Nuclear models

Models we will consider…• Independent particle shell model

• Look at data that motivates the model• Construct a model• Make and test predictions from the model

Our approach…

• Collective models

• Fermi gas model

Shell Model - data2p separation energy (between isotones)

Becomes much smaller after 8, 20, 28, 50, 82, 126

2n separation energy (between isotopes)

Shell Model - dataARn + A-4Po

T

Sudden rise at N = 126

Neutron capture cross section

Very small at N = 28, 50, 82, 126

Abrupt change in nuclear radius at N = 20, 28, 50, 82, 126R

Ravg

Shell Model - data

And, the observation of discrete photon energies E emitted from nuclear de-excitation

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

Shell ModelAssume 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 non-relativistic Schrodinger quantum mechanics.

Assume that the potential, V, is spherically symmetric and therefore only a function of r, V(r)

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V r( ) = −Vo r ≤ R

V r( ) = 0 r > R

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V r( )∝ r 2 €

V r( ) = −Vo1+ exp r − R( ) /a[ ]

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Vso r( )r L ⋅r s Spin-orbit potential

Shell Model

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Vso r( )r L ⋅r s

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ˆ H , ˆ J 2[ ] = 0 ˆ H , ˆ J z[ ] = 0

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J 2 =r J ⋅

r J

r J =

r L +

r s

J 2 =r L ⋅

r L + 2

r L ⋅r s +

r s ⋅r s

J 2 = L2 + s 2 + 2r L ⋅r s

r L ⋅r s = J 2 − L2 − s 2

2

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J 2 = j j +1( )h2

L2 = l l +1( )h2

s 2 = s s +1( )h2

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r L ⋅ r s = 1

2j j +1( ) − l l +1( ) − s s +1( )[ ] h2

Q.M.

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ψn,l ,s, j ,m j

good quantum numbers

Shell Model

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2 ⋅ 2l +1( )

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r L ⋅ r

s j=l +1/2

−r L ⋅ r

s j=l −1/ 2

=2l +1( )

2h2

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ψn,l ,s, j ,m j

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2 j +1( )Multiplicities --

2 spin states

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ml

different states

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m j

different states

=

Energy difference (splitting) increases with

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l

Shell Model energy levels

Energy splitting increases with

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l

Spectroscopic state

multiplicity

Systematics…

A Z NNumber

of known stable

nucleonsStable Radio-

active μ

I Odd Odd Even 50 50 11 Uullψ lre & po.

II Odd Even Odd 55 36 4 Uullψ μll, ne.

III Even Odd Odd 4 4 9 Uullψ poitive

IV Even Even Even 165 12 1 Indeterμinte

Nucleon Clifiction Nucler μoμent

Nuclear magnetic moments

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μN = 3.152451×18−18 MeV /gauss

μ β = 5.788378 ×18−15 MeV /gauss

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μNμ β

= meM N

≈ 11836€

μ p = 2.7928 μ N

μ n = −1.9131μ NIntrinsic (measured) dipole magnetic moments

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μL* = e

2Mr L

μ l* = eh

2Ml l +1( )

μ l* = μ N l l +1( )

L is orbital angular momentum for single nucleon

M is nucleon mass

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μl ≡lμN max z-axis projection

Nuclear magnetic moments

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μ p* = 2μ N s s +1( )

μ p = ±μ N

From electron case, you expect to have for this fermion --

Does not agree with measurement

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μ p = 2.7928 μ N

μ n = −1.9131μ NMeasured dipole magnetic moments

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μ p* = gpμ N s s +1( ) = gpμ N

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μ p = ± gp μ N12

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gp=μ p

*

μ N s s +1( )

gp=2μ p

μ N ; gp= 2 2.7928( ) = 5.5856

Nuclear magnetic momentsAnd, by the same analysis, one gets --

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μ p = 2.7928 μ N

μ n = −1.9131μ NMeasured dipole magnetic moments€

μn* = gn μ N s s +1( ) = gn μ N

32

μ n = ± gn μ N12

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gn = μ n*

μ N s s +1( )

gn =2μ nμ N

; gn = 2 −1.9131( ) = -3.8262

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gp = 5.5856

gn = −3.8262

Nuclear magnetic momentsConsider 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

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l*

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s*

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j*

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s*

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l*

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s*

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j*

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s*

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j = l − s

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j = l + s

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gl

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gl

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gs

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g

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gs

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μn* = gn μ N s s +1( ) = gn μ N

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μ n = ± gn μ N12

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μ p* = gpμ N s s +1( ) = gpμ N

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μ p = ± gp μ N12

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μl* = gl μN l * = gl μN l l +1( )

μ l = ± gl μN l

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μ j* = gμN j* = gμN j j +1( )

μ j = ± gμN j

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proton : gl =1 neutron : gl = 0

Nuclear magnetic moments

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cos l * j*( ) =

l *( )

2+ j*

( )2

− s*( )

2

2l * j*

cos s* j*( ) =

s*( )

2+ j*

( )2

− l *( )

2

2s* j*

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μ j* = μ l * cos l * j*( ) + μ s* cos s* j*

( )

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g j*( ) = gl l *

( )cos l * j*( ) + gs s*

( )cos s* j*( )[ ]μ N

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g( ) = gl

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟1+

l *( )

2− s*

( )2

j*( )

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥+ gs

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟1−

l *( )

2− s*

( )2

j*( )

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

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l*

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s*

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j*

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s*

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l*

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s*

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j*

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s*

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j = l − s

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j = l + s

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gl

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gl

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gs

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g

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g€

gs

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gs

Nuclear magnetic moments

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g( ) = gl

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟1+

l *( )

2− s*

( )2

j*( )

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥+ gs

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟1−

l *( )

2− s*

( )2

j*( )

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

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j*( )

2= j j +1( ) = l + 1

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟⋅ l + 1

2+1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

j*( )

2= l l + 2( ) + 3

4

s*( )

2= 3

4

Consider the case:

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j = l + s( )

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g( ) = gllj

+ gs1/2

j…some algebra happens here…

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l*

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s*

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j*

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s*

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l*

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s*

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j*

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s*

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j = l − s

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j = l + s

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gl

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gl

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gs

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g

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gs

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gs

Nuclear magnetic momentsConsider the case:

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j = l + s( )

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g( ) = gllj

+ gs1/2

j

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l*

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s*

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j*

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s*

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l*

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s*

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j*

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s*

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j = l − s

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j = l + s

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gl

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gl

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gs

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g

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gs

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μ = g( ) j = g( )Iμ = gl l + gssμ = gl l + μ s

But, if

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I = l + s( )

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μ =gl (I −1/2) + μ s

Nuclear magnetic momentsConsider the case:

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j = l − s( )

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g( ) = gll +1j +1

− gs1/2j +1

Four cases to consider: both cases shown here for odd proton & odd neutron

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l*

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s*

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j*

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s*

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l*

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s*

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j*

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s*

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j = l − s

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j = l + s

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gl

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gl

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gs

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g

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μ = gl l + gl + μ s[ ]I

I +1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

But, if

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I = l − s( )

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μ =gl I − μ s − gl

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟ II +1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

Nuclear magnetic moments

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l*

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s*

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j*

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s*

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l*

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s*

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j*

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s*

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j = l − s

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j = l + s

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gl

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gl

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gs

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g

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gs

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gs

Proton: Neutron:

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gl =1, μ s = μ p

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gl = 0, μ s = μ n

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μ =gl I − μ s − gl

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟ II +1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

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j = l − s( )

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μ =gl (I −1/2) + μ s

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j = l + s( )