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

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#$

%

&'=12A−D( )

sostituendo in:

N 5 3 + Z 5 3

A2 3!

"#

$

%&=

A+D2

!

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$

%&5 3

+A−D2

!

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$

%&5 3

A2 3

!

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&&&&

e sviluppando i binomi arrestandosi al secondo ordine:

N 5 3 + Z 5 3

A2 3!

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$

%&=

12!

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53 A5 3 + 5

3A2 3D+10

18A−1 3D2 + A5 3 − 5

3A2 3D+10

18A−1 3D2

A2 3

!

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$

%

&&&

N 5 3 + Z 5 3

A2 3!

"#

$

%&=

12!

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532A+10

9D2

A!

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$

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12!

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23A+ 5

9D2

A!

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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