A Review of
The Nuclear Shell Model
By Febdian Rusydi
Why We Need the Model?
To describe and predict nuclear
properties associated with the
structure.
This Review will focus on:
Angular Momentum & parity, J
Ground and excited state configuration
Magnetic moment, 
Presentation Overview
1. Historical development
2. Why Shell Model: The Evidences
3. How to develop the model
4. How to explain the ground and excited
state configuration of an nucleus
5. How to determine the magnetic
moment of the nucleus
Historical Development
1927-28: Statistical Law of Fermions developed
by Fermi
1932-33: Magic Number 2, 8, 20, 28, 50, 82,
126 pointed out by Barlett & Elsasser
1934: The nuclear structure model begun to
discuss. Fermi Gas Model developed, then
applied to nuclear structure.
1935: Liquid Drop Model by Weizsäcker
1936: Bohr applied LDM to nuclear structure
The magic number remained mystery…
Binding Energy per Nuclear
Particle
4He and 12C  -cluster
Solid Red  Experimental
Dash Black  Semi-empirical
Why Shell Model?
Old-fashioned thought:
nucleons collide with each
other. No way for shell model.
Nuclear scattering result:
that thought doesn’t fit the
data!
Magic number even doesn’t
look to support shell model!
BUT
Indication that nuclear potential
can be approached by a
Potential-Well
 Experiment evidence
Atomic physics  electron orbits
around the core
?
But, how is inside the core???
The Evidence #1:
Excitation Energy of First 2+State
N/Z=20/20
Review Physics Letter 28 (1950) page 432
N/Z=50/40 N/Z=82/60
Z=50
N/Z=126/82
Z=70
Z=30
The Evidence #2:
Neutron Absorption X-section
E. B. Paul, “Nuclear & Particle Physics”, North Holland Pub. Comp., 1969, page 259
(Logarithmic)
The Evidence #3:
Neutron Separation Energy
Frauenfelder & Henley, “Subatomic Physics”, Prentice Hall, 1991, page 488
Conclusion so far…
Nuclear structure BEHAVES alike
electron structure
Magic number  a Closed Shell
Properties:
1. Spherical symmetric
2. Total angular momentum = 0
3. Specially stable
Presentation Overview
1. Historical development
2. Why Shell Model: The Evidences
3. How to develop the model
4. How to explain the ground and excited
state configuration of an nuclei
5. How the determine the magnetic
moment of the nuclei
to
Let’s Develop the Theory!
Keyword:
Explain the magic number
Steps:
1. Find the potential well that
resembles the nuclear density
2. Consider the spin-orbit coupling
Shell Model Theory:
The Fundamental Assumption
The Single Particle Model
1. Interactions between nucleons are
neglected
2. Each nucleon can move
independently in the nuclear
potential
Various forms of the
Potential Well
1. Square Well
2. Harmonic Oscillation
3. Woods - Saxon
Potential
R
r
V(r)
V0
a
  








 


i
i
ij
i
i
i r
V
r
v
r
V
T
H )
(
)
(
)
(
' 
Residual potential
Central potential
Cent. Pot >> Resd. Pot,
then we can set   0.
Finally we have 3 well
potential candidates!
Full math. Treatment:
Kris L. G. Heyde, Basic Ideas and Concepts in Nuclear Physics, IoP, 1994, Chapter 9
The Closed Shell:
Square Well Potential
The closed shell  magic
number
 
0
2
1
)
(
2
2
2
2
2
2






 


 nl
nl
nl R
Mr
l
l
r
V
E
r
M
R
dr
d 

The Closed Shell:
Harmonic Potential
The closed shell
 magic number
2
2
0
2
1
)
( r
M
U
r
V 



The Closed Shell:
Woods - Saxon Potential
The closed shell 
magic number
 
a
R
r
Vo
r
V 


exp
1
)
(
But…
This potential
resembles with
nuclear density from
nuclear scattering
The Closed Shell:
Spin-Orbit Coupling Contribution
Maria Mayer (Physical Review 78 (1950),
p16) suggested:,
1.There should be a non-central
potential component
2.And it should have a magnitude
which depends on the S & L
Hazel, Jensen, and Suess also came to the
same conclusion.
The Closed Shell:
Spin-Orbit Coupling Calculation
The non-central Pot.
2
)
(
)
(
'

ls
V
r
V
r
V ls


     
 




















2
1
2
1
2
2
1
2
2
1
1
1
l
j
l
l
j
l
s
s
l
l
j
j
ls

)
(
2
1
2
r
V
l
E ls
ls



Energy splitting
Experiment: Vls = negative
 Energy for spin up < spin down
j = l +/- ½
j = l - ½
j = l + ½
Delta Els
Full math. Treatment:
Kris L. G. Heyde, Basic Ideas and
Concepts in Nuclear Physics, IoP, 1994,
Chapter 9
SMT:
The
Closed
Shell
Povh,
Particle
&
Nuclei
(3
rd
edition),
Springer
1995,
pg
255
SMT: The Ground State
How to determine the Quantum Number J ?[1]
1. J (Double Magic number or double closed
shell) = 0+. If only 1 magic number, then J
determined by the non-magic number
configuration.
2. J determined from the nucleon in outermost
shell (i.e., the highest energy) or hole if
exist.
3.  determined by (-1)l, where l(s,p,d,f,g,…) =
(0, 1, 2, 3, 4, …). To choose l: consider
hole first, then if no hole  nucleon in
outermost shell.
SMT: The Ground State
(example)
How to configure ground state of nucleus
Nuclide Z and N
number
Orbit assignment Shell
Model
J Note
6He Z= 2
N= 2
(1s1/2)2
(1s1/2)2
s1/2 0+ Double magic number
11B Z= 5
N= 6
(1s1/2)2 (1p3/2)-1
(1s1/2)2 (1p3/2)4
p3/2 3/2- 1 hole @ 1p3/2
Closed shell
12C Z= 6
N= 6
(1s1/2)2 (1p3/2)4
(1s1/2)2 (1p3/2)4
p3/2 0+ Double Closed shell
15N Z= 7
N= 8
(1s1/2)2 (1p3/2)4 (1p1/2)-1
(2nd mg.#)
p1/2 1/2- 1 hole @ 1p1/2
16O Z= 8
N= 8
(2nd mg.#)
(2nd mg.#)
p1/2 0+ Double magic number
17F Z= 9
N= 8
(1s1/2)2 (1p3/2)4 (1p1/2)2 (1d2)1
(2nd mg.#)
d5/2 5/2+ 1 proton in outer shell
27Mg Z= 12
N= 15
(2nd mg.#) (1d5/2)4
(2nd mg.#) (1d5/2)6 (2s1/2)-1
s1/2 1/2+ 4 proton coupled @ 1d5/2
1 hole @ 2s1/2
37Sr Z= 38
N= 49
(3rd mg.#) (2p3/2)4 (1f5/2)6
(3rd mg.#) (2p3/2)4 (1f5/2)6 (2p3/2)4(1g9/2)-1
g9/2 9/2+ Closed shell @ f5/2
1 hole @ g9/2
SMT: Excited State
Some conditions must be known: energy
available, gap, the magic number exists,
the outermost shell (pair, hole, single
nucleon).
Otherwise, it is quite difficult to predict
precisely what is the next state.
SMT: Excited State (example)
Let’s take an example 18O with ground state
configuration:
– Z= 8 – the magic number
– N=10 – (1s1/2)2 (1p3/2)4 (1p1/2)2 (1d5/2)2 or (d5/2)2
If with E ~ 2 [MeV], one can excite neutron to (d5/2)
(d3/2), then with E ~ 4 [MeV], some possible excite
states are:
– Bring 2 neutron from 1p1/2 to 2d5/2  (d5/2)4 0  J  5
– Bring 2 neutron from 2d5/2 to 2d3/2  (d3/2)2 0  J  3
– Bring 1 neutron from 2d5/2 to 1f7/2  (f7/2)1 1  J  6
– Some other probabilities still also exist
SMT: Mirror & Discrepancy
Mirror Nuclei
15NZ=7  15OZ=8
If we swap protons &
neutrons, the strong
force essentially does
not notice it
Discrepancy
The prediction of SMT
fail when dealing with
deformed nuclei.
Example: 167Er
Theory 7/2 -
Exprm  7/2 +
Collective Model! 
SMT: Mirror Nuclei (Example)
Povh,
Particle
&
Nuclei
(3
rd
edition),
Springer
1995,
pg
256
SMT: The Magnetic Moment
Since L-S
Coupling 
associated to
each individual
nucleon
SO   sum over
the nucleonic
magnetic moment
 




A
i
s
l
N
nucleus g
s
g
l
1
1
1
1



 
1
2
1








l
g
g
g
g
J
g
Jg
s
g
l
g
l
s
l
nucleus
N
nucleus
j
s
l
N
nucleus








values of gl and gs
proton Neutron
gl 5.586 -3.826
gs 1 0
Full math. Treatment:
A. Shalit & I. Talmi, Nuclear Shell Model, page 53-59
Conclusions
1. How to develop the model
- Explain the magic number
- Single particle model
- Woods – Saxon Potential
- LS Coupling Contribution
2. Theory for Ground & Excited State
- Treat like in electron configuration
- J can be determined by using the guide
3. Theory for Magnetic Moment
-  is sum over the nucleonic magnetic moment
Some More Left…
Some aspects in shell Model Theory that are
not treated in this discussion are:
1. Quadruple Moment – the bridge of Shell
Model Theory and Collective Model Theory.
2. Generalization of the Shell Model Theory –
what happen when we remove the
fundamental assumption “the nucleons
move in a spherical fixed potential,
interactions among the particles are
negligible, and only the last odd particle
contributes to the level properties”.

The Nuclear Shell Model

  • 1.
    A Review of TheNuclear Shell Model By Febdian Rusydi
  • 2.
    Why We Needthe Model? To describe and predict nuclear properties associated with the structure. This Review will focus on: Angular Momentum & parity, J Ground and excited state configuration Magnetic moment, 
  • 3.
    Presentation Overview 1. Historicaldevelopment 2. Why Shell Model: The Evidences 3. How to develop the model 4. How to explain the ground and excited state configuration of an nucleus 5. How to determine the magnetic moment of the nucleus
  • 4.
    Historical Development 1927-28: StatisticalLaw of Fermions developed by Fermi 1932-33: Magic Number 2, 8, 20, 28, 50, 82, 126 pointed out by Barlett & Elsasser 1934: The nuclear structure model begun to discuss. Fermi Gas Model developed, then applied to nuclear structure. 1935: Liquid Drop Model by Weizsäcker 1936: Bohr applied LDM to nuclear structure The magic number remained mystery…
  • 5.
    Binding Energy perNuclear Particle 4He and 12C  -cluster Solid Red  Experimental Dash Black  Semi-empirical
  • 6.
    Why Shell Model? Old-fashionedthought: nucleons collide with each other. No way for shell model. Nuclear scattering result: that thought doesn’t fit the data! Magic number even doesn’t look to support shell model! BUT Indication that nuclear potential can be approached by a Potential-Well  Experiment evidence Atomic physics  electron orbits around the core ? But, how is inside the core???
  • 7.
    The Evidence #1: ExcitationEnergy of First 2+State N/Z=20/20 Review Physics Letter 28 (1950) page 432 N/Z=50/40 N/Z=82/60 Z=50 N/Z=126/82 Z=70 Z=30
  • 8.
    The Evidence #2: NeutronAbsorption X-section E. B. Paul, “Nuclear & Particle Physics”, North Holland Pub. Comp., 1969, page 259 (Logarithmic)
  • 9.
    The Evidence #3: NeutronSeparation Energy Frauenfelder & Henley, “Subatomic Physics”, Prentice Hall, 1991, page 488
  • 10.
    Conclusion so far… Nuclearstructure BEHAVES alike electron structure Magic number  a Closed Shell Properties: 1. Spherical symmetric 2. Total angular momentum = 0 3. Specially stable
  • 11.
    Presentation Overview 1. Historicaldevelopment 2. Why Shell Model: The Evidences 3. How to develop the model 4. How to explain the ground and excited state configuration of an nuclei 5. How the determine the magnetic moment of the nuclei to
  • 12.
    Let’s Develop theTheory! Keyword: Explain the magic number Steps: 1. Find the potential well that resembles the nuclear density 2. Consider the spin-orbit coupling
  • 13.
    Shell Model Theory: TheFundamental Assumption The Single Particle Model 1. Interactions between nucleons are neglected 2. Each nucleon can move independently in the nuclear potential
  • 14.
    Various forms ofthe Potential Well 1. Square Well 2. Harmonic Oscillation 3. Woods - Saxon Potential R r V(r) V0 a                i i ij i i i r V r v r V T H ) ( ) ( ) ( '  Residual potential Central potential Cent. Pot >> Resd. Pot, then we can set   0. Finally we have 3 well potential candidates! Full math. Treatment: Kris L. G. Heyde, Basic Ideas and Concepts in Nuclear Physics, IoP, 1994, Chapter 9
  • 15.
    The Closed Shell: SquareWell Potential The closed shell  magic number   0 2 1 ) ( 2 2 2 2 2 2            nl nl nl R Mr l l r V E r M R dr d  
  • 16.
    The Closed Shell: HarmonicPotential The closed shell  magic number 2 2 0 2 1 ) ( r M U r V    
  • 17.
    The Closed Shell: Woods- Saxon Potential The closed shell  magic number   a R r Vo r V    exp 1 ) ( But… This potential resembles with nuclear density from nuclear scattering
  • 18.
    The Closed Shell: Spin-OrbitCoupling Contribution Maria Mayer (Physical Review 78 (1950), p16) suggested:, 1.There should be a non-central potential component 2.And it should have a magnitude which depends on the S & L Hazel, Jensen, and Suess also came to the same conclusion.
  • 19.
    The Closed Shell: Spin-OrbitCoupling Calculation The non-central Pot. 2 ) ( ) ( '  ls V r V r V ls                               2 1 2 1 2 2 1 2 2 1 1 1 l j l l j l s s l l j j ls  ) ( 2 1 2 r V l E ls ls    Energy splitting Experiment: Vls = negative  Energy for spin up < spin down j = l +/- ½ j = l - ½ j = l + ½ Delta Els Full math. Treatment: Kris L. G. Heyde, Basic Ideas and Concepts in Nuclear Physics, IoP, 1994, Chapter 9
  • 20.
  • 21.
    SMT: The GroundState How to determine the Quantum Number J ?[1] 1. J (Double Magic number or double closed shell) = 0+. If only 1 magic number, then J determined by the non-magic number configuration. 2. J determined from the nucleon in outermost shell (i.e., the highest energy) or hole if exist. 3.  determined by (-1)l, where l(s,p,d,f,g,…) = (0, 1, 2, 3, 4, …). To choose l: consider hole first, then if no hole  nucleon in outermost shell.
  • 22.
    SMT: The GroundState (example) How to configure ground state of nucleus Nuclide Z and N number Orbit assignment Shell Model J Note 6He Z= 2 N= 2 (1s1/2)2 (1s1/2)2 s1/2 0+ Double magic number 11B Z= 5 N= 6 (1s1/2)2 (1p3/2)-1 (1s1/2)2 (1p3/2)4 p3/2 3/2- 1 hole @ 1p3/2 Closed shell 12C Z= 6 N= 6 (1s1/2)2 (1p3/2)4 (1s1/2)2 (1p3/2)4 p3/2 0+ Double Closed shell 15N Z= 7 N= 8 (1s1/2)2 (1p3/2)4 (1p1/2)-1 (2nd mg.#) p1/2 1/2- 1 hole @ 1p1/2 16O Z= 8 N= 8 (2nd mg.#) (2nd mg.#) p1/2 0+ Double magic number 17F Z= 9 N= 8 (1s1/2)2 (1p3/2)4 (1p1/2)2 (1d2)1 (2nd mg.#) d5/2 5/2+ 1 proton in outer shell 27Mg Z= 12 N= 15 (2nd mg.#) (1d5/2)4 (2nd mg.#) (1d5/2)6 (2s1/2)-1 s1/2 1/2+ 4 proton coupled @ 1d5/2 1 hole @ 2s1/2 37Sr Z= 38 N= 49 (3rd mg.#) (2p3/2)4 (1f5/2)6 (3rd mg.#) (2p3/2)4 (1f5/2)6 (2p3/2)4(1g9/2)-1 g9/2 9/2+ Closed shell @ f5/2 1 hole @ g9/2
  • 23.
    SMT: Excited State Someconditions must be known: energy available, gap, the magic number exists, the outermost shell (pair, hole, single nucleon). Otherwise, it is quite difficult to predict precisely what is the next state.
  • 24.
    SMT: Excited State(example) Let’s take an example 18O with ground state configuration: – Z= 8 – the magic number – N=10 – (1s1/2)2 (1p3/2)4 (1p1/2)2 (1d5/2)2 or (d5/2)2 If with E ~ 2 [MeV], one can excite neutron to (d5/2) (d3/2), then with E ~ 4 [MeV], some possible excite states are: – Bring 2 neutron from 1p1/2 to 2d5/2  (d5/2)4 0  J  5 – Bring 2 neutron from 2d5/2 to 2d3/2  (d3/2)2 0  J  3 – Bring 1 neutron from 2d5/2 to 1f7/2  (f7/2)1 1  J  6 – Some other probabilities still also exist
  • 25.
    SMT: Mirror &Discrepancy Mirror Nuclei 15NZ=7  15OZ=8 If we swap protons & neutrons, the strong force essentially does not notice it Discrepancy The prediction of SMT fail when dealing with deformed nuclei. Example: 167Er Theory 7/2 - Exprm  7/2 + Collective Model! 
  • 26.
    SMT: Mirror Nuclei(Example) Povh, Particle & Nuclei (3 rd edition), Springer 1995, pg 256
  • 27.
    SMT: The MagneticMoment Since L-S Coupling  associated to each individual nucleon SO   sum over the nucleonic magnetic moment       A i s l N nucleus g s g l 1 1 1 1      1 2 1         l g g g g J g Jg s g l g l s l nucleus N nucleus j s l N nucleus         values of gl and gs proton Neutron gl 5.586 -3.826 gs 1 0 Full math. Treatment: A. Shalit & I. Talmi, Nuclear Shell Model, page 53-59
  • 28.
    Conclusions 1. How todevelop the model - Explain the magic number - Single particle model - Woods – Saxon Potential - LS Coupling Contribution 2. Theory for Ground & Excited State - Treat like in electron configuration - J can be determined by using the guide 3. Theory for Magnetic Moment -  is sum over the nucleonic magnetic moment
  • 29.
    Some More Left… Someaspects in shell Model Theory that are not treated in this discussion are: 1. Quadruple Moment – the bridge of Shell Model Theory and Collective Model Theory. 2. Generalization of the Shell Model Theory – what happen when we remove the fundamental assumption “the nucleons move in a spherical fixed potential, interactions among the particles are negligible, and only the last odd particle contributes to the level properties”.