protons, and an electron mass is 1/1860 that of a proton, then the body will contain 40,000/1860 = 21.5 g of electrons. A tiny fraction of our mass, but there are lots of them – around 24 billion billion billion or so. The entire chemistry of our life processes, our thoughts and feelings, are mediated by the workings of these electrons.

..., 23/04/11,

The NUCLEUS programme, based on the NUBASE data, lists 3179 nuclides, with up to 118 protons, and ~260 nucleons. The figure of ~300 stable nuclides is taken from a CERN publication… http://cerncourier.com/cws/article/cern/28587 (accessed 29 March 2011)

..., 23/04/11,

A small, maybe wayward, thought…we dress differently for work, relaxation, sport, formal occasions, and so on. We are the same individual, but engaged in different social functions, revealing a different facet of our human individuality. In the next chapter we will see that the electronic orbitals around an atom reflect the nature of the chemical relationship it’s in at the time.


ReferencesFred Adams and Greg Laughlin, "The Five Ages of the Universe", Touchstone, 2000

Jim Al-Khalili, “Everything and Nothing”, broadcast on BBC4; programme 1, “Everything”,

21 March2011, and programme 2, “Nothing”, 28 March 2011. http://www.bbc.co.uk/bbcfour/

Jonathan Allday, "Quarks, Leptons and the Big Bang", Institute of Physics publishing,

2003

AME 2003a, “The AME2003 atomic mass evaluation (I): Evaluation of input data,

adjustment procedures”, A.H. Wapstra, G. Audi and C. Thibault, Nuclear Physics A,

volume 729 (2003) p.129–336

The paper is available in the file Ame2003a.pdf from…


AME 2003b, “The AME2003 atomic mass evaluation (II): Tables, graphs and references”,

G. Audi, A.H. Wapstra and C. Thibault, Nuclear Physics A, volume 729 (2003) p.337–676

The paper is available in file Ame2003b.pdf from…


This data is the basis for the “mass.mas03” table, available as an ASCII file from…

http://www.nndc.bnl.gov/amdc/masstables/Ame2003/filel.html or…

http://www.nndc.bnl.gov/amdc/web/masseval.html

The mass.mas03 table gives values for mass excess, binding energy/nucleon and atomic

mass (amu), and can be opened in Notepad, and converted to EXCEL format.

Arnold B. Arons, "Development of Concepts of Physics", Addison-Wesley, 1965.

Philip Ball, "The Elements - a very short introduction", Oxford, 2004

John Barrow and Frank Tipler, “The Anthropic Cosmological Principle”, Oxford University

Press, 1986.

Carlos Bertulani, "Nuclear Physics in a Nutshell", Princeton University Press, 2007.

John Tyler Bonner, "Why size matters", Princeton University Press, 2006

Bill Bryson, "A short history of nearly everything", Doubleday, 2003

Margaret Burbidge, Geoffrey Burbidge, William Fowler and Fred Hoyle, “Synthesis of the

elements in the stars”, Reviews of Modern Physics, vol. 29, p.547-650, 1957./tt/file_convert/577cc1bd1a28aba71193cafd/document.doc of 169 28/04/2023


Wikipedia article at…http://en.wikipedia.org/wiki/B%C2%B2FH (accessed 24 March 2011),

and the full paper available from…http://prola.aps.org/pdf/RMP/v29/i4/p547_1 (Accessed 5

March 2011).

Deborah Cadbury, “The Dinosaur Hunters”, Fourth Estate, 2001

Marcus Chown, "Afterglow of Creation", Arrow, 1993

Marcus Chown, "The Magic Furnace", Jonathan Cape, 1999

Donald Clayton, “Handbook of Isotopes of the Cosmos”, Cambridge, 2003.

Frank Close, "Particle Physics - a very short introduction", Oxford, 2004

Frank Close, "The New Cosmic Onion", Taylor and Francis, 2007

Frank Close, Michael Marten and Christine Sutton, "The Particle Odyssey", Oxford, 2004

Ken Croswell, "The Alchemy of the Heavens", Oxford University Press, 1996

Paul Davies, "The Goldilocks Enigma", Allen Lane, 2006

Armand Delsemme, "Our Cosmic Origins", Cambridge, 1998

Ken Dobson, David Grace and David Lovett, "Physics", Colllins Educational, 1997

John Emsley, "Nature's Building Blocks", Oxford, 2001

Timothy Ferris, "Coming of Age in the Milky Way", The Bodley Head, 1988

Timothy Ferris, "The Whole Shebang", Weidenfeld and Nicolson, 1997

M. P. Fewell, “The atomic nuclide with the highest mean binding energy”, American

Journal of Physics, volume 63, p.653 ,1995, accessed through Google Scholar, August

2010, also referenced in http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin2.html#c1

and http://en.wikipedia.org/wiki/Type_II_supernova

Richard Feynman, "QED - the strange theory of light and matter", Penguin, 1990

Richard Feynman, "The pleasure of finding things out", Penguin, 2000

Richard Feynman, Robert Leyton and Matthew Sands, “Lectures on Physics”, Volumes I

to III, Addison-Wesley, 1963.

W. A. Fowler, Nobel Prize address, at…

http://nobelprize.org/nobel_prizes/physics/laureates/1983/fowler-lecture.pdf

Harald Fritzsch, “Quarks - the stuff of matter”, Penguin, 1992


http://nobelprize.org/nobel_prizes/physics/laureates/1983/fowler-lecture.pdf


Harald Fritzsch, “Elementary Particles - building blocks of matter”, World Scientific, 2005

Robert Gilmore, "The Wizard of Quarks", Copernicus books, 2001

George Greenstein, “Frozen Star”, Futura, 1986

John Gribbin, "Q is for Quantum - an Encyclopedia of Particle Physics", The Free Press,

1998

John Gribbin, "Stardust", Allen Lane, 2000

John Gribbin, "The Universe - a biography", Penguin, 2006

John Gribbin, "Famous for 14 minutes 49 seconds", New Scientist, vol. 137, 13 March

1993, p. 14

Moo-Young Han, "Quarks and Gluons: a century of particle charges", World Scientific,

1999.

Roger Harrison, ed., "Book of Data", published for the Nuffield Foundation by Penguin

Books, 1973

Stephen Hawking, "A brief History of Time", Bantam Press, 1988

Stephen Hawking, "The Universe in a Nutshell", Bantam Press, 2001

Tony Hey and Patrick Walters, "The New Quantum Universe", Cambridge, 2003

Kris Heyde, "From Nucleons to the atomic nucleus - perspectives in Nuclear Physics",

Springer, 1998

Ben Jonson, “The Alchemist”, 1610.

http://www.gutenberg.org/dirs/etext03/lchms10.txt

http://en.wikipedia.org/wiki/The_Alchemist_(play ) (both accessed 20 April 2011)

William J. Kaufman III and Roger A. Freedman, "Universe", W.H.Freeman, 5th edition,

1999

Michael Kent, "Advanced Biology", Oxford, 2000

Lawrence Krauss, "Atom", Abacus, 2002

Don Lincoln, “The Quantum Frontier: the large hadron collider”, The Johns Hopkins

University Press, 2009.

Lucretius, “On the Nature of Things”, Translated by Ronald Latham, Penguin, 1951


http://en.wikipedia.org/wiki/The_Alchemist_(play

http://www.gutenberg.org/dirs/etext03/lchms10.txt


Katharina Lodders, "Solar System Abundances and Condensation Temperatures of the

Elements", The Astrophysical Journal, volume 591, p. 1220–1247, 2003.

The full paper is available at…http://solarsystem.wustl.edu/publications/2000-2009/p2003/

and the graph is available at…

http://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements (both accessed 6 March

2011)

Laurence Marschall, "The Supernova story", Princeton University Press, 1988

Stephen Mason, "Chemical Evolution", Clarendon Press, 1991

J.MacDonald and D.J.Mullan, “Big Bang Nucleosynthesis: The Strong Nuclear Force

meets the Weak Anthropic Principle”, Physical Review D, volume 80, p. 043507, 2009;

available at http://arxiv.org/PS_cache/arxiv/pdf/0904/0904.1807v2.pdf (accessed 6 April 2011).

Reference 4 in http://en.wikipedia.org/wiki/Dineutron (accessed 6 April 2011)

Ray Mackintosh, Jim Al-Khalili, Björn Jonson, and Teresa Peña, “Nucleus: A trip into the

heart of matter”, The Johns Hopkins University Press, 2001

George Marx, “Life in the Nuclear Valley”, Physics Education, 2001, vol. 36, p.375

Michael Nelkon and Philip Parker, "Advanced Level Physics", Heinemann, 4 th edition,

1977

NUBASE 2003, “The NUBASE evaluation of nuclear and decay properties”, G. Audi, O.

Bersillon, J. Blachot and A.H. Wapstra, Nuclear Physics A volume 729 (2003) p.3–128

The full paper is available in Nubase2003.pdf, or just the data in ASCII format, in

nubtab03.asc, from…

http://www.nndc.bnl.gov/amdc/nubase/filel.html or from….

http://www.nndc.bnl.gov/amdc/web/nubase_en.html

The table nubtab03 gives values of mass excess, and can be opened in Notepad, and

converted to EXCEL format. The table nubtab03 and mass.mas03 give the same mass

excess values of nuclides.

Oliver Sacks, “Uncle Tungsten: Memories of a Chemical Boyhood”, Picador, 2002.

Carl Sagan, "Cosmos", Macdonald Futura, 1980

Joseph Silk, "On the shores of the unknown", Cambridge, 2005



Simon Singh, "Big Bang", Fourth Estate, 2004

Timothy Paul Smith, “Hidden Worlds”, Princeton University Press, 2003

George Smoot, "Wrinkles in Time - the imprint of creation", Little, Brown and company,

1993

George Smoot, Nobel Prize address, “Cosmic Microwave Background Radiation

Anisotropies: Their discovery and utilisation”, published in Reviews of Modern Physics,

vol. 79, 1349, 2007, and also available at: http://rmp.aps.org/pdf/RMP/v79/i4/p1349_1

(accessed 7 Feb 2011)

Edwin F. Taylor and John Archibald Wheeler, “Spacetime Physics”, W.H.Freeman, 1966

Stuart Ross Taylor, "Destiny or Chance", Cambridge, 1998

Robert Turnbull, "The Structure of Matter", Blackie, 1979

George Wallerstein (editor), "Synthesis of the elements in stars: forty years of progress"

http://cococubed.asu.edu/papers/wallerstein97.pdf (accessed 21 Jan. 2011)

Steven Weinberg, "The First Three Minutes", Basic Books, 2nd edition, 1993

Walt Whitman, Song of Myself, Leaves of Grass (1881-82)

http://www.whitmanarchive.org/published/LG/1881/poems/27 (accessed 20 April 2011)

Chandra Wickramasinghe, “Cosmic Dragons: life and death on our planet”, Souvenir

Press, 2001

William S. C. Williams, "Nuclear and Particle Physics", Clarendon Press, 1991,

updated1997

WHW, or Stan Woosley, A. Heger and Thomas Weaver, “The evolution and explosion of

massive stars”, Reviews of Modern Physics, vol.74, October 2002, p 1015-1071

(shortened to ‘WHW’ in the text)

Stan Woosley and Thomas Janka, "The physics of core-collapse supernovae", Nature

Physics, volume 1, p. 147-154, 2005; available at

http://arxiv.org/ftp/astro-ph/papers/0601/0601261.pdf, and also as an external link in

http://en.wikipedia.org/wiki/Silicon_burning (both accessed 21 Jan. 2011)

Stan Woosley and Thomas Weaver, "The great supernova of 1987", in "Stars and

Galaxies: Citizens of the Universe", readings from Scientific American, Edited by Donald

E. Osterbrock, W. H. Freeman, 1990.


http://en.wikipedia.org/wiki/Silicon_burning

http://arxiv.org/ftp/astro-ph/papers/0601/0601261.pdf

http://www.whitmanarchive.org/published/LG/1881/poems/27

http://cococubed.asu.edu/papers/wallerstein97.pdf

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