Transcript A Review of The Nuclear Shell Model

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.
2.
3.
4.
Historical development
Why Shell Model: The Evidences
How to develop the model
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?
Atomic physics  electron orbits
around the core
But, how is inside the core???
?
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
The Evidence #1:
Excitation Energy of First 2+State
N/Z=20/20
N/Z=126/82
Z=30
N/Z=50/40
N/Z=82/60
Z=50
Z=70
Review Physics Letter 28 (1950) page 432
The Evidence #2:
(Logarithmic)
Neutron Absorption X-section
E. B. Paul, “Nuclear & Particle Physics”, North Holland Pub. Comp., 1969, page 259
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.
2.
3.
4.
Historical development
Why Shell Model: The Evidences
How to develop the model
How to explain the ground and excited
state configuration of an nuclei
5. How the
to determine the magnetic
moment of the nuclei
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


H '   Ti  V (ri )    v(rij )  V (ri ) 


i
i
Central potential
V(r)
R
Residual potential
1. Square Well
Cent. Pot >> Resd. Pot,
then we can set   0.
Finally we have 3 well
potential candidates!
r
2. Harmonic Oscillation
3. Woods - Saxon
Potential
V0
a
Full math. Treatment:
Kris L. G. Heyde, Basic Ideas and Concepts in Nuclear Physics, IoP, 1994, Chapter 9
The Closed Shell:
Square Well Potential
d2
2M 
l  l  1  2 

Rnl  0
2 Rnl 
2 E nl  V ( r ) 
2
dr
r 
2Mr 
The closed shell  magic
number
The Closed Shell:
Harmonic Potential
V (r)  U  12 M02 r 2
The closed shell
 magic number
The Closed Shell:
Woods - Saxon Potential
Vo
V (r ) 
1  exp r aR 
But…
This potential
resembles with
nuclear density from
nuclear scattering
The closed shell 
magic number
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
V ' (r )  V (r )  Vls
ls
2
 l
 j  l  12
ls







j
j

1

l
l

1

s
s

1
The non-central Pot.

 2
2
 l  1

2

 j  l  12
 2
Energy splitting
2l  1
Els 
Vls (r )
2
Experiment: Vls = negative
 Energy for spin up < spin down
Full math. Treatment:
Kris L. G. Heyde, Basic Ideas and
Concepts in Nuclear Physics, IoP, 1994,
Chapter 9
j=l-½
j = l +/- ½
Delta Els
j=l+½
Povh, Particle & Nuclei (3rd edition), Springer 1995, pg 255
SMT: The Closed Shell
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+
11B
Z= 5
N= 6
(1s1/2)2 (1p3/2)-1
(1s1/2)2 (1p3/2)4
p3/2
3/2-
12C
Z= 6
N= 6
(1s1/2)2 (1p3/2)4
(1s1/2)2 (1p3/2)4
p3/2
0+
15N
Z= 7
N= 8
(1s1/2)2 (1p3/2)4 (1p1/2)-1
(2nd mg.#)
p1/2
1/2-
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
Double magic number
1 hole @ 1p3/2
Closed shell
Double Closed shell
1 hole @ 1p1/2
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
15N
15O

Z=7
Z=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! 
Povh, Particle & Nuclei (3rd edition), Springer 1995, pg 256
SMT: Mirror Nuclei (Example)
SMT: The Magnetic Moment
Since L-S
Coupling 
associated to
each individual
nucleon
SO   sum over
the nucleonic
magnetic moment
 nucleus
1 A
  N l1 g l  s1 g s 
 i1
1
J
 N   g l l  g s s    Jg j  g nucleus N


g  gl
 g nucleus g l  s
2l  1
 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”.