1. Chapter 6: Nuclear Structure
Abby Bickley
University of Colorado
NCSS ‘99
Additional References:
Choppin (CLR), Radiochemistry and Nuclear Chemistry, 2nd Edition, Chapter 11
Friedlander (FKMM), Nuclear and Radiochemistry, 3rd Edition, Chapter 10

2. 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

3. 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, 105B, 147N, 5023Va (could have large t1/2)
Nuclei with an even number of protons have a large number of stable isotopes
Even # protons Odd # protons
50Sn:10 (isotopes) 47Ag: 2 (isotopes)
48Cd: Sb:2
52Te: Rh:1
49In:1
53I: 1
Roughly equal numbers of stable even-odd and odd-even nuclei

4. Implications for Nuclear Models I
Proton-proton and neutron-neutron pairing must result in energy stabilization of bound state nuclei
Pairing of protons with protons and neutrons with neutrons results in the same degree of stabilization
Pairing of protons with neutrons does not occur (nor translate into stabilization)

5. 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.

6. Implications for Nuclear Models II
4. 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
Eq. 1

7. General Nuclear Properties I
Binding energy per nucleon approximately constant for all stable nuclei
8.9
7.4
Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.

8. 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
dl = skin thickness
Rl = Half density radius
Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.

9. 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: Eq. 3

10. Liquid Drop Model Components I
Volume Energy
Surface Energy
c1 = MeV, c2 = MeV, c3 = MeV, c4 = MeV, k = 1.79
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

11. Liquid Drop Model Components II
Coulomb
Energy
Pairing
Energy
c1 = MeV, c2 = MeV, c3 = MeV, c4 = MeV, k = 1.79
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

12. Problem 1
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.

13. Problem 1: Answers 15N = 6.87 MeV/nucleon 148Gd = 8.88 MeV/nucleon

14. 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)
Eq. 4
Eq. 5

15. Mass Parabolas Example 1 A = 75 or 157
Parabola Vertex:
ZA=[-f2 / 2f1] Eq. 6
Minimum mass & Maximum EB
Used to find mass and EB difference between isobars
Nuclear charge of minimum mass is derivative of Eq. 5 => 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.

16. Mass Parabolas Example 2 A = 156
Eq. 5 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.

17. Problem 2
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 15 & 16.

18. Problem 2: 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?
15765Tb, 15664Gd, 7533As
Compare your results with the mass parabolas on slides 15 & 16.

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

20. 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

21. Atomic Orbitals Schrodinger equation solution reveals quantum numbers
n = principal, describes energy level
l = angular momentum, 0n-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 = 1s22s22px12py1 2s 3s 2p 1s

22. Shell Structure of Nucleus Historical 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

23. 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

24. 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

25. Pairing
Just as electrons tend to pair up to form a stable bond, so do like-nucleons; pairing results in increased stability
Even-Z and even=N nuclides are the most abundant stable nuclides in nature (165/275)
From 15O to 35Cl all odd-Z elements have one stable isotope while all even-Z elements have three
The heaviest stable natural nuclide is 20983Bi (N=126)
The stable end product of all naturally occurring radioactive series of elements is Pb with Z=82

26. 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

27. Shell Model Evidence - Abundances

28. Shell Model Evidence - Stable Isotopes
Stable Isobars
# of Isobars
# of Neutrons
The number of stable isotopes of a given element is a reflection of the relative stability of that element. Plot of number of isotopes vs N shows peaks at
N = 20, 28, 50, 82
A similar effect is observed as a function of Z

29. 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 

30. 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
N = 19
N = 20
N = 21
Sc 40 +
0.86s
Sc 41
0.60s
Sc 42
0.68s
Ca 39
Ca 40
96.94
Ca 41 EC
1.03E6 y
K 38
7.63m
K 39
93.25
K 40 -
1.28E9y
Z = 21
Z = 20
Z = 19

31. 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-capture
8838Sr(50n)
5.8E-3
8738Sr(50n) 16
13654Xe(82n)
0.16
15654Xe(81n)
2.65E6

32. 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

33. 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!!!

34. 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

35. Nuclear Potential Well
Nucleon orbit = nucleon quantum state
Similar to quantum state of valence electron
BUT
Nucleon “feels” 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)

36. 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)

37. 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

38. Nuclear Potential Well
Depth of well represents
binding energy

39. Nuclear Potential Functions
Square Well Potential
Harmonic Oscillator Potential
Woods-Saxon
Exponential Potential
Gaussian Potential
Yukawa Potential:
Note: R = nuclear radius
r = distance from center of nucleus

40. Nuclear Potential Functions
Exact shape of well is
uncertain and depends
on mathematical function
assumed for the interaction
Yukawa
Exponential
Gaussian
Square Well

41. Neutron vs Proton Potential Wells
Coulomb repulsion
prevents potential well
from being as deep for
protons as for neutrons

42. 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

43. 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

44. 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

45. { Spin Angular Momentum No External B-field Applied External B-field
Project spin angular momentum onto the field axes
Allowed values are units of hbar
ps(z) = hbar ms

46. 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

47. 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

48. Orbital Angular Momentum
3 h
2 h
1 h
0 h
-1 h
-2 h
-3 h
Orientations of ml in a magnetic field
B-field axis pl
pl(z)
Project orbital angular momentum onto the field axes
Allowed values are units of hbar
pl(z) = hbar ml

49. 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
Electrons: 0  l < (n-1)
For nucleons (but not electrons) l can exceed n

50. 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 1 2 3 4 5 6 7
symbol s p d f g h i j

51. Quantum States

52. Energy Level Diagram Isotropic Harmonic Oscillator Levels1f 2s 1d 1p 2d 1g 2p 2f 1h 3s 1i 3p
nl nucleons [total] 2 6 14 10 22 18 26 40 34 20 8 92 70 68 58
138
112
106
Isotropic Harmonic Oscillator Levels

53. 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

54. 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

55. 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…

56. Total Angular Momentum Example
What are the allowed values of j for a nucleon with
l = 1, s = 1/2

57. Total Angular Momentum Example
What are the allowed values of j for a nucleon with
l = 1, s = 1/2
Answer: j = 1/2 or 3/2

58. 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????

59. 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

60. Example
What is the level ordering for
94Be
3115P
5927Co

61. Example What is the level ordering for 94Be: protons (1s21/2 2p23/2)
neutrons (1s21/2 2p33/2)
3115P: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s11/2)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2)
5927Co: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f77/2)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f87/2 2p43/2)

62. 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

63. 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
mj = -j, -j+1, -j+2….0…. j-2, j-1, j

64. 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

65. Summary: Single Particle Quantum Numbers
Schrodinger Equation Solutions (H = E)
Fundamental Quantum Numbers
n - 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)

66. 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

67. 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

68. 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

69. 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!)

70. 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)
Nuclear spin of odd-A nuclei is determined by the single unpaired nucleon
Odd-odd nuclei:
I(odd-odd) = jp + jn = (lp  1/2) + (ln  1/2)

71. LS Coupling - Application
Steps to determine nuclear spin state
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)
Fill energy level diagram for odd nucleon
If odd-odd remember to fill the levels independently for protons and neutrons
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

72. Example What is the nuclear spin for 94Be: protons (1s21/2 2p23/2)
neutrons (1s21/2 2p33/2)
3115P: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s11/2)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2)
5927Co: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f77/2)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f87/2 2p43/2)

73. Example What is the nuclear spin for: (I)
94Be: protons (1s21/2 2p23/2) (0)
neutrons (1s21/2 2p33/2) (3/2)
3115P: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s11/2) (1/2)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2) (0)
5927Co: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f77/2) (7/2)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f87/2 2p43/2) (0)

74. jj Coupling Complementary to LS coupling
Considers affect of strong spin-orbit coupling for individual nucleons in a nucleus
Total nuclear spin: (vector sum)
I = {j1+j2+j3…..}

75. 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

76. 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 => (-)
If resultant solution remains the same => (+)
Parity rules for combining wave functions
++ = +
+- = -
-- = +

77. 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
Odd-odd nuclei => parity is the product of the wave functions of the unpaired nucleons l 1 2 3 4 5 6 7
symbol s p d f g h i j
parity + -

78. Example What is the parity for: (I) 94Be: protons (1s21/2 2p23/2) (0)
neutrons (1s21/2 2p33/2) (3/2)
3115P: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s11/2) (1/2)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2) (0)
5927Co: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f77/2) (7/2)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f87/2 2p43/2) (0)

79. Example What is the parity for: (I) ()
94Be: protons (1s21/2 2p23/2) (0)
neutrons (1s21/2 2p33/2) (3/2) (-)
3115P: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s11/2) (1/2) (+)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2) (0)
5927Co: protons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f77/2) (7/2) (-)
neutrons (1s21/2 2p43/2 1p21/2 1d65/2 2s21/2 1d43/2 1f87/2 2p43/2) (0)

80. 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 178O has a spin and parity of 5/2+

81. Example What is the standard spin and parity notation for: 94Be: 3/2-
5927Co: 7/2-

82. 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)

83. 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 84)
Two types of deformed nuclei depending on which axis is compressed
Prolate:  > 0
Oblate:  < 0
Maximum observed value of  =  0.6

84. Deformed Nuclei z Oblate: Spins on short axis -x y Prolate: Spins on
long axis

85. 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: a b Z

86. 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)

87. Collective Nuclear Model Levels

88. 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

89. 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 2 nucleons
Some undeformed levels also reverse order (1f5/22p3/2)
Nilsson levels predict energies, angular momenta, quantum numbers, etc better for deformed nuclei than the shell model

90. Nilsson Diagram (Z or N 50)

91. Nilsson Diagram (50  Z or N 82)

92. Example
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?

93. Example
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+