Transcript Nuclear models

Nuclear models
Models we will consider…
• Independent particle shell model
• Collective models
• Fermi gas model
Our approach…
• Look at data that motivates the model
• Construct a model
• Make and test predictions from the model
Shell Model - data
2p separation energy
(between isotones)
Becomes much smaller after
8, 20, 28, 50, 82, 126
2n separation energy
(between isotopes)
T
Shell Model - data
  + A-4Po
Sudden rise at N = 126
ARn

Neutron capture cross section 
Very small  at N = 28, 50, 82, 126
R
Ravg
Abrupt change in nuclear radius
at N = 20, 28, 50, 82, 126
Shell Model - data
T show sharp discontinuities near N,Z of 28, 50, 82, 126
BE for last n added: sharp discontinuities near, 50, 82, 126
e.g., (d,p), (n,), ( ,n), (d,t) reactions
And, the observation of discrete photon energies E
emitted from nuclear de-excitation
Shell Model
Assume that the nucleons move (independently) in a
potential, V, created by the other nucleons in the nucleus.
Assume that the problem can be addressed by the nonrelativistic Schrodinger quantum mechanics.
Assume that the potential, V, is spherically symmetric and
therefore only a function of r, V(r)
V r   Vo r  R
V r   0
V r  
rR
V r  r 2


Vo
1  exp r  R/ a 
Spin-orbit potential Vso rL  s



Shell Model
Vso rL  s


Hˆ , Jˆ 2  0
Hˆ , Jˆz  0
J2  J  J
JLs
J 2  L  L  2L  s  s  s
Q.M.
J 2  L2  s 2
Ls 
2
  1  s s  1
2
  1
2
s 2  s s  1
2
L2 
J 2  L2  s 2  2L  s
1
L  s  j  j  1 
2
J 2  j  j  1
 n,
2

good quantum numbers
,s, j,m j
Shell Model
 n,
,s, j,m j
2  2  1
Multiplicities --
=
2 j  1
2 spin states

different m states

Ls
j 1/ 2

 Ls
j 1/ 2

2  1
different m j states
2
2

Energy difference (splitting) increases with

Shell Model
energy levels
Energy splitting
increases with
Spectroscopic state

multiplicity
Systematics…
Nucleon Classification
Nuclear moment s
Number
of known
St able
stable
nucleons
Radioact ive

A
Z
N
I
O dd
O dd
E ven
50
50
11
U s ually large & pos .
II
O dd
E ven
O dd
55
36
4
U s ually s mall, neg.
III
E ven
O dd
O dd
4
4
9
U s ually pos itive
IV
E ven
E ven
E ven
165
12
1
I ndeterminate
Nuclear magnetic moments
e
L
L is orbital angular momentum for single nucleon
2M
e
*
 
  1 M is nucleon mass
2M
   N max z-axis projection
 *   N   1
 L* 
 p  2.7928  N
 n  1.9131  
N
Intrinsic (measured) dipole magnetic moments
 N  3.152451 18 18 MeV / gauss
   5.788378 18 15 MeV / gauss
N
m
1
 e 
  M N 1836
Nuclear magnetic moments
From electron case, you expect to have for this fermion --
 *p  2 N s s  1
Does not agree with measurement
 p  N
 p  2.7928  N
 n  1.9131  N
 *p
Measured dipole magnetic moments
 g p  N s s  1  g p  N
 p   gp N
1
2
3
2
gp
gp
 *p
 N s s  1
2 p
N
; g p  22.7928  = 5.5856
Nuclear magnetic moments
And, by the same analysis, one gets --
 n*
 gn  N s s  1  gn  N
 n   gn  N
1
2
 p  2.7928  N
 n  1.9131  N
3
2
gn 
 n*
 N s s  1
2 n
gn 
; gn  21.9131 = -3.8262
N
Measured dipole magnetic moments

g p  5.5856
gn  3.8262
Nuclear magnetic moments
Consider nuclei with odd A.
Assume that the pairing interaction causes the “core” of paired
nucleons to have net I = 0.
Assume that the induced magnetic dipole moment is due to the
last unpaired nucleon.
Use this to estimate the nuclear magnetic dipole moment within this model.
Nuclear magnetic moments
 *p
 g p  N s s  1  g p  N
3
2
gs s *
1
 p   gp N
2
 n*

 gn  N s s  1  gn  N
3
2
1
2
 n   gn  N


 

  g N
*
*
 g N

 1
proton : g  1
gj
*
neutron : g  0
gj *
*g
 
*
gs
s


   
j  s
 
*g
s *gs
j s

*j  gN j *  gN j  j  1
 j   g N j
   g N
s *gs
Nuclear magnetic moments
 j   s 


* *
cos  j 
2 * j*
* 2
* 2
* 2
gs s *
s   j    

* *
cos s j 
2 s* j*
* 2
 j*   * cos
* 2
* 2



 j   s* coss j 
* *
* *
*

g 
g    1
 2 


* *
*
 
j  s
 
 
 gs s cos s * j *  N
*

2

* 2 
 s  g 
s


1
2

 2 
j*




  
 
*
gj
gj *
*g
*
gs
s


   
g j  g cos j     
*
s *gs

* 2 
   s  

* 2
j
  
* 2
*g
s *gs
j s
Nuclear magnetic moments
Consider the case: j    s
s *gs
gs s *
*g g j *
*g
 1   1 
*
j
 j  j  1        1
g j  
 2   2  
 3
*
gs
s
* 2
*
j
   2 
s gs

 
4
  


2
3
*
j  s j  s
s


4

 


* 2
* 2 
* 2
* 2 
 s  g 
 s 
g 
s
  
1
g    1

2



*
* 2
 2 


2
j
j








 
* 2
 
 
  
 
…some algebra happens here…
  
 
g   g
j
 gs
1/2
j
Nuclear magnetic moments
Consider the case: j    s
g   g


But, if I    s
  g (I 1/2)   s

gs s *
1/2
 gs
j
j

  g  j  g I
  g  gs s
  g  s
s *gs



 

gj
*
gj *
*g
 
*
gs
s



  
j  s
 

*g
s *gs
j s
Nuclear magnetic moments
Consider the case: j    s
g   g
1
1/2
 gs
j 1
j 1

 I 
  g  g   s 

I  1
But, if I    s

g  I 
  g I   s  


2 I  1

s *gs
gs s *



 
gj
*
gj *
*g
 
*
gs
s



  
j  s
 
*g
s *gs
j s
Four cases to consider:


both cases shown here for
odd proton & odd neutron
Nuclear magnetic moments
Proton:
g 1,  s   p
Neutron:
g  0,  s   n
gs s *

j
  s
   g (I 1/2)   s

j    s

g  I 
  g I   s  


2 I  1
s *gs


 

gj
*
gj *
*g
 
*
gs
s



  
j  s
 

*g
s *gs
j s