Chapters 5-6: Strong Force & Nuclear Structure

Chapters 5-6: Strong Force & Nuclear Structure

Abby Bickley

University of Colorado

NCSS ‘99

Additional References:Choppin (CLR), Radiochemistry and Nuclear Chemistry, 2nd Edition, Chapter 11Friedlander (FKMM), Nuclear and Radiochemistry, 3rd Edition, Chapter 10

2

Particle Classifications• Fermions:

– Spin 1/2 particles

– Obey Pauli Exclusion Principle

• No more than one fermion can occupy the same quantum state

– Antisymmetric wave functions

– Two classifications

• Hadrons - interact via strong nuclear force; composed of quarks; neutrons, protons, etc.

• Leptons - fundamental particles, ie no substructure; electron, muon, tau, neutrino.

• Bosons:

– Integer spins

– Obey Bose-Einstein Statistics

• Any number of bosons can share the same quantum state

– Symmetric wave functions

– Force mediators- Photons, gluons, etc.

3

Fundamental Forces

• Virtual particles - exchange particles that exist for a short period of time to convey force; existence must obey Heisenberg Uncertainty Principle

Relative Strength

Range Exchange Particle

Gravity 10-38 ∞ Graviton

Weak 10-13 <10-18 m Z0,W+,W-

Electromagnetic 10-2 ∞ photon

Strong 1 10-14 m gluon

(pion)

4

Heisenberg’s Uncertainty Principle

• Conservation of energy can only be violated by E and t as long as the uncertainty principle remains valid.

• How far can an exchange particle travel without violating HUP?

• Assume exchange particle moves at the speed of light, c.

• Insert HUP for t and E=mc2

€

E ⋅Δt ≥ h

€

t ≤ h /ΔE

€

R = c ⋅Δt

€

R ≤ h /mc

5

Problem

• If the effective range of the weak force is 10-18 m, what is the maximum allowed mass of the exchange particle?

Answer: (GeV/c2)

6

Problem

• If the effective range of the weak force is 10-18 m, what is the maximum allowed mass of the exchange particle?

Answer: (GeV/c2)

€

R ≤h

mc ⇒ m ≤

h

Rc

m ≤4.136 ×10−15eV ⋅s

2π ⋅10−18m ⋅2.998 ×108m /s

m ≤ 2.20 ×10−6eV ⋅s2 /m2

m ≤198GeV /c 2

7

Properties of the Strong Force

• Range: less than nuclear radius, <1.4fm• Attractive: on the distance scale of 1fm, overcomes

coulombic force to hold charged protons together• Repulsive: on the distance scale of <0.5fm• Charge Independent: interaction is independent of

nucleon electrical charge, ie p-n = p-p, n-n

r

V(r)

8

Charge Independence

A Nucleus Total B.E.

(MeV)

Coulomb Energy (MeV)

Net Nuclear B.E. (MeV)

3 3H -8.486 0 -8.4863He -7.723 0.829 -8.552

13 13C -97.10 7.631 -104.73413N -94.10 10.683 -104.770

23 23Na -186.54 23.13 -209.6723Mg -181.67 27.75 -209.42

41 41Ca -350.53 65.91 -416.4441Sc -343.79 72.84 -416.63

9

The Nucleus• As chemist’s what do we already know

about the nucleus of an atom?

10

The Nucleus• As chemist’s what do we already know

about the nucleus of an atom?– Composed of protons and neutrons– Carries an electric charge equivalent to the

number of protons & atomic number of the element

– Protons and neutrons within nucleus held together by the strong force

• Any model of nuclear structure must account for both Coulombic repulsion of protons and Strong force attraction between nucleons

11

Two Nucleon Systems• Combinatorics gives us three

possible states, but only one occurs in nature!– nn

• Unbound and comes apart easily• Free neutrons decay on the time scale of

~10min

– pp• More unstable than nn due to Coulomb

repulsion

– pn - deuteron• Stable and naturally occuring• Spins of n and p align parallel in ground

state configuration• Non-spherical structure

12

Chart of the Nuclides

13

Empirical Observations• Chart of the nuclides:

– 275 stable nuclei– 60% even-even – 40% even-odd or odd-even– Only 5 stable odd-odd nuclei

21

H, 63Li, 10

5B, 147N, 50

23Va (could have large t1/2)

• Nuclei with an even number of protons have a large number of stable isotopesEven # protons Odd # protons50Sn:10 (isotopes) 47Ag: 2 (isotopes)48Cd: 8 51Sb:252Te: 8 45Rh:1

49In:153I: 1

• Roughly equal numbers of stable even-odd and odd-even nuclei

14

Implications for Nuclear Models I

1. Proton-proton and neutron-neutron pairing must result in energy stabilization of bound state nuclei

2. Pairing of protons with protons and neutrons with neutrons results in the same degree of stabilization

3. Pairing of protons with neutrons does not occur (nor translate into stabilization)

15

Problem• Which nucleus is stable?

12B 12C 12N

p n p np n

16

Problem• Which nucleus is stable?

17

Chart of the nuclides• Light elements: N/Z = 1• Heavy elements: N/Z 1.6

– Implies simple pairing not sufficient for stability

• Neutron Rich: (N>Z)– N>Z: nucleus will - decay to

stability– N>>Z: neutron drip line

• Proton Rich: (N<Z)– N<Z: nucleus will + decay or

electron capture to achieve stability

– N<<Z: proton drip line (very rare)

Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.

18

Implications for Nuclear Models II4. Pairing not sufficient to achieve stability

Why?Coulomb repulsion of protons grows with Z2:

Nuclear attractive force must compensate all stable nuclei with Z > 20 contain more neutrons than protons

€

ECoul =3e2Z 2

5r0A1/ 3

ECoul =1.44Z1Z2

R1 + R2( )

19

General Nuclear Properties I• Binding energy per nucleon ~ constant for all stable nuclei

• Implies all nucleons in the nucleus do not interact with one another • If they did the BE per nucleon would be proportional to the mass

number


8.9

7.4Mass Number

MeV

/u

20

General Nuclear Properties II• Nuclear radius is proportional to the cube root of the mass

r = r0 A1/3 Eq. 2

• Experimental studies show ~uniform distribution of the charge and mass throughout the volume of the nucleus


Rl = Half density radius

dl = skin thickness

21

Simple Nuclear Potential Well

• Potential energy of nucleons as they approach the nucleus• Neutron - feels no effective force until reaches surface of nucleus; constant

attractive force at interior of nucleus.• Proton - coulombic repulsion as approaches nucleus; constant attractive

force at interior of nucleus; coulombic repulsion from other protons decreases depth of potential well relative to neutron well.

22

Liquid Drop Model (1935)• Treats nucleus as a statistical assembly of neutrons

and protons with an effective surface tension - similar to a drop of liquid

• Rationale:– Volume of nucleus number of nucleons

• Implies nuclear matter is incompressible

– Binding energy of nucleus number of nucleons• Implies nuclear force must have a saturation character, ie each

nucleon only interacts with nearest neighbors

• Mathematical Representation:– Treats binding energy as sum of volume, surface and

Coulomb energies:

€

EB (MeV ) = c1A 1− kN − Z

A

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎡

⎣ ⎢

⎤

⎦ ⎥− c2A

2 / 3 1− kN − Z

A

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎡

⎣ ⎢

⎤

⎦ ⎥+ −

c3Z2

A1/ 3+c4Z

2

A

⎡

⎣ ⎢

⎤

⎦ ⎥+ δ

23

Liquid Drop Model Components I

• Volume Energy:– Binding energy of nucleus number of nucleons– Correction factor accounts for symmetry energy (for a given A the

binding energy due to only nuclear forces is greatest for nuclei with equal numbers of protons and neutrons)

• Surface Energy:– Nucleon at surface are unsaturated reduce binding energy

surface area– Surface-to-volume ratio decreases with increasing nuclear size

term is less important for large nuclei

Volume Energy Surface Energy

c1 = 15.677 MeV, c2 = 18.56 MeV, c3 = 0.717 MeV, c4 = 1.211 MeV, k = 1.79 €


A

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎡

⎣ ⎢

⎤

⎦ ⎥− c2A

2 / 3 1− kN − Z

A

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎡

⎣ ⎢

⎤

⎦ ⎥+ −

c3Z2

A1/ 3+c4Z

2

A

⎡

⎣ ⎢

⎤

⎦ ⎥+ δ

24

Liquid Drop Model Components II

• Coulomb Energy:– Electrostatic energy due to Coulomb repulsion between protons – Correction factor accounts for diffuse boundary of nucleus

(accounts for skin thickness of nucleus)

• Pairing Energy:– Accounts for added stability due to nucleon pairing

– Even-even: = +11/A1/2

– Even-odd & odd-even: = 0– Odd-odd: = -11/A1/2

€


A

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎡

⎣ ⎢

⎤

⎦ ⎥− c2A

2 / 3 1− kN − Z

A

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎡

⎣ ⎢

⎤

⎦ ⎥+ −

c3Z2

A1/ 3+c4Z

2

A

⎡

⎣ ⎢

⎤

⎦ ⎥+ δ

Coulomb Energy

c1 = 15.677 MeV, c2 = 18.56 MeV, c3 = 0.717 MeV, c4 = 1.211 MeV, k = 1.79

Pairing Energy

25

Problem

• Using the binding energy equation for the liquid drop model, calculate the binding energy per nucleon for 15N and 148Gd.

• Compare these results with those obtained by calculating the binding energy per nucleon from the atomic mass and the masses of the constituent nucleons.

26

Problem: Answers

• 15N = 6.87 MeV/nucleon

• 148Gd = 8.88 MeV/nucleon

• 15N = 7.699 MeV/nucleon

• 148Gd = 8.25 MeV/nucleon

27

Mass Parabolas• Represent mass of atom as difference between sum of

constituents and total binding energy:

• Substitute binding energy equation for EB and group terms by power of Z:

• For a given number of nucleons (A) f1, f2 and f3 are constants

• Functional form represents a mass-energy parabola– Single parabola for odd A nuclei ( = 0)

– Double parabola for even A nuclei ( = ±11/A1/2)

€

M = ZMH + (A − Z)Mn − EB

€

M = f1Z2 + f2Z + f3 + δ

f1 = 0.717A−1/ 3 +111.036A−1 −132.89A−4 / 3

f2 =132.89A−1/ 3 −113.029

f3 = 951.958A −14.66A2 / 3

28

Mass Parabolas Example 1A = 75 or 157

• Parabola Vertex:– ZA=[-f2 / 2f1]– Minimum mass & Maximum EB

• Used to find mass and EB difference between isobars

• Nuclear charge of minimum mass is derivative of M Eqn => not necessarily integral

• Comparison of Z = 75 and Z = 157– Valley of stability broadens with

increasing A

• For a given value of odd-A only one stable nuclide exists

• In odd-A isobaric decay chains the -decay energy increases monotonically Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.

29

Mass Parabolas Example 2A = 156

• Eqn results in two mass parabola for a given even value of A

• For a given value of even-A their exist 2 (or 3) stable nuclides

• In this figure both 156Gd and 156Dy are stable

• In even-A isobaric decay chains the -decay energies alternate between small and large values

• This model successfully reproduces experimentally observed energy levels

• BUT…….Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.

30

Problem

• Find the nuclear charge (ZA) corresponding to the maximum binding energy for:A = 157, 156 and 75

• To which isotopes do these values correspond?

• Compare your results with the mass parabolas on slides 28 & 29.

31

Problem: Answers

• Find the nuclear charge (ZA) corresponding to the maximum binding energy for:

A = 157, 156 and 75

ZA = 64.69, 64.32, 33.13

• To which isotopes do these values most closely correspond?157

65Tb, 15664Gd, 75

33As

• Compare your results with the mass parabolas on slides 28 & 29.

32

Fermi Gas Model• Model emphasizes free particle character of nucleons & allows average behavior of lg

nuclei to be described by thermodynamics• Assume nucleus is composed of a degenerate Fermi gas of p & n

– Degenerate - particles occupy lowest possible energy states– Fermi gas - all particles obey Pauli Principle

• Fermi Wavenumber - highest state occupied by nucleons• Fermi Energy - gas is characterized by kinetic energy of highest filled state

• When # n>p then must calculate Fermi energies of neutrons and protons separately

€

k f =π

r0

2Z

3πA

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 3

, E f =k f

2h2

2m , KE =

3

5E f

€

E fproton ≅ 53⋅

Z

A

⎛

⎝ ⎜

⎞

⎠ ⎟2 / 3

MeV , E fneutron ≅ 53⋅

A − Z

A

⎛

⎝ ⎜

⎞

⎠ ⎟2 / 3

MeV

33

Fermi Gas Potential Well

• EF,p - Fermi energy of proton• EF,n - Fermi energy of neutron• EC - coulombic energy• B - binding energy• U0 - depth of potential well• Fermi level - uppermost filled energy level, approximately -8MeV.

Excitedstates

34

Problem

• What is the average Fermi energy of a neutron in a 208Pb nucleus?

35

Problem

• What is the average Fermi energy of a neutron in a 208Pb nucleus?

€

k f =π

r0

2Z

3πA

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 3

=π

1.2 fm

2 ⋅89

3π ⋅208

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 3

=1.145 fm−1

E f =k f h( )

2

2m=

1.18 fm−1 ⋅197.33MeV ⋅ fm( )2

2 ⋅939.566MeV= 27.2MeV

36

Magic Numbers

• Nuclides with “magic numbers” of protons and/or neutrons exhibit an unusual degree of stability

• 2, 8, 20, 28, 50, 82, 126

• Suggestive of closed shells as observed in atomic orbitals

• Analogous to noble gases

• Much empirical evidence was amassed before a model capable of explaining this phenomenon was proposed

• Result = Shell ModelFriedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.

37

Atomic Orbitals History• Plum Pudding Model: (Thomson, 1897)

– Each atom has an integral number of electrons whose charge is exactly balanced by a jelly-like fluid of positive charge

• Nuclear Model: (Rutherford, 1911)– Electrons arranged around a small massive core of protons and

neutrons* (added later)

• Planetary Atomic Orbitals: (Bohr, 1913)– Assume electrons move in a circular orbit of a given radius around

a fixed nucleus

– Assume quantized energy levels to account for observed atomic spectra

– Fails for multi-electron systems

• Schrodinger Equation: (1925)– Express electron as a probability distribution in the form of a

standing wave function

38

Atomic Orbitals• Schrodinger equation solution reveals quantum numbers

– n = principal, describes energy level

– l = angular momentum, 0n-1 (s,p,d,f,g,h…)

– m = magnetic, - l l, describes behavior of atom in external B field

– ms = spin, -1/2 or 1/2

• Pauli Exclusion Principle: e-’s are fermions no two e-’s can have the same set of quantum numbers

• Hund’s Rule: when electrons are added to orbitals of equal energy a single electron enters each orbital before a second enters any orbital; the spins remain parallel if possible.

• Example: C = 1s22s22px12py

1

1s

2s

3s

2p

39

Shell Structure of NucleusHistorical Evolution

• Throughout 1930’s and early 1940’s evidence of deviation from liquid drop model accumulates

• 1949: Mayer & Jenson – Independently propose single-particle orbits

• Long mean free path of nucleons within nucleus supports model of independent movement of nucleons

• Using harmonic oscillator model can fill first three levels before results deviate from experiment (2,8,20 only)

– Include spin-orbit coupling to account for magic numbers• Orbital angular momentum (l) and nucleon spin (±1/2) interact • Total angular momentum must be considered • (l+1/2) state lies at significantly lower energy than (l-1/2) state• Large energy gaps appear above 28, 50, 82 & 126

40

Single Particle Shell Model

• Assumes nucleons are distributed in a series of discrete energy levels that satisfy quantum mechanics (analogous to atomic electrons)

• As each energy level is filled a closed shell forms• Protons and neutrons fill shells and energy levels

independently• Mainly applicable to ground state nuclei• Only considers motion of individual nucleons

41

Shell Model and Magic Numbers

• Magic numbers represent closed shells

• Elements in periodic table exhibit trends in chemical properties based on number of valence electrons (Noble gases:2,10,18,36..)

• Nuclear properties also vary periodically based on outer shell nucleons

42

Pairing

• Just as electrons tend to pair up to form a stable bond, so do like-nucleons; pairing results in increased stability

a) Even-Z and even=N nuclides are the most abundant stable nuclides in nature (165/275)

b) From 15O to 35Cl all odd-Z elements have one stable isotope while all even-Z elements have three

c) The heaviest stable natural nuclide is 20983Bi (N=126)

d) The stable end product of all naturally occurring radioactive series of elements is Pb with Z=82

43

Shell Model Evidence - Abundances

• The most abundantly occurring nuclides in the universe (terrestrial and cosmogenic) have a magic number of protons and/or neutrons

• Large fluctuations in natural abundances of elements below 19F are attributed to their use in thermonuclear reactions in the prestellar stage

€

816O(8p,8n), 14

28Si(14 p,14n), 3888Sr(38p,50n),

3989Y (39p,50n), 40

90Zr(40p,50n), 50118Sn(50p,68n),

56138Ba(56p,82n), 57

139La(57p,82n), 58140Ce(58p,82n), 82

208Pb(82p,126n)

44

Shell Model Evidence - Abundances

45

Shell Model Evidence - Stable Isotopes

• The number of stable isotopes of a given element is a reflection of the relative stability of that element. Plot of number of isobars vs N shows peaks at – N = 20, 28, 50, 82

• A similar effect is observed as a function of Z

Stable Isobars

# of

Iso

bars

# of Neutrons

46

Shell Model Evidence - Alpha Decay

• Shell Model predictions:– Nuclides with 128 neutrons =>

• short half life• Emit energetic

– Nuclides with 126 neutrons =>• Long half life• Emit low energy

€

zAX→2

4He+ z−2A−4X

€

84212Po(128n)→ 82

208Pb(126n)+24He, t1/2 = 46s, Eα = 8.78MeV

84210Po(126n)→ 82

206Pb(124n)+24He, t1/2 =138.4d, Eα = 5.31MeV

85213At(128n)→ 83

209Bi(126n)+24He, t1/2 =11μs, Eα = 9.08MeV

85211At(128n)→ 83

207Bi(124n)+24He, t1/2 = 7.2h, Eα = 5.87MeV

47

Shell Model Evidence - Beta Decay

• If product contains a magic number of protons or neutrons the half-life will be short and the energy of the emitted will be high

€

zAX → β −+z−1

AX, zAX → β ++z+1

AX

Sc 40+

0.86s

Sc 41 +

0.60s

Sc 42 +

0.68s

Ca 39+

0.86s

Ca 4096.94

Ca 41EC

1.03E6 y

K 38+

7.63m

K 3993.25

K 40 -

1.28E9y

N = 19 N = 20 N = 21

Z = 21

Z = 20

Z = 19

48

Shell Model Evidence - Neutrons

• Neutrons do not experience Coulomb barrier even thermal neutrons (low kinetic energy) can penetrate the nucleus

• Inside nucleus neutron experiences attractive strong force and becomes bound

• To escape the nucleus a neutron’s KE must be greater than or equal to the nuclear potential at the surface of the nucleus

• Observation: the absorption cross section for 1.0MeV neutrons is much lower for nuclides containing 20, 50, 82, 126 neutrons compared to those containing 19, 49, 81, 125 neutrons

Nuclide (barns)

n-capture88

38Sr(50n) 5.8E-3

8738Sr(49n) 16

13654Xe(82n) 0.16

15654Xe(81n) 2.65E6

49

Shell Model Evidence - Energy

• The energy needed to extract the last neutron from a nucleus is much higher if it happens to be a magic number neutron

• Energy needed to remove a neutron – 126th neutron from 208Pb = 7.38 MeV– 127th neutron from 209Pb = 3.87 MeV

50

Shell Model Evidence - Nucleon Interactions

• Every nucleon is assumed to move in its own orbit independent of the other nucleons, but governed by a common potential due to the interaction of all of the nucleons

• Implication: in ground state nucleus nucleon-nucleon interactions are negligible

• Implication: mean free path of ground state nucleon is approximately equal to the nuclear diameter

• Experimental data does not support this conclusion!!!

51

Shell Model Evidence - Nucleon Interactions

• Scattering experiments show frequent elastic collisions

• Implication: mean free path << nuclear radius• Explanation: Pauli exclusion principle prohibits

more than two protons or neutrons from occupying the same orbit (protons and neutrons are fermions)

• Why Pauli?: nucleon-nucleon collisions result in momentum transfer between the participants BUT all lower energy quantum states are filled occurrence forbidden

• Severely limits nucleon-nucleon collision rate

52

Nuclear Potential Well• Nucleon orbit = nucleon quantum state

– Similar to quantum state of valence electron

BUT

– Nucleon “feels” average total effect of interactions of all nucleons

– Implication: nuclear potential is the same for all nucleons

• Strong Force– All nucleons (regardless of their electrical charge)

attract one another

– Attractive force is short range and falls rapidly to zero outside of the nuclear boundary (~1 fm)

53

Nuclear Potential - Protons

• Protons do experience a Coulomb barrier a proton must have kinetic energy equal or greater than ECoul to penetrate the nucleus

• If Eproton< ECoul proton will back scatter

• Inside nucleus proton experiences attractive strong force and becomes bound

• To escape the nucleus a proton’s kinetic energy must be greater than or equal to ECoul (in the absence of quantum tunneling)

54

Nuclear Potential - Neutrons

• Neutrons do not experience Coulomb barrier even thermal neutrons (low kinetic energy) can penetrate the nucleus

• Inside nucleus neutron experiences attractive strong force and becomes bound

• To escape the nucleus a neutron’s kinetic energy must be greater than or equal to the nuclear potential at the surface of the nucleus

55

Nuclear Potential WellDepth of well representsbinding energy

56

Nuclear Potential Functions• Square Well Potential

• Harmonic Oscillator Potential

• Woods-Saxon

• Exponential Potential

• Gaussian Potential

• Yukawa Potential:

€

V (r) = −V0 (r ≤ R), V (r) = 0 (r ≥ R)

€

V (r) = −V0 1−r

R

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎡

⎣ ⎢

⎤

⎦ ⎥

€

V (r) = −V0e−r

R

€

V (r) = −V0e−r 2

R 2

€

V (r) = −V0

e−r

R

r +1.0

R

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

Note: R = nuclear radiusr = distance from center of nucleus

€

V (r) =V0

1+ er−R

a

⎛

⎝ ⎜

⎞

⎠ ⎟

57

Nuclear Potential Functions

Exact shape of well is uncertain and dependson mathematical functionassumed for the interaction

Square Well

Gaussian

Exponential

Yukawa

58

Neutron vs Proton Potential Wells

Coulomb repulsion prevents potential well from being as deep for protons as for neutrons

59

Quantized Energy Levels• Schrodinger Equation: developed to find wave functions

and energies of molecules; also can be applied to the nucleus

• Choose functional form of nuclear potential well and solve Schrodinger Equation:

H = E • Wave equation allows only certain energy states defined

by quantum numbers

• n = principal quantum number, related to total energy of the system

• l = azimuthal (radial) quantum number, related to rotational motion of nucleus

• ms = spin quantum number, intrinsic rotation of a body around its own axis

60

Angular Momentum• Associated with the rotational motion of an

object • Like linear motion, rotational motion also

has an associated momentum • Orbital angular momentum:

pl = mvrr• Spin angular momentum:

ps = Irot

• A vector quantity always has a distinct orientation in space

61

Magnetic Quantum Effects - Spin• A rotating charge gives rise to a magnetic moment

(s). • Electrons and protons can be conceptualized as

small magnets• Neutrons have internal charge structure and can

also be treated as magnets• In the absence of a B-field magnets are disoriented

in space (can point any direction)• In the presence of a B-field the electron, proton

and neutron spins are oriented in specific directions based upon quantum mechanical rules

62

Spin Angular MomentumNo External B-field Applied External B-field

Project spin angular momentum onto the field axesAllowed values are units of hbarps(z) = hbar ms

{

63

Spin Angular Momentum• Quantum mechanics requires that the spin

angular momentum of electrons, protons and neutrons must have the magnitude

• s is the spin quantum number

• For protons and neutrons (just like for electrons) spin is always 1/2

€

ps = h s(s+1)

64

Magnetic Quantum Effects - Orbitals• The orbital movement of an atomic electron or a

nucleon gives rise to another magnetic moment (l) • This magnetic moment also interacts with an

external B-field in a similar manner to the spin magnetic moment

• Quantum mechanics governs how the orbital plane may be oriented in relation to the external field

• The orbital angular momentum vector (pl) can only be oriented such that its projection onto the z-axis (field axis) has values

pl (z) = hbar ml • Where ml = magnetic orbital quantum number

ml = -l, -l+1, -l+2….0…. l-2, l-1, l

65

Orbital Angular Momentum

• Project orbital angular momentum onto the field axes

• Allowed values are units of hbar• pl(z) = hbar ml

3 h

2 h

1 h

0 h

-1 h

-2 h

-3 h

Orientations of ml in a magnetic field

B-field axispl

pl(z)

66

Orbital Angular Momentum• Quantum mechanics requires that the orbital

angular momentum of electrons, protons and neutrons must have the magnitude

• l is the orbital quantum number • Allowed values of l:

– Nucleons: 0 l (≤ or ≥) n– Electrons: 0 l < (n-1)

• For nucleons (but not electrons) l can exceed n

€

pl = h l(l +1)

67

Orbital Angular Momentum

• The numerical values of the orbital angular momentum quantum number (l) are designated by the familiar spectroscopic notation

• Remember: l can only have positive integral values (including 0)

l 0 1 2 3 4 5 6 7

symbol s p d f g h i j

68

Quantum States

69

Energy Level Diagram

Isotropic Harmonic Oscillator Levels

1s

1f

2s1d

1p

2d1g

2p

2f

1h

3s

1i3p

nl nucleons [total]

2

614

210

6

22

21018

266

14

2

4034

2018

8

92

706858

138112106

70

Problem

• Using the harmonic oscillator energy levels what is the level ordering for – 9

4Be

– 3115P

– 5927Co

71

Problem

• Using the harmonic oscillator energy levels what is the level ordering for – 9

4Be: protons (1s2 1p2)

neutrons (1s2 1p3)

– 3115P: protons (1s2

1p6 1d7 )

neutrons (1s2 1p6 1d8 )

– 5927Co: protons (1s2

1p6 1d10 2s2 1f7 )

neutrons (1s2 1p6 1d10 2s2 1f12 )

72

Spin Orbit Coupling• Spin and orbital angular

momenta are vector quantities vector coupling occurs to form a resultant vector pj

• Total angular momentum:

pj = pl + ps

• Coupling splits degeneracy of orbital angular momentum states

73

Spin Orbit Coupling• For nucleons and electrons the orbital and spin angular

momenta add vectorially to form a resultant vector (pj)

pj = pl + ps

• The resultant is oriented towards an external magnetic field so that the projections on the field axis are

pj(z) = hbar mj

• The magnitude of pj is

pj = hbar [j(j+1)]1/2

j = l s• j is the total angular quantum number of the particle

74

Total Angular Momentum

• Total angular quantum number (j) can have two different values for each orbital quantum number (l)

• However, j can only have positive values!!

• Implication: when l = 0, only j=1/2 is allowed

• All allowed values of j are half-integers

j = 1/2, 3/2, 5/2, 7/2…

75

Problem

• What are the allowed values of j for a nucleon with

l = 1, s = 1/2

76

Problem

• What are the allowed values of j for a nucleon with

l = 1, s = 1/2

• Answer: j = 1/2 or 3/2

77

Total Angular Momentum• Example: n = 1, l = 1, s = 1/2• Standard atomic notation:

– Electron in 1p1 state– leading 1=principal quantum number– p = orbital angular quantum number– superscript 1 = spin quantum number

• Standard nuclear notation:– Nucleon in 1p1/2 state or 1p3/2 state– But we know that spin orbit coupling splits the degeneracy of

the 6 existing p states – This results in a two-fold degenerate 1p1/2 state and a four-

fold degenerate 1p3/2 state– Which is lower in energy????

78

Level Ordering of States• Split degenerate

states with higher j are always more stable than those with lower j– Energetically:

1p1/2 > 1p3/2

• Neutrons and protons fill levels independently

79

Problem

• Including spin orbit coupling what is the level ordering for – 9

4Be

– 3115P

– 5927Co

80

Problem

• Including spin orbit coupling what is the level ordering for – 9

4Be: protons (1s21/2 1p2

3/2)

neutrons (1s21/2 1p3

3/2)


1/2 1p43/2 1p2

1/2 1d65/2 2s1

1/2)


3/2 1p21/2 1d6

5/2 2s21/2)


1/2 1p43/2 1p2

1/2 1d65/2 2s2

1/2 1d43/2 1f7

7/2)


3/2 1p21/2 1d6

5/2 2s21/2 1d4

3/2 1f87/2

2p43/2)

81

Magnetic Quantum Numbers• For each of the angular momentum quantum

numbers (l, s, j) there exists a magnetic analogue (ml, ms, mj) representing the resolved component of the original quantum number along the axis of the applied magnetic field

• Each magnetic quantum number can be derived from the related quantum numbers

• ms: – Magnetic spin angular momentum quantum number

– has only 2 allowed values (s)

– For protons, neutrons and electrons s = 1/2

– ms = +1/2 or -1/2

82

Magnetic Quantum Numbers• ml:

– Magnetic orbital angular momentum quantum number– Has (2l+1) possible values– Can be positive or negative integral values– ml = -l, -l+1, -l+2….0…. l-2, l-1, l– Example: l = 3 or p orbital

• (2*3+1) = 7 allowed values• ml:= -3, -2, -1, 0, 1, 2, 3

• mj:– Magnetic total angular momentum quantum number– Has (2j+1) possible values– Can be positive or negative integral values– mj = -j, -j+1, -j+2….0…. j-2, j-1, j

83

Nucleon Total Angular Momentum• Each nucleon has an associated orbital

angular momentum and an associated spin angular momentum

• The total angular momentum quantum number of the nucleon is given by:

j = l + s• The total angular momentum is:

pj = hbar[j(j+1)]1/2

• The observable maximum total angular momentum is

pj = hbar x j

84

Summary: Single Particle Quantum Numbers

• Schrodinger Equation Solutions (H = E)

• Fundamental Quantum Numbersn - principal s - spin (1/2)

l - orbital j - total (l s)

• Magnetic Quantum Numbers

ml - magnetic orbital (-l, -l+1….0…. l-1, l)

ms - magnetic spin (+1/2 or -1/2)

mj - magnetic total (-j, -j+1….0….j-1, j)

85

Nucleus: Total Angular Momentum• When two or more nucleons come together to

form a nucleus the momentum components of the individual particles interact to give a resultant total angular momentum characteristic of the nucleus

• The energy level of the nucleus as a whole is represented by the resultant (I)

• I is historically referred to (inappropriately) as the spin of the nucleus BUT do not confuse it with the spin quantum number of a nucleon (s)

pI = hbar [I(I+1)]1/2

• The observable maximum value of the total nuclear angular momentum is pI = hbar x I

86

Nucleus: Total Angular Momentum• Nucleon-nucleon coupling of the spin and

orbital motions of the individual nucleons is not clearly understood

• Two limiting coupling modes exist:– LS coupling– jj coupling (dominant)

• In reality the coupling probably lies in between the two models

87

LS Coupling• Also known as Russell-Saunders coupling• The interaction of the orbital motion of a nucleon

with its own spin is considered to be weak or negligible

• The orbital motions of different nucleons interact strongly with each other

• The resultant total orbital angular momentum of the nucleus is represented by L and is the vector sum of the individual nucleons

L = li

• Allowed values of L = 0, 1, 2, 3…• Common symbol notation S, P, D, F

Upper Case

88

LS Coupling• The spin motions of different nucleons interact

strongly with each other• The resultant total spin angular momentum of the

nucleus is represented by S and is the vector sum of the spins of the individual nucleons

S = si

• The total angular momentum of the nucleus is represented by I (or sometimes J)

I = L S (hence the name LS coupling!)

• The sign of S is determined by whether S and L are parallel or antiparallel

89

LS Coupling - Application• The individual orbital and spin angular momenta

of paired nucleons cancel each other and do not contribute to the total nuclear angular momentum

• Even-even nuclei have zero nuclear spin (I=0)• Even-Odd or Odd-Even:

– Nuclear spin determined by the single unpaired nucleon

– I (even-odd) = jp or jn = 1/2, 3/2, 5/2, 7/2, 9/2……

• Odd-odd nuclei:– Nuclear spin determined by combination of unpaired

nucleons

– I(odd-odd) = |j1 - j2| j1 + j2

– I(odd-odd) = 1, 2, 3, 4, 5….

90

LS Coupling - ApplicationSteps to determine nuclear spin state

1. Determine if nucleus is even-even, even-odd, odd-even or odd-odd

• If even-even I=0• If N and/or Z are odd continue with step 2 for the odd

nucleon(s)

2. Fill energy level diagram for odd nucleon • If odd-odd remember to fill the levels independently for

protons and neutrons

3. Find the value of j for the energy level occupied by the unpaired nucleon

• For an even-odd or odd-even system this value is the total nuclear spin

• For an odd-odd nucleus the model can not predict the overall state; nuclear spins can range in value from

|j1 - j2| to j1 + j2

91

Example

• What is the nuclear spin for – 9


3/2)


3/2)


1/2 2p43/2 1p2

1/2 1d65/2 2s1

1/2)


3/2 1p21/2 1d6

5/2 2s21/2)


1/2 2p43/2 1p2

1/2 1d65/2 2s2

1/2 1d43/2 1f7

7/2)


3/2 1p21/2 1d6

5/2 2s21/2 1d4

3/2 1f87/2

2p43/2)

92

Example

• What is the nuclear spin for: (I)– 9


3/2) (0)


3/2) (3/2)


1/2 2p43/2 1p2

1/2 1d65/2 2s1

1/2) (1/2)


3/2 1p21/2 1d6

5/2 2s21/2) (0)


1/2 2p43/2 1p2

1/2 1d65/2 2s2

1/2 1d43/2 1f7

7/2) (7/2)


3/2 1p21/2 1d6

5/2 2s21/2 1d4

3/2 1f87/2 2p4

3/2) (0)

93

jj Coupling • Complementary to LS coupling

• Considers affect of strong spin-orbit coupling for individual nucleons in a nucleus

• The orbital and spin motions of the same nucleon may interact strongly

• Total nuclear spin: (vector sum)

I = {j1+j2+j3…..}

where ji = li ± si

94

LS vs jj Coupling

• These models represent two extremes of a coupling that in reality is most accurately represented as a continuum

• In general,– jj coupling preferred for very heavy nuclei– Light nuclei are a mixture

95

Nordheim’s Rules:(spins of odd-odd nuclei)

• Only 5 stable odd-odd nuclei2

1 H, 6

3Li, 105B, 14

7N, 5023Va (could have large t1/2)

• When dealing with two unpaired nucleons predicting the spin of the nucleus is less certain

• Nordheim’s Rules generally give the correct estimate

96

Nordheim’s Rules: (Spins of odd-odd nuclei)

• Given and unpaired proton and neutron having orbital angular momenta l1 and l2 and total angular momenta j1 and j2

• If (j1 + j2 + l1 + l2) = even:

I = |j1 - j2|

• If (j1 + j2 + l1 + l2) = odd:

I = j1 + j2

97

Problem

• What is the spin of the 76As nucleus?

98

Problem

• What is the spin of the 76As nucleus?

• Answer:76As (Z=33, N=43)

33rd proton = f5/2

43rd neutron = g9/2

(j1 + j2 + l1 + l2) = (5/2+9/2+3+4) = 14

I = |j1 - j2| = | 5/2 - 9/2 | = 2

99

What is Parity?

• In physics parity is the name of the symmetry of interactions under spatial inversion.

• For all phenomena the principle of left-right or top-bottom symmetry as mirror images exits

• For any event the mirror image is also possible and would be governed by the same physical laws

• ie the laws of nature are invariant under reflection

100

Parity• A conserved quantity in nuclear reactions involving the

emission of photons and nucleons• Nuclear property related to the symmetry properties of the

wave function• Parity is odd(-) or even(+) based on whether or not the

wave function is symmetric • To test for symmetry reverse the signs of all of the spatial

coordinates in the function– If resultant solution changes sign => parity = -

(-x,-y,-z) = -(x,y,z) – If resultant solution remains the same => parity = +

(x,y,z) = (-x,-y,-z)

• Parity rules for combining wave functions– ++ = +– +- = -– -- = +

101

Parity• Orbital angular momentum states result from

solutions of the Schrodinger equation they are wave functions with an associated parity

• Simple rules:– Even-even nuclei have a ground state (+) parity– Even-odd and odd-even nuclei have a parity equal to

that of the wave function of the unpaired nucleon P = (-1)l

– Odd-odd nuclei => parity is the product of the wave functions of the unpaired nucleons (slide 96)

l 0 1 2 3 4 5 6 7

symbol s p d f g h i j

parity + - + - + - + -

102

Example

• What is the parity for: (I)– 9


3/2) (0)


3/2) (3/2)


1/2 2p43/2 1p2

1/2 1d65/2 2s1

1/2) (1/2)


3/2 1p21/2 1d6

5/2 2s21/2) (0)


1/2 2p43/2 1p2

1/2 1d65/2 2s2

1/2 1d43/2 1f7

7/2) (7/2)


3/2 1p21/2 1d6

5/2 2s21/2 1d4

3/2 1f87/2 2p4

3/2) (0)

103

Example

• What is the parity for: (I) ()– 9


3/2) (0)


3/2) (3/2) (-)


1/2 2p43/2 1p2

1/2 1d65/2 2s1

1/2) (1/2) (+)


3/2 1p21/2 1d6

5/2 2s21/2) (0)


1/2 2p43/2 1p2

1/2 1d65/2 2s2

1/2 1d43/2 1f7

7/2) (7/2) (-)


3/2 1p21/2 1d6

5/2 2s21/2 1d4

3/2 1f87/2 2p4

3/2) (0)

104

Nuclear Spin and Parity

• It is customary to represent the nuclear spin and parity together

• For example in the nuclide 178O the odd

nucleon is the 9th neutron which occupies the 1d5/2 state– Spin = 5/2– Parity = +– Standard notation to say ground state of 17

8O has a spin and parity of 5/2+

105

Example

• What is the standard spin and parity notation for:– 9

4Be: 3/2-

– 3115P: 1/2+

– 5927Co: 7/2-

106

Parity Conservation

• Parity is found to be conserved in all strong interactions involving the emission of photons and nucleons

• In weak interactions involving electrons, neutrinos and mesons there is evidence of non-conservation of parity

• The - decay of 60Co in a strong magnetic field at low temperature is found to emit particle anisotropically => ie no symmetry

• Except in weak interactions parity is always conserved

107

Parity Violation

• Symmetric (isotropic) emission => parity conserved

• Asymmetric (anisotropic) emission => parity violated

Mirror Image

isotropic

anisotropic

108

Magnetic Total Nuclear Momentum Quantum Number

• Analogous to the magnetic quantum numbers of the individual nucleons, the nucleus as a whole has a magnetic total nuclear momentum quantum number

• Represented by mI

• Is the projection of the total nuclear momentum on the field axis

• mI has 2I+1 allowed values:

mI = -I, -(I-1).…-1, 0, 1….(I-1), I

109

Magnetic Moments

• Magnetic Moment (i):

– A nuclear parameter dependent upon the underlying nuclear structure

– a measure of the response of the nucleus to an external magnetic field

– Net effect of the motion of the protons plus the intrinsic spins of the protons and neutrons

– Can be measured experimentally

110

Magnetic Moments

• Magnetic Moment:

• gl and gs are the gyromagnetic ratios:

€

i = glLi + gsSi

€

Protons :

gl = lμ0 , gs = 5.5845μ0

Neutrons :

gl = 0 , gs = −3.8263μ0

Nuclear Magneton :

μ0 =eh

2mpc

111

Magnetic Moments

• Application:– The strong coupling of nucleons due to filled orbitals

and paired spins results in the cancellation of the spins and angular momenta

– Consequently, the magnetic moments will be small and strongly dependent upon the unpaired nucleons

– Even-even => magnetic moment is 0

– Odd-even and even-odd nuclei => use Schmidt Limit• j = l + s: = l gl + 1/2gs

• j = l - s: = (j/j+1)[(l+1)gl - 1/2gs]

– Odd-odd => no prediction available

112

Neutron Magnetic Moment

• The existence of a magnetic moment for the neutron suggests a complex structure for the particle

• The neutron is believed to be composed of equal amounts of positive and negative charges

• The negative charge is on average further from the spin axis than the positive charge thus leading to a large negative magnetic moment

• The large gs factors for the proton and neutron support the theory of a complex distribution of charge within the nucleons

113

Nucleon Magnetic Moments

• The proton and neutron differ only in the middle and outermost regions of the nucleus

• In the outermost diffuse region the proton is positively charged and the neutron is negatively charged

• The inner core of all nucleons is the same• In contrast, the electron is an elementary

particle with the same center of mass and charge

114

Nuclear Charge Distribution• To understand the structure of the nucleus it is

important to know how the protons are spatially distributed

• The presence of a single proton displaced from the center of the nucleus is important because it can result in an effective potential experienced beyond the walls of the nucleus

• Spherical nuclei act as magnetic monopoles• Deformed nuclei can have the properties of dipoles,

quadrupoles, octupoles, etc• These electrical moments can be predicted if the

nuclear spin (I) of the nucleus is known• Monopole (I=0), dipole (I=1/2), quadrupole (I=1)

(Most common)

115

Deformed Nuclei• Liquid-drop model and shell model both assume a

spherically symmetric nucleus– This is a reasonable assumption for magic number

nuclei– But other nuclei have distorted shapes

• Magnitude of deviation from spherical is quantized by = 2(a-c)/(a+c)

where a and c are the radius with respect to the z and x axes (see diagram slide 116)

• Two types of deformed nuclei depending on which axis is compressed– Prolate: > 0– Oblate: < 0

• Maximum observed value of = 0.6

116

Deformed Nucleiz

y

-xOblate: Spins on short axis

Prolate: Spins on long axis

117

Nuclear Quadrupole Moment• Nuclei with quadrupole moments (I=1) are common• Let a nucleus with finite quadrupole moment be represented

by an ellipsoid with the semi-axis (b) parallel to the symmetry axis (Z) and the semi-axis (a) perpendicular

• If the total charge of the nucleus (Ze) is assumed to be uniformly distributed throughout the ellipsoid, the quadrupole moment of the nucleus in the Z direction can be calculated using:

€

Q =2

5Z(b2 − a2)

Z

ab

118

Collective Nuclear Model• Proposed by Bohr & Mottelsen in 1953• Models nucleus as a highly compressed liquid that can

experience internal rotations and vibrations• Includes 4 discrete modes of collective motion

– Rotate around z-axis– Rotate around y-axis– Oscillate between prolate and oblate– Vibrate along x-axis

• Used to calculate allowed vibrational and rotational levels between standard nuclear levels

• These new levels are equivalent to excited states• Validity depends on whether each mode of collective motion

can be treated independently• Works best for strongly deformed nuclei (238U)

119

Collective Nuclear Model Levels

120

Unified Model of Deformed Nuclei• Shell model suffers from discrepancies

between experimental and theoretical spin states for certain nuclei

• Angular momentum of odd-A deformed nuclei contains components from deformed core and unpaired nucleon

• Results in a modification of the energy levels that changes their ordering

121

Nilsson Levels• Nilsson calculated energy levels of odd-A nuclei

as a function of the nuclear deformation () • Each j state from the shell model is split into j+1/2

levels and may contain a maximum of 2 nucleons• Some undeformed levels also reverse order

(1f5/22p3/2)

• Nilsson levels predict energies, angular momenta, quantum numbers, etc better for deformed nuclei than the shell model

122

Nilsson Diagram (Z or N 50)

123

Nilsson Diagram (50 Z or N 82)

124

Problem• Given a nuclear deformation of 0.11, find

the spin and parity of 23Na using the shell model and the Nilsson diagram.

• Which state does the chart of the nuclides confirm exists in nature?

125

Problem• Given a nuclear deformation of 0.11, find

the spin and parity of 23Na using the shell model and the Nilsson diagram.

• Shell model: 5/2+• Nilsson: 3/2+

• Which state does the chart of the nuclides confirm exists in nature? 3/2+

126

Standard Model: Particles

• Provides a description of the fundamental particles and forces that govern matter

• Quarks and leptons as identified as the elementary particles

• Each quark and lepton has an antimatter partner which is referred to as an antiquark or antilepton

127

The Quarks• Standard model describes fundamental

particles and forces• Three families of quarks• Spin 1/2 fermions• Carry electric and color charge • Exist in the bound state as hadrons

– Baryon: 3 bound quarks– Meson: 2 bound quarks

• Never observed in isolation• Naturally occur in three families

Proton

128

Elementary Particles: Quarks• We know that the nucleus of an atom is composed

of nucleons (protons & neutrons)• But these nucleons also have a quark substructure

– Proton = uud

– Neutron = udd

• The antimatter equivalent to the proton is the antiproton (uud)

• The most common mesons are pions and kaons

+: ud, -: ud, K+: us, K-: us

129

Elementary Particles: Leptons• Spin 1/2 fermions

• Point-like => no substructure

• Never bound

130

Standard Model - Forces• Standard model includes the forces that govern the

interactions between matter• Each force is conveyed by a mediating (or exchange)

particle• Weak force governs radioactive decay• Strong force binds quarks in hadrons and nucleons in the

nucleus• Gravitational force has not yet been incorporated into the

standard model

131

Quantum Chromodynamics• Theory that describes the properties of the

strong force• Color = property associated with interaction

(analogous to electric charge)• Every quark carries a color charge of red or

green or blue• Every gluon (exchange particle) also carries

a color charge

132

Quantum Chromodynamics• Coupling between color carriers

INCREASES with distance– (opposite behavior to the more familiar

electromagnetic force)

• Confinement:– At large distances the QCD potential is large

and confines quarks inside bound state it is not possible to separate bound quarks

• Asymptotic Freedom:– At very small distances the QCD potential is

weak and quarks behave as if they are unbound

133

r

V(r)

What is confinement?

• QCD is a “confining” gauge theory,with an effective potential.

• Energy required to separate quarks is greater than the pair rest mass.

• No one has ever seen a free quark.

Confinement

• QCD potential between color carriers increases linearly with distance

€

V (r) = −A(r)

r+Kr

QCD Potential

134

r

V(r)

What is confinement?

• QCD is a “confining” gauge theory,with an effective potential.

• Energy required to separate quarks is greater than the pair rest mass.

• No one has ever seen a free quark.

Confinement

q

qq

q

q

q

qq

• QCD potential between color carriers increases linearly with distance

€

V (r) = −A(r)

r+Kr

QCD Potential

135

M. Schmelling, hep-ex/9701002

What is asymptotic freedom?

• QCD potential weakens at small distances (or lg momentum transfers Q)

• This allows perturbative calculations to be used to predict the system behavior

€

V (r) = −A(r)

r+Kr

QCD Potential

136

What is asymptotic freedom?

• QCD potential weakens at small distances (or lg momentum transfers Q)

• This allows perturbative calculations to be used to predict the system behavior

€

V (r) = −A(r)

r+Kr

QCD Potential Asymptotic Freedom

• Quarks behave as if they are unbound

137

Asymptotic Freedom• Quarks behave as if they are unbound or free when

separated by only very small distances

• Theory tells us that it might be possible to achieve this state in systems of extreme temperature and/or density

• It is this deconfined state that is known as the Quark Gluon Plasma (QGP)

• Conceptually the QGP can be visualized as a soup of freely moving quarks and gluons

138

Isospin

Chapters 5-6: Strong Force & Nuclear Structure

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Transcript of Chapters 5-6: Strong Force & Nuclear Structure