7. nuclear models - istituto nazionale di fisica...
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7. Nuclear models
7.1 Fermi gas model
7.2 Shell model
• Free electron gas: protons and neutrons moving quasi-freely within thenuclear volume
• 2 different potentials wells for protons and neutrons.
• Spherical square well potentials with the same radius
Fermi Gas Model
Statistics of the Fermi distribution
Given a volume V the numer of states dn goes like:
dn =4 p
2dpV
(2 h)3
n =4 p
2dpV
(2 h)3
=VpF
3
62h3
0
pF
If T=0 the nucleus is in the ground state andpF (Fermi Momentum) is the maximum possiblemomentum of the ground state.
N = 2n =VpF ,n
3
32h3
Z =VpF ,p
3
32h3
Spin 1/2 particles
N= number of neutrons
Z= number of protons
V =4
3R3
=4
3R0
3A R
0=1.21 fm
Z = N = A /2A
2=4
3
Ro
3ApF
3
32h3
pF = pF ,n = pF ,p =h
R0
9
8
1
3
250MeV /c
The nucleon moves in thenucleus with a large momentum
Fermi Energy
Binding Energy: BE/A= 7-8 MeV V0=EF+B/A~ 40MeV
EF =pF2
2mN
33 MeV
Nucleons are rather weakly bound inthe nucleus
Potential well in the Fermi-gas model
The neutron potential well is deeper that the proton well because of themissing Coulomb repulsion. The Fermi Energy is the same, otherwise the p-->ndecay would happen spontaneously. This implies that they are more neutronsstates available and hence N>Z the heavier the nuclei become.
< EKin >=
EKin p2dp
0
pF
p2dp
0
pF=3
5
pF2
2mN
20MeV
EKin (N,Z) = N < EKin,N > +Z < EKin,Z >=3
10mN
(NpF ,N2
+ ZpFZ2)
pFN =N 3
2h3
V
=
N 92h3
4 R03A
1
3
pFZ =Z 3
2h3
V
=
Z 92h3
4 R03A
1
3
EKin =3
19mN
h2
R0
92h3
4 R03
2
3 Z
5
3 + N
5
3
A
2
3
EKin =h2
R02
3
10mN
9
4
2
3 N
5
3 + Z
5
3
A
2
3
=3
10mN
h2
R02
9
4
2
3
A +5
9
(N Z)2
A+ ...
N-Z
EKin Decay
L’espressione precedente si ottiene da: D = N − Z
N= A2+
N − Z2
"
#$
%
&'=12A+D( )
Z= A2−
N − Z2
"
#$
%
&'=12A−D( )
sostituendo in:
N 5 3 + Z 5 3
A2 3!
"#
$
%&=
A+D2
!
"#
$
%&5 3
+A−D2
!
"#
$
%&5 3
A2 3
!
"
####
$
%
&&&&
e sviluppando i binomi arrestandosi al secondo ordine:
N 5 3 + Z 5 3
A2 3!
"#
$
%&=
12!
"#$
%&
53 A5 3 + 5
3A2 3D+10
18A−1 3D2 + A5 3 − 5
3A2 3D+10
18A−1 3D2
A2 3
!
"
###
$
%
&&&
N 5 3 + Z 5 3
A2 3!
"#
$
%&=
12!
"#$
%&
532A+10
9D2
A!
"#
$
%&=
12!
"#$
%&
23A+ 5
9D2
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
EB = T −V = −cVolA+ c
Sym
N − Z( )2
A
Calculation of the state density
h2
2m=
h2
2m
2
x2
+2
y2
+2
z2
= E
h2
2m
2X(x)
x2
= EX X(x)..... (r) = X(x)Y (y)Z(z)
YZEX X + XZEyY + XYEZZ = EXYZ
2X(x)
x2
= kX k =1
h2mE
X = A eik x
+ B eik x
Boundary Conditions : at x = y = z =a
2: X =Y = Z = 0
X+
=2
2acosk
+x X =
2i
2asink x k
+=
+
a, =1,3,5.. k =
a, = 0,2,4...
EX =h2
2mk2
=1
2m
h
a
2
E =1
2m
h
a
2
( x
2+ y
2+ z
2) p
2= 2mE =
h
a
2
( x
2+ y
2+ z
2)
3a3p3
a
Volume inwhich thestate densityis calculated
a
Considering the approximation that the nuclear potential shold have a sharpedge in correspondence of the nuclear radius, one can approximate that toparticles trapped in the pot potential
We count how many states we have in a spherical volume of radius between and +
2= x
2+ y
2+ z
2d = 4
2
dn =1
8d =
2
2=a
hp d =
a
hdp
dn =2
a2
2h2p2 a
hdp =
a3
22h3p2dp =
4 p2dpv
(2 h)3
Momentum volume
Space volume
Phase-space occupied by one state
p2
= 2mE p2dp = 2m
3EdE
dn = m
3
2 ( 22h3)1v dE E = C
1EdE
n = 2dn
dEdE = 2C
1v EdE = ( 8m
3
2 )(32h3)1v EF
2
3
0
EF
0
EF
Number of spin 1/2 particles which sit in the well:
EF= maximal energy of theparticle in the well
dn/dE
EF
E
EF
=1
2m
3
2
3
4
3h2 n
v
2 / 3
30MeV
in Cu atoms :n
v= 6 10
23 9
64= 8 10
22cm
3EF
7eV
Hierarchy of energy eigenstatesof 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
< ls >
h2
=1
2( j( j +1) l(l +1) s(s+1)) =
l /2 j = l +1/2
(l +1) /2 j = l 1/2
Els =2(l +1)
2<Vls(r) >
Moving belowMoving above
The energy levels transform into nlj levels.
Shell model
Single particle level calculated in theshell model.
Spin-orbitcoupling
Saxon-Wood Potential
1g58
1s
1p
2s
2p
1d
1f
40
92
2d3s
1h
2(2l+1)
2
6
102
14
18
6
10
Single-Particle-Spectrum of the Wood-Saxon Potentialfor a heavy nucleus
Energy of first excited nuclear level in gg-nuclei
Energy levels of some nucleiExcita
tion
Ene
rgy
PbCaCaOHe208
82
48
20
40
20
16
8
4
2
Nuclear Magnetic Moments in Shell Model
μnucl = μN (gl li + gssi) /hi=1
A
gl =1 for p
0 for n
gs =5.58 for p
3.83 for n
< μnucl >= μN < nucl gl li + gssi nucl > /hi=1
A
Wigner-Eckart Theorem: The expectation value of any vector operator of a systemis equal to the projection onto ist angular momentum
< μnucl >= gnuclμN< J >
hgnucl =
< JMJ gl li + gssi JMJ >
< JMJ J2JMJ >i=1
A
In the case of a single nucleon in addition to the close shell the angularmomentum of the closed shell nuclei couples to 0. The nuclear magneticMomentum of is equal to the nuclear magnetic mom. of the valence nucleon.
gnucl =gl ( j( j +1) + l(l +1) s(s+1)) + gs( j( j +1) + s(s+1) l(l +1))
2 j( j +1)
μnuclμN
= gnuclJ = gl ±gs gl
2l +1
J = l ±1/2
Magnetic moments:Shell model calculations and experimental values