Physics 2018: Great Ideas in Science:The Physics Module

Nuclear Physics Lecture Notes

Dr. Donald G. Luttermoser

East Tennessee State University

Edition 1.0

Abstract

These class notes are designed for use of the instructor and students of the course Physics 2018:Great Ideas in Science. This edition was last modified for the Fall 2007 semester.

III. Nuclear Physics

A. The Structure of Matter.

1. Matter exists in 3 different thermodynamic states:

a) A solid is a rigid body −→ takes a lot of energy to change

its shape. Solids can be classified into two types:

i) Crystalline solids have atoms that are struc-

tured in an orderly fashion.

ii) Amorphous solids have randomly arranged atoms.

b) A liquid is fluid in nature −→ moderate energy required

to change its shape.

c) A gas is also fluid in nature −→ little energy required to

change its shape.

2. If a gas gets hot enough, electrons circling the nucleus of the

atoms in the gas are “ripped” away from the nucleus =⇒ the gas

becomes ionized =⇒ ionized gas is called a plasma.

3. Matter consists of a distribution of particles (atoms and molecules).

a) Atoms consist of a nucleus surrounded by electrons (which

are negatively [–] charged). The nucleus consists of pro-

tons (positive [+] charge) and neutrons (no [0] charge).

b) H, He, Li, Be, B, C, N, O, etc. are the elements of the

periodic table = atoms. The number of protons in the

nucleus defines each atom.

c) For elements heavier than H, typically the number of neu-

trons is equal to the number of protons. Isotopes of

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atoms contain different numbers of neutrons in the nu-

cleus (e.g., 12C, 13C, and 14C are isotopes of carbon).

d) Neutral Atoms: # protons = # electrons.

i) If electrons are taken away from the atom such

that the number of protons exceeds the number of

electrons, the atom becomes a positive ion (e.g.,

H+ = H II = singly ionized hydrogen).

ii) If the number of electrons exceeds the number of

protons in the nucleus, the atom becomes a neg-

ative ion (e.g., H− = negative hydrogen ion, a

hydrogen atom with two electrons instead of one).

iii) The ionization stage of an atom can be labeled

in a variety of different ways:

=⇒ Roman numerals: I = neutral atom (e.g.,He I, neutral helium), II = singly ionized

(e.g., Fe II), III = doubly ionized (e.g., C III),

etc.

=⇒ ‘+’ exponents (positive ions): if no expo-

nent appears, then we have a neutral atom

(e.g., He, neutral helium), ‘+’ = singly ion-ized (e.g., Fe+), ‘3+’ = triply ionized (e.g.,

C 3+), etc.

=⇒ ‘−’ exponents (negative ions): ‘−’ = one ex-

tra electron (e.g., H−), ‘−−’ = two extra elec-

trons (e.g., C−−), etc.

e) Molecules are a collection of atoms that are bound to-

gether by molecular bonds.

i) Salt: NaCl (1 sodium atom + 1 chlorine atom).

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ii) Water: H2O (2 hydrogen atoms + 1 oxygen atom).

iii) Methane: CH4 (1 carbon atom + 4 hydrogen

atoms).

f) Molecules can adhere to each other through chemical

bonds making a structured lattice =⇒ solids.

B. An Overview of Nuclear Reactions.

1. In order to get nuclear particles to interact with each other, one

needs to collide these particles with a very high kinetic energy

(i.e., high speed).

a) Since the nucleus is composed of both neutrons (zero charge)

and protons (positive charge), two nuclei close together

will experience a strong repulsive force due to Coulomb’s

electric force law.

b) Remember that forces are derivatives of potential energy.

Hence if one wishes to bring two nuclei closer and closer

together, one must increase the kinetic energy of said nu-

clei to high enough values to overcome the electric poten-

tial field.

c) From thermodynamics heat is nothing more than the av-

erage kinetic energy of the particles that make up a system

(i.e., matter).

d) Hence, one way to get nuclei moving fast is to heat them

up.

2. Reactions in chemistry and physics can be one of two types:

a) Endothermic reactions absorb energy.

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b) Exothermic reactions release energy.

3. The element iron (Fe) has a nucleus that is the most stable of all

the atomic nuclei.

a) Elements lighter (i.e., less massive) than iron will produce

exothermic nuclear reactions when they fuse together to

make heavier elements =⇒ nuclear fusion.

b) Elements heavier (i.e., more massive) than iron will pro-

duce exothermic nuclear reactions when they break apart

to make lighter elements =⇒ nuclear fission.

4. The amount of energy released duing these exothermic reactions

is proportional to the amount of the mass difference between

the parent (original) and daughter (offspring) particles through

Einstein’s famous equation

E = mc2 . (III-1)

5. Conservation Laws.

a) If we assign a baryon number B of +1 to each baryon

(nucleon or hyperon) and –1 to an antinucleon or antihy-

peron, then in a closed system

∑B = constant. (III-2)

b) Similarly, if we assign a lepton number L of +1 to each

lepton (i.e., e−, µ, ν, etc.) and of –1 to antileptons (i.e.,

e+, µ, ν, etc.), then in a closed system

∑L = constant. (III-3)

c) A similar conservation law does not exist for bosons —

the mesons and field particles (i.e., photons).

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d) Charge must be conserved in a nuclear reaction.

e) Mass-energy, via E = m c2, must be conserved in a nuclear

reaction.

f) Momentum must be conserved. Hence a matter-antimatter

reaction must create two photons to conserve momentum

(e.g., e− + e+ → 2γ).

C. Thermonuclear Reactions.

1. In 1938, it became clear that the long-term energy source for

stars must be thermonuclear fusion reactions. In these reactions,

lighter elements burn to form heavier elements =⇒ nucleosyn-

thesis.

2. Two nuclei will fuse to form one nuclei if they come within

10−13 cm of each other — but they must be moving fast enough to

overcome the Coulomb repulsion that exists between like charged

particles.

a) Particles must be at a high temperature to be moving fast.

b) This high temp completely ionizes all of the nuclei.

c) Temps must build even more to get the kinetic energy to

overcome the Coulomb barrier.

3. In main sequence stars, H is fused into He. Since H is composed

of 1 baryon and He, 4 baryons (2p + 2n), 4 H nuclei must be

used to construct one He nuclei.4×mH = 4 × 1.0078 amu = 4.0312 amu

−mHe = 4.0026 amu

∆m = 0.0286 amu

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a) This mass deficit, ∆m, is converted into energy:

E = ∆m c2 = (0.0286) × (1.66 × 10−24 gm)

×(9.00 × 1020 cm2/s2)

= 4.3 × 10−5 erg

b) From this calculation, we see that the energy release effi-

ciency, η, of this H→He reaction is 0.0286/4.0312 = 0.0071

=⇒ only 0.71% of the original mass of H is converted to

energy!

c) With this in mind, we can rewrite Einstein’s famous equa-

tion as

E = η m c2, (III-4)

where η is the efficiency of the reaction and m is the initial

mass in the reaction.

d) We can calculate the total energy the Sun will release

during its main sequence lifetime. Since the reactions are

only being carried out in the core of the Sun and this core

contains about 10% of the Sun’s mass, the total energy

release will be

Etot = 0.1M� η c2

= 1.28 × 1051 ergs.

e) The present luminosity of the Sun is 3.90 × 1033 erg/s. If

the Sun’s luminosity remains somewhat constant while on

the main sequence, we can determine its main sequence

lifetime:

tMS(�) = Etot/L� = 3.28 × 1017 sec = 1.04 × 1010 yrs,

the Sun’s main sequence lifetime is about 10 billion years.

Since the Sun is currently 5 billion years old, it is at mid-

life.

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4. As previously mentioned, high temperatures are needed to over-

come the Coulomb repulsion of the charged nuclei. But how high

a temperature?

a) From classical physics, we can set the average kinetic en-

ergy of the particles involved equal to the thermal energy

of the particles and solve for the temperature (see page

334 of Carroll and Ostlie). This gives

Tclassical =2Z1Z2e

2

3kBr, (III-5)

where Z1e is the charge on particle 1, Z2e is the charge on

particle 2, and r is the distance where a nuclear reaction

will occur (about the size of a nucleus, 10−13 cm).

b) For 2 protons coming together (Z1 = Z2 = 1), this gives

a temperature of 1010 K, whereas the Sun’s central tem-

perature is only 1.58 × 107 K.

c) We could also investigate this in terms of energy. For

the Sun’s central temperature, each proton will have a

thermal energy of 1 keV, whereas the Coulomb potential

barrier is 1000 keV (1 MeV)! Not all the particles have

this energy, some are moving at much quicker velocities,

hence have higher thermal energies and temperatures fol-

lowing the Maxwell-Boltzmann distribution of velocities.

Unfortunately, the number of particles in the tail of this

Maxwellian is insufficient to account for the Sun’s lumi-

nosity.

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d) We can now turn to quantum mechanics to solve the prob-

lem. As discussed in the subsection on particle spin, in

reality elementary particles are not little billiard balls col-

liding with each other as a result of following trajectories.

Instead, they follow probability distributions described by

their wave functions. In quantum mechanics, there is a

small probability that wave functions can penetrate en-

ergy barriers that are higher than the energy of the wave

function. This effect is known as quantum tunneling.

e) Using quantum mechanics, we can describe a tempera-

ture needed to produce a sufficient number of tunneling

events to sustain a nuclear reaction (see page 335 of the

textbook) as

Tquantum =4µmZ2

1Z22e

4

3kBh2 , (III-6)

where µm is the reduced mass of the colliding “particles”

and h is Planck’s constant.

f) In this equation, two protons can come together (i.e., fuse)

at a temperature of 107 K, which is consistent with the

central temperature deduced for the Sun.

g) A more detailed calculation from statistical mechanics

shows that the bulk of the energy is being liberated by

reactions involving particles in the high energy tail of a

Maxwellian distribution.

i) Particles with energies at the Gamow Peak will

be the ones that supply most of the energy through

thermonuclear reactions.

ii) The Gamow Peak corresponds to a local maxi-

mum in the two probability functions: the e−E/kBT

III–8

Maxwell-Boltzmann distribution term and the e−bE−1/2

quantum tunneling penetration term, where,

b ≡23/2π2µ1/2

m Z1Z2e2

h.

iii) The Gamow Peak for a given temperature will

occur at the energy of

E◦ =

(bkBT

2

)2/3

, (III-7)

for the Sun, the Gamow Peak is at 6 keV.

5. By making use of statistical mechanics in conjunction with quan-

tum mechanics, stellar interior modelers set up power laws that

describe the energy production rate per unit mass of the form

ε = ε◦XiXxραT β , (III-8)

where the X ’s are the mass fractions of the fusing particles, and

ε◦, α. and β are constants that depend upon the reactions in-

volved (more to come on this).

D. Various Reaction Chains

1. Two different fusion processes convert H into He, the first is

important for stars with Tc<∼ 1.8 × 107 K (M <

∼ 1.3M�, ∼F5 V

star) and is called the proton-proton chain.

a) The first of this reaction chain is called the PP I chain:

Energy Reaction

Reaction Released Time

(MeV)1H + 1H −→ 2H + e+ + νe 1.442 1.4 × 109 yr1H + 2H −→ 3He + γ 5.493 6 sec

3He + 3He −→ 4He + 1H + 1H 12.859 106 yr

III–9

i) 1H = hydrogen atom (1 proton).

ii) 2H = heavy hydrogen (1 proton + 1 neutron) =

deuterium.

iii) 3He = light helium (2 protons + 1 neutron).

iv) 4He = helium (2 protons + 2 neutron) = alpha

particle.

v) γ = Gamma ray photon.

vi) e+ = positron (positive charge) = anti-electron

(antimatter). This positron interacts with the free

electrons in the core virtually immediately which

produces 2 additional gamma ray photons.

vii) νe = electron neutrino (neutral particle). The

neutrino’s absorption cross section is negligible and

leaves the stellar core (and star) immediately with-

out further interaction. The energy loss from the

neutrino is 0.263 MeV which has not been included

in the Energy Released column.

viii) In this PP I change, please note that the follow-

ing reaction can take place 1.4% of the time that

the first reaction takes place

1H + e− + 1H −→ 2H + νe

the so-called “pep” (proton-electron-proton) reac-

tion which releases 1.4 MeV and loses an additional

1.4 MeV in energy loss from the neutrino.

III–10

ix) Note that the Energy Released column is a com-

bination of the energy of any gamma rays created,

the kinetic energy = thermal energy picked up by

the resulting nuclei as a result of this reaction, and

the energy gained by the positron annihilation.

x) The times listed for the Reaction Time column is

that for the Sun’s central temperature.

xi) Note that the first two reaction steps must occur

twice before the last one can take place.

xii) The last step of this reaction chain is occurs

69% of the time in comparison to the other two

PP chains in the production of 4He in the Sun.

xiii) This PP I chain dominates the other PP chains

in stars with central temperatures of T <∼ 1.6 ×

107 K.

xiv) Note the long average time it takes for the first

and third reaction steps to take place for a single

particle. However when normalized by the total

number of particles in the Sun’s (or star’s) core,

about 9.0 × 1037 of these reactions take place per

second!

b) A second chain, called PP II, also can occur (31% of the

time in the Sun) in the production of 4He once the first

two steps of the PP I chain occur.

III–11

Energy ReactionReaction Released Time

(MeV)3He + 4He −→ 7Be + γ 1.586 1.0 × 106 yr

7Be + e− −→ 7Li + νe 0.861 0.4 yr7Li + 1H −→ 4He + 4He 17.347 104 yr

i) 7Be = beryllium-7 atom (4 protons + 3 neutrons).

ii) 7Li = lithium-7 atom (3 protons + 4 neutrons).

iii) The neutrino energy loss in this PP chain is 0.80 MeV.

iv) This PP II chain dominates the other PP chains

in stars with central temperatures of 1.6×107 <∼ T <

∼2.5 × 107 K.

v) The Energy Released and the Reaction Time have

the same meaning as they did for the PP I chain.

c) A third and final PP chain, called PP III, occurs only

0.3% of the time in the Sun in the production of 4He once

the first two steps of the PP I chain occurs and the first

step of the PP II chain occurs.

Energy ReactionReaction Released Time

(MeV)7Be + 1H −→ 8B + γ 0.135 70 yr

8B −→ 8Be + e+ + νe 17.98 1 sec8Be −→ 4He + 4He 0.095 1 sec

III–12

i) 8B = boron-8 atom (5 protons + 3 neutrons).

ii) 8Be = beryllium atom (4 protons + 4 neutrons).

iii) The neutrino energy loss in this PP chain is 7.2 MeV.

The Davis solar neutrino experiment, which de-

tected only 1/3-rd of the predicted solar neutrinos

was most sensitive to these 8B beta decay neutrinos.

Recently, neutrinos have been found to oscillate be-

tween the 3 known neutrino states which accounts

for the low detection rate of the Davis experiment.

iv) This PP III chain dominates the other PP chains

in stars with central temperatures of T >∼ 2.5×107 K

in the production of helium.

v) The Energy Released and the Reaction Time have

the same meaning as they did for the PP I chain.

2. For more massive stars (Tc>∼ 1.8 × 107 K, M >

∼ 1.3M�,

∼F5 V star), the CNO cycle is the dominant reaction chain.

a) This reaction chain uses carbon as a catalyst:

Energy ReactionReaction Released Time

(MeV)12C + 1H −→ 13N + γ 1.95 1.3 × 107 yr

13N −→ 13C + e+ + νe 2.22 7 min13C + 1H −→ 14N + γ 7.54 2.7 × 106 yr14N + 1H −→ 15O + γ 7.35 3.2 × 108 yr

15O −→ 15N + e+ + νe 2.71 82 sec15N + 1H −→ 12C + 4He 4.96 1.1 × 105 yr

i) 12C = carbon-12 (6 protons + 6 neutrons).

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ii) 13C = carbon-13 (6 protons + 7 neutrons).

iii) 13N = nitrogen-13 (7 protons + 6 neutrons) [ra-

dioactive].

iv) 14N = nitrogen-14 (7 protons + 7 neutrons).

v) 15N = nitrogen-15 (7 protons + 8 neutrons).

vi) 15O = oxygen-15 (8 protons + 7 neutrons) [ra-

dioactive].

vii) 16O = oxygen-16 (8 protons + 8 neutrons).

viii) The neutrino energy loss in the 13N beta de-

cay is 0.710 MeV and the neutrino energy loss is

1.000 MeV for the second 15O beta decay.

ix) The Energy Released and the Reaction Time have

the same meaning as they did for the PP I chain.

b) Note that this reaction sequence does not make any new

elements other than He!

c) For the last step in the CNO cycle, an additional set of

reactions can take place:

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Energy ReactionReaction Released Time

(MeV)15N + 1H −→ 16O + γ 12.126 1.0 × 107 yr16O + 1H −→ 17F + γ 0.601 3.0 × 1010 yr

17F −→ 17O + e+ + ν 2.762 3 min17O + 1H −→ 14N + 4He 1.193 2.0 × 1011 yr

i) 17F = fluorine-17 (9 protons + 8 neutrons).

ii) 18O = oxygen-16 (8 protons + 10 neutrons).

iii) The neutrino energy loss in the 17F beta decay is

0.94 MeV.

iv) The resulting 14N isotope can then be used back

in the

14N + 1H −→ 15O + γ

reaction in the primary CNO cycle.

3. The thermonuclear reaction rate, ε (in erg/gm/s), is very sen-

sitive to temperature. For the two hydrogen to helium reaction

chains (i.e., the proton-proton chain and CNO cycle), we can

write Eq. (IV-48) as two separate equations:

εpp = ε◦ ρ X2(

T

106

)β

(III-9)

εcc = ε◦ ρ XXCN

(T

106

)β

, (III-10)

where X is the mass fraction of hydrogen (as defined on page IV-8

of the course notes), XCN is the weighted average of the combined

mass fraction of carbon and nitrogen (since these species are the

III–15

two lead-off species of the two CNO cycles), note that typically

XCN =1

3Z , (III-11)

with the Z being the metalicity mass fraction. Finally, ε◦ and α

are temperature dependent constants given in the following table

(from Bosmas-Crespin, Fowler, Humblet 1954, Bull. Soc. Royale

Sciences Liege, No. 9-10, 327).

εpp εcc

T/106 log ε◦ α T/106 log ε◦ α

4 – 6 –6.84 6 12 – 16 –22.2 206 – 10 –6.04 5 16 – 24 –19.8 18

9 – 13 –5.56 4.5 21 – 31 –17.1 16

11 – 17 –5.02 4 24 – 36 –15.6 1516 – 24 –4.40 3.5 36 – 50 –12.5 13

In this table, for overlapping temperatures, a weighted average

is used to get the final rate.

4. We will see shortly, that when a main sequence uses up all of its

H fuel in its core, the now He-rich core will contract and heat

up. When temperatures exceed 108 K, helium fusion can begin.

Helium fuses via the triple-α process. The ash of this reaction

is carbon.

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a) The reaction is as follows:

Energy

Reaction Released(MeV)

4He + 4He ←→ 8Be + γ -0.09218Be + 4He −→ 12C + γ 7.37

i) Since this reaction chain is not occurring in the

Sun, we have not reported on any reaction times

here since they are very temperature dependent.

ii) As can be seen, three α-particles (i.e., He nuclei)

fuse to become one carbon nuclei.

iii) 8Be is unstable and quickly decays, so there is

not much berylium around for the 2nd chain to take

place. For every 1 berylium nuclei, there are 1010

α-particles, however this ratio is more than enough

to release enough energy to power a red giant star.

Note that this reaction actually drains energy away

from the gas =⇒ it is an endothermic reaction.

All reactions we have mentioned up to now have

been exothermic =⇒ they release energy.

b) The 3α process has an even greater sensitivity to temper-

ature than the CNO cycle:

ε3α ≈ 10−8ρ Y 3(

T

106

)30

erg/gm/s, (III-12)

where Y is the fractional abundance of helium per unit

mass.

c) This is the way the Universe makes carbon. As such,

the C atoms that make up our DNA were created in an

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ancient red giant star that no longer exists. To quote Carl

Sagan, we are star stuff!

d) Should the core of a red giant obtain temperatures that

exceed a few hundred million Kelvins, another reaction

can take place via an alpha (α) capture:

12C + 4He −→ 16O + γ

which releases 7.161 MeV of energy.

e) Most of the 16O in the Universe is made in this fashion.

5. Finally, if somewhat higher temperatures are ever encountered

inside a star, which happens during stellar evolution of massive

stars, even heavier elements can be created from the fusion of

additional α-particles and α-particle by-products:

Energy Minimum

Reaction Released Temperature(MeV) Required (106 K)

14O + 4He −→ 20Ne + γ 4.730 70020Ne + 4He −→ 24Mg + γ 9.317 150024Mg + 4He −→ 28Si + γ 9.981 1800

28Si + 4He −→ 32S + γ 6.948 250032S + 4He −→ 36Ar + γ 6.645 350012C + 12C −→ 24Mg + γ 13.930 80016O + 16O −→ 32S + γ 16.539 2000

a) Various silicon burning reactions can occur at tempera-

tures exceeding 3 × 109 K. Silicon burning produces the

iron (Fe) group elements.

b) Once Fe is formed, reactions that produce heavier ele-

ments are all endothemic and have a tough time forming

via the standard thermonuclear burning. Such elements,

III–18

and the elements not built upon α-particles, are created

via the r- (for rapid neutron capture) and s- (for slow

neutron capture) processes. These processes will be dis-

cussed in the supernovae section of the course notes (i.e.,

§VII).

c) The reaction times of this heavy element nucleosynthesis

will be discussed in the stellar evolution sections of the

notes.

III–19