7. Nuclear models
7.1
Fermi gas model
7.2
Shell model
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 p 2 dpV
dn =
(2h) 3
pF
n=
0
VpF3
4 p 2 dpV
= 2 3
(2h) 3
6 h
VpF3 ,n
N = 2n = 2 3
3 h
Spin 1/2 particles
Z=
If T=0 the nucleus is in the ground state and
pF (Fermi Momentum) is the maximum possible
momentum of the ground state.
3
VpF,
p
3 2 h 3
N= number of neutrons
Z= number of protons
V=
4 3 4 3
R = R0 A R0 = 1.21 fm
3
3
Z = N = A /2
3
o
3
F
2 3
1
3
A 4 R Ap
h 9 =
pF = pF,n = pF , p = 250MeV /c
2 3 3 h
R0 8 The nucleon moves in the
nucleus with a large momentum
pF2
33 MeV
Fermi Energy E F =
2mN
Binding Energy: BE/A= 7-8 MeV
V0=EF+B/A~ 40MeV
Nucleons are rather weakly bound in
the nucleus
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.
pF
E
< E Kin >=
Kin
p 2 dp
0
pF
2
p dp
3 pF2
=
20MeV
5 2mN
0
E Kin (N,Z) = N < E Kin,N > +Z < E Kin,Z >=
3
2
2
(NpF,N
+ ZpFZ
)
10mN
1
pFN
N 3 2 h 3 N 9 2 h 3 3
=
=
3
V
4 R0 A EKin
Decay
1
Z 3 h Z 9 2 h 3 3
pFZ = =
3
V
4 R0 A 5
2 5
2 2 3 3
3 h 9 h Z 3 + N 3 E Kin =
2
19mN R0 4 R03 3
A
2 3
E Kin
2
3
5
3
5
3
h 2 3 9 N + Z
3 h 2 9 = 2
=
2
2
10m
R0 10mN 4 R
4
N
0
A3
2
3
N-Z
5 (N Z) 2
A
+
+
...
9
A
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
h 2 2 2 2 = +
+
= E
2m
2m x 2 y 2 z 2 h 2 2 X(x)
= E X X(x)..... (r) = X(x)Y (y)Z(z)
2m x 2
YZE X X + XZE yY + XYE Z Z = EXYZ
2 X(x)
= kX
x 2
k=
1
2mE
h
X = Ae ik x + Beik x
X + =
2
cos k+ x
2a
1 h h2 2
k =
EX =
2m a 2m
2i
sin k x
2a
2
2
1 h 2
2
2
E=
( x + y + z )
2m a 3 a 3 p 3
h 2 2
p = 2mE = ( x + 2y + 2z )
a
2
a
a
: X =Y = Z =0
2
+
+
k =
, = 1,3,5.. k =
, = 0,2,4...
a
a
Boundary Conditions : at x = y = z =
X =
a
Volume in
which the
state density
is calculated
We count how many states we have in a spherical volume of radius between and +
2 = 2x + 2y + 2z
d = 4 2
1
a
a
dn = d = 2
=
p
d =
dp
8
2
h
h
a3
4 p 2 dpv
a2 2 a
2
dn =
p
dp = 2 3 p dp =
(2h) 3
2 2 h 2 h
2 h
2
p = 2mE
2
3
p dp = 2m E dE
Momentum volume
Space volume
Phase-space occupied by one state
3
2
dn = m ( 2 2 h 3 )1 v dE E = C1 E dE
Number of spin 1/2 particles which sit in the well:
EF
EF
3
2
dn
2
3
1
n= 2
dE = 2C1v E dE = ( 8m 2 )(3 h ) v E F3
dE
0
0
dn/dE
2/3
E
EF
EF= maximal energy of the
particle in the well
1 23 43 2 n E F = 3 h 30MeV
2m v
n
9
in Cu atoms : = 6 10 23 = 8 10 22 cm 3 E F 7eV
v
64
Hierarchy of energy eigenstates
of the harmonic oscillator potential
N = 2(n-1) + l
E = h(N + 3/2)
Shell Model Potential
Skin thickness
Spin Orbit Coupling
It changes the hierarchy of the energy levels:
V (r) = Vcentral (r) + Vls (r)
< ls >
h2
l /2 j = l + 1/2
Moving below
< ls > 1
=
(
j(
j
+
1)
l(l
+
1)
s(s
+
1))
=
Moving above
2
h2
(l + 1) /2 j = l 1/2
E ls =
2(l + 1)
< Vls (r) >
2
The energy levels transform into nlj levels.
Saxon-Wood Potential
Shell model
2(2l+1)
3s
2d
1g
1f
92
1h
2p
2s
1d
10
18
6
14
58
40
2
10
1p
6
1s
2
Spin-orbit
coupling
Single particle level calculated in the
shell model.
Single-Particle-Spectrum of the Wood-Saxon Potential
for a heavy nucleus
Energy of first excited nuclear level in gg-nuclei
Energy levels of some nuclei
Excitation Energy
4
2
He
16
8
O
40
20
Ca
48
20
Ca
208
82
Pb
Nuclear Magnetic Moments in Shell Model
A
μnucl = μN (gl li + gs si ) /h
i=1
1 for p
gl = 0 for n
5.58 for p
gs = 3.83 for n
A
< μnucl >= μN < 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
< μnucl
<J>
>= gnucl μN
h
A
gnucl = i=1
< JM J gl li + gs si JM J >
< JM J J 2 JM 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.
gl ( j( j + 1) + l(l + 1) s(s + 1)) + gs( j( j + 1) + s(s + 1) l(l + 1))
2 j( j + 1)
g g = gnucl J = gl ± s l J = l ± 1/2
2l + 1 gnucl =
μnucl
μN
Magnetic moments:
Shell model calculations and experimental values