1. 7. Nuclear models
Fermi gas model
Shell model

2. Statistics of the Fermi distribution
Fermi Gas Model
Free electron gas: protons and neutrons moving quasi-freely within the nuclear volume
2 different potentials wells for protons and neutrons.
Spherical square well potentials with the same radius
Statistics of the Fermi distribution
Given a volume V the numer of states dn goes like:
4p2dpV
dn 
If T=0 the nucleus is in the ground state and
(2h)3
p (Fermi Momentum) is the maximum possible F
momentum of the ground state.
n  
pF 4p2dpV
Vp3 
(2h)3 6 2h3 F
N= number of neutrons Z= number of protons
Vp3
Vp3
Z F, p
N  2n F ,n
3 2h3
3 2h3
Spin 1/2 particles

3. p2 o F  p Fermi Energy E F  33 MeV 2m 4 4 V  R  R A
3 3
R  1.21 fm 3 3 1
A 4 R3 Ap3
h  9  3
o F  p
Z  N  A /2
 p  p 
 
 250MeV /c
2 3 3 2h3 F
F ,n
F , p
R  8 
The nucleon moves in the nucleus with a large momentum p2
Fermi Energy E F  33 MeV F 2m N
Binding Energy: BE/A= 7-8 MeV V0=EF+B/A~ 40MeV
Nucleons are rather weakly bound in the nucleus

4. Potential well in the Fermi-gas model
The neutron potential well is deeper that the proton well because of the missing Coulomb repulsion. The Fermi Energy is the same, otherwise the p-->n decay would happen spontaneously. This implies that they are more neutrons states available and hence N>Z the heavier the nuclei become.

5.  E p2dp  p2dp p2  N  3 2h3   N  9 2h3 3 pFN   EKin  
 E  0 F  20MeV
Kin pF
 p2dp
5 2m N 3
E (N,Z)  N  E  Z  E 
(Np 2
Zp ) 2
Kin
Kin,N
Kin,Z
10m
F,N FZ N 1
 N  3 2h3   N  9 2h3 3
pFN  
    1
EKin
 Decay  V
 
4R A 3
 Z  3 2h3   Z  9 2h3 3
pFZ   
  
   V
4R A 3
2  5 5 
3 h2  9 2h3  3  Z 3  N 3 
EKin 
1190m
 
R 4R
0  3
   5 2
A 3 5  N 
N-Z h2 2 2
3  9  3 N 3  Z 3
5 (N  Z)2
10mN R0  84   9 A
3 h2  9  3  
 ...
EKin 
  
2 
  A 
R0 10mN  4  2 2
A 3
p2, se Z = N, di ciascun nucleone F

6. N  Z   A  D 5 3  A  D 5 3   N 5 3  Z 5 3     
L’espressione precedente si ottiene da:
D  N  Z
N= A  N  Z   1 A  D
 
2  2  2
Z = A  N  Z   1 A  D
 
2  2  2
sostituendo in:
 A  D 5 3  A  D 5 3 
 N 5 3  Z 5 3 
    
   2   2    
A2 3  
A2 3   
e sviluppando i binomi arrestandosi al secondo ordine:
5  A5 3
2 
 N 5 3  Z 5 3 
A D  A D  A  A D  A D 
2 3  1 3
 1 3  
 
      A
2 3
 2    5 A
2 3
 N 5 3  Z 5 3  2
 1 3 
10 D2 
 1 3 
5 D2  
 
 
 2A 
  
  A 
A  2  
2 3
9 A   2   9 A 
NOTA BENE:
A) Il risultato prova che l’energia cinetica totale ha un MINIMO quando N = Z, e dato che E(totale) = T –V, con T proporzionale ad A, E(totale), che e’ negativa ed e’ anche la BINDING ENERGY totale, e’ proporzionale ad A (primo termine dell’equazione semiempirica delle masse)
B) Aggiungendo il secondo termine, come nell’equazione semiempirica delle masse, la BINDING ENERGY diminuisce in modulo. A parte I valori dei coefficienti, il modello a gas di Fermi predice correttamente 2
N  Z  A
EB  T  V  c A  c
Vol
Sym

7. Calculation of the state density
Considering the approximation that the nuclear potential shold have a sharp edge in correspondence of the nuclear radius, one can approximate that to particles trapped in the pot potential h2
h2 2 2 2 
    
  
 E 2m
2m x y 2 z 
2 2
Volume in which the state density is calculated h2
2 X (x) 
2m x 2
 EX X (x)..... (r)  X (x)Y (y)Z(z)
YZEX X  XZEyY  XYEZ Z  EXYZ a a
2 X (x) 1
 kX k  2mE h
x 2
X  A e  B e
ik x 
 ik x 
Boundary Conditions : at x  y  z  a : X  Y  Z  0 2
   2
2i  
X  
cos k  x
X   sin k x k   ,   1,3,5.. k 
,   0,2,4...  2a
  2a 
 
a a h2 2
1  h
EX  k  2
  2m 
2m  a  2
1  h 2
 h
E 
 
(     )
2 2 2
p2  2mE   
(     ) 2 2 2
2m  a  x y z
 a  x y z
 3  a3 p3

8. a2 a3 33  3 h   2m   v         d  42 1  2 a a
We count how many states we have in a spherical volume of radius between 
and 
  
2 2
   
2 2
d  42
x y z 1  2 a a
dn  d  8
  2
  p
d  dp h h
Momentum volume a2 a3 
2 a
h 2 h
4p dpv 2
dn  p
2  2h2
dp 
p dp  2
Space volume
Phase-space occupied by one state
2 3
(2h)3
p2  2mE p2dp  3
2m3 E dE
dn  m 2 ( 2 2h3 )1v dE E  C E dE 1
Number of spin 1/2 particles which sit in the well: EF
n   2 dE  2C v  dn EF
3 2
E = maximal energy of the
E dE  ( 8m )(3 h ) vE
2 3
2 3 1 F dE 1 F
particle in the well
dn/dE
 E
 1 
2 4
2/ 3
2 n 
EF  
33  3 h  
 30MeV
 2m   v 
in Cu atoms : n  6 1023  9  8 1022 cm3  E
v 64
 7eV F E F

9. Hierarchy of energy eigenstates of the harmonic oscillator potential
N = 2(n-1) + l E = h(N + 3/2)

10. Shell Model Potential
Skin thickness

11. Spin Orbit Coupling h2 (l  1) /2 V (r)  V (r)  V (r)  ls 
It changes the hierarchy of the energy levels:
V (r)  V (r)  V (r)  ls  h2
central ls
 ls  1 
l /2 j  l  1/2
Moving below Moving above
 ( j( j  1)  l(l  1)  s(s  1))  
h 2 2
(l  1) /2
j  l 1/2
E  2(l  1)  V (r) 
ls ls 2
The energy levels transform into nlj levels.

12. Saxon-Wood Potential Shell model 2 2(2l+1) Spin-orbit coupling
1h 2d 92 3s 10 58 1g 18 2p 6 40 1f 14
2s 1d 2 10 1p
Spin-orbit coupling 6 1s 2
Single particle level calculated in the shell model.

13. Single-Particle-Spectrum of the Wood-Saxon Potential for a heavy nucleus

14. Energy of first excited nuclear level in gg-nuclei

15. Energy levels of some nuclei
16O 40Ca
8 20
48Ca
208Pb
4 He 2 20 82
Excitation Energy

16. nucl  N (gl li  gs si )/ h
Nuclear Magnetic Moments in Shell Model A
nucl  N (gl li  gs si )/ h
i1 1
for p 5.58
for p
g  
g   l 
0 for n A s 
3.83 for n
 nucl  N  nucl gl li  gs si nucl  / h
i1
Wigner-Eckart Theorem: The expectation value of any vector operator of a system is equal to the projection onto ist angular momentum
 J  h A
gnucl  
i1
 JMJ gl li  gs si JMJ 
 nucl  gnuclN
 JM J 2 JM 
J J
In the case of a single nucleon in addition to the close shell the angular momentum of the closed shell nuclei couples to 0. The nuclear magnetic Momentum of is equal to the nuclear magnetic mom. of the valence nucleon.
g  gl ( j( j  1)  l(l  1)  s(s  1))  gs( j( j  1)  s(s  1)  l(l  1))
nucl
2 j( j  1) 
nucl  g
gs  gl 
J  l  1/2
nucl J  gl   N 
2l  1 

17. Magnetic moments:
Shell model calculations and experimental values