Structure of nuclei
3224
Nuclear and Particle Physics
Ruben Saakyan
UCL
Fermi gas model. Assumptions
• The potential that an individual nucleon feels is the superposition of
the potentials of other nucleons. This potential has the shape of a
sphere of radius R=R0A1/3 fm, equivalent to a 3-D square potential
well with radius R
• Nucleons move freely (like gas) inside the nucleus, i.e. inside the
sphere of radius R.
• Nucleons fill energy levels in the well up to the “Fermi energy” EF
• Potential wells for protons and neutrons can be different
– If the Fermi energy were different for protons and neutrons, the nucleus would
undergo b-decay into an energetically more favourable state
– Generally stable heavy nuclei have a surplus of neutrons
– Therefore the well for the neutron gas has to be deeper than for the proton gas
– Protons are therefore on average less strongly bound than neutrons (Coulomb
repulsion)
• 2 protons/2 neutrons per energy level, since spins can be
Fermi momentum and Fermi
energy
• The number of possible states available to a nucleon
inside a volume V and a momentum region dp is
4 V 2
n( p)dp  dn 
p dp
3
(2 )
density of states factor
• In the nuclear ground state all states up to a maximum
momentum, the Fermi momentum pF, will be occupied.
Integration leads to the following number of states n.
Since every state can contain two fermions, the
number of protons Z and neutrons N are also given:
V ( pF ) 3
n
6 2 3
V ( pFp )3
Z
3 2 3
The nuclear volume V is given as
V ( pFn )3
N
3 2 3
4
4
V   R 3   R03 A
R0  1.21 fm
3
3
from electron scattering
Fermi momentum and Fermi
energy
• Assuming the depths of the neutron and proton wells are
the same and Z = N = A/2, the Fermi momentum
 9 
n
p
pF  p F  p F   
R0  8 
1/ 3
 250 MeV / c
• The energy of the highest occupied state, the Fermi
energy is
pF2
EF 
 33 MeV
2M
M is the nucleon mass
Fermi gas model. Potential
• The difference between the Fermi energy and the top of the
potential well is the binding energy B’ = 7-8 MeV/nucleon that we
already know from the liquid drop model
• The depth of the potential well V0 is to a good extent
independent of the mass number A:
V0  EF  B '  40 MeV
Derivation of symmetry term
Derivation of symmetry term (ctd)
Nuclear Models
• The liquid drop model allows reasonably good
descriptions of the binding energy. It also gives a
qualitative explanation for spontaneous fission.
• The Fermi gas model, assuming a simple 3D
well potential (different for protons and neutrons)
explained the terms in SEMF that were not
derived from the liquid drop model.
• Nucleons can move freely inside the nucleus.
This agrees with the idea that they experience
an overall effective potential created by the sum
of the other nucleons
• There are things which the Fermi gas model can
not explain. This will lead us to the Shell Model.
The Shell Model
Basics
• The Shell Model is based very closely on the ideas from
atomic physics: orbital structure of atomic electrons
• Atomic energy levels n = 1, 2, 3,… In nuclear physics
we are not dealing with the same simple Coulomb
potential: radial node quantum number n
• Atomic Physics: for any n there are energy-degenerate
levels with orbital angular momentum l = 0,1,2,…,(n-1)
• For any l there are (2l+1) sub-states with different
values of the projection of l along any chosen axis
ml = -l, -l+1,…,0,1,…,l-1,l – magnetic quantum number
Due to rotational symmetry of Coulomb potential
these sub-states will be degenerate in energy
Basics
• Since electrons have spin-1/2, each of the states
above can be occupied by 2 electrons with ,
corresponding to the spin-projection number
ms=1/2. Again both states will have the same
energy.
• Summarizing, any energy eigenstate in, say, H2
atom has quantum numbers (n, l, ml , ms ) and for
any n there will be nd degenerate states
n -1
nd  2 (2l  1)  2n 2
l 0
Basics
• This degeneracy can be broken if there is a preferred
direction in space (magnetic field). Recall spin-orbit
coupling and fine structure.
• Going beyond H2 atom one has to introduce electronelectron Coulomb interaction.
• This introduces splitting to any level n according to l.
The degeneracies in ml and ms are unchanged.
• If a shell or sub-shell is filled, then  m  0 and  m  0
s
l
In this case Pauli principle implies L = S = 0 and J = L + S = 0
Such atoms (with paired off electrons) are chemically inert
Z = 2, 10, 18, 36, 54
Nuclear Shell Structure Evidence
Neutrons
Magic numbers
Binding energy curve revisited
Infinite Spherical Well
Spherical Harmonics
Shell structure
Infinite Well/Harmonic oscillator
Shell Model Potential
Spin-Orbit Potential
Shell Model – Energy Levels
Shell Model – Energy Levels
Observed magic
numbers
2
8
20
28
50
82
126
…
Spins in the Shell Model
• Shell model can be used to make predictions
about the spins of ground states
• A filled sub-shell must have J=0
• This means that, since magic number nuclides
have closed sub-shells, the contribution to the
nuclear spin from protons/neutrons with magic
number must be zero
• Hence doubly magic nuclei are predicted to have
zero nuclear spin (observed experimentally)
Spins in the Shell Model
• All even-even nuclei have zero nuclear spin
• Pairing hypothesis: For ground state nuclei,
pairs of n and p in a given sub-shell always
couple to give a combined angular momentum
of zero, even when the sub-shell is not filled.
• Last neutron/proton determines the net nuclear
spin.
– In odd-A there is only one unpaired nucleon. Net spin
can be determined precisely
– In even-A odd-Z/odd-N nuclides we have an unpaired
p and an unpaired n. Hence the nuclear spin will lie in
the range
|jp-jn| to (jp+jn)
Parities in the Shell Model
• The parity of a single-particle quantum state
depends exclusively on l with P = (-1)l
• P =  Pi . A pair of particles with the same l will
always have P = +1
• From pairing hypothesis we have:
• Pnucleus = Plast_p  Plast_n
• The parity of any nuclide (including odd-odd)
can be predicted (confirmed by experiment)
Magnetic moments in the Shell
Model
 m = gj j mN, mN – nuclear magneton, gj – Lande ge
m

factor
2M
• For odd-odd nuclei we have to consider an
unpaired n and an unpaired p
• For even-odd nuclei one has to “only” find out
orbital and intrinsic components of magnetic
moment of the single unpaired nucleon
N
p
Magnetic moments in the Shell Model
We need to combine gs s and gl l
j ( j  1)  l (l  1) - s ( s  1)
j ( j  1) - l (l  1)  s ( s  1)
gl 
gs
2 j ( j  1)
2 j ( j  1)
IF j  l  1/ 2
jg j  gl l  g s / 2
for j  l  1/ 2
gj 
1 

 1 
jg j  gl j 1 
 - gs j 

 2l  1 
 2l  1 
for j  l - 1/ 2
Magnetic moments in the Shell Model
Since gl = 1 for p and gl = 0 for n, gs  +5.6 for p and gs  -3.8 for n
1
 j  2.8
for j  l  1/ 2
2
1 
2.3

 1 
jg proton  j 1 
5.6

j

1
for j  l - 1/ 2



j 1
 2l  1 
 2l  1 
1
jg neutron  -3.8   -1.9
for j  l  1/ 2
2
 1  1.9 j
jg neutron  3.8  j 
for j  l - 1/ 2

 2l  1  j  1
jg proton  l  5.6 
For a given j the measured moments lie between j = l -1/2 and j = l+1/2
but beyond that the model does not predict the moments accurately
Unlike spin and parities the Shell Model does not predict
magnetic moments very well
Excited states in the Shell
Model
• First one or two excited states can be predicted relatively
easily
• Consider 178O
protons: (1s1/2)2 (1p3/2)4 (1p1/2)2
neutrons: (1s1/2)2 (1p3/2)4 (1p1/2)2 (1d5/2)1
• 3 possibilities for 1st excited state
– One of the 1p1/2 protons to 1d5/2, giving (1p1/2)-1 (1d5/2)1
– One of the 1p1/2 neutrons to 1d5/2, giving (1p1/2)-1 (1d5/2)2
– 1d5/2 neutron to next level, 2s1/2 or 1d3/2 giving (2s1/2)1 or (1d3/2)1
• The 3d possibility corresponds to the smallest energy shift and
therefore it is favourable
Excited states in the Shell
Model
• Comparing the above predictions with
experimental results it was found that the
expected excited states do exist but not always
in precisely the order anticipated
• Higher excited states calculation is much more
complicated
• Collective model is an attempt to bring together
shell and liquid drop models
• Recent encouraging developments in nuclear
calculations due to progress in computing power
b-decay. Fermi theory
• W, Z, quarks were not known. Theory based on
general principles and analogy with QED
A
Z
X
Y  e-  e
A
Z 1
• Fermi’s Second Golden Rule

2
2
M n( E )
•  - transition rate, |M| - matrix element, n(E) – density
of states (phase space determined by the decay’s
kinematics)
b-decay. Fermi theory
M   ( gO) i dV
*
f
• g – dimensionless coupling constant, O five basic classes of Lorentz invariant
interaction operators
– scalar S, pseudo-scalar P, vector V, axialvector A, tensor T
– The main difference is the effect on the spin
states of the particles
• Fermi guessed that O should be of vector
type (EM interaction transmitted by photon
with spin-1)
Fermi coupling constant
• If we do not consider particle spins matrix element can
be thought in terms of a classical weak interaction
potential, like the Yukawa potential
• Point-like interaction. Matrix element in this case is just
a constant M = GF/V
• GF – Fermi coupling constant
• Can be applied to any weak process provided the
energy is not too great
• Extracted from muon decay GF = 90 eV fm3
• Often quoted as
G /( c)3  1.166 10-5 GeV -2
F
b-decay. Electron momentum
distribution
E  Ee  E
d 
2
M n ( E - Ee )ne ( Ee )dEe
2
Recall that n( pe ) dpe 
V
 2 
2
4

p
and the same for n
e dpe
3
by changing variables using dp / dE  E / pc 2
4 V
n( Ee )dEe 
pe Ee dEe
3 2
(2 ) c
Since M  GF / V
GF2
d
 3 7 4 pe Ee p E
dEe 2 c
and similar for n( E )
where p c  E2 - m2c 4 
 E - Ee 
2
- m2c 4
b-decay. Electron momentum
distribution
Fermi screening factors F(Z, Ee)
GF2
d dEe d

 3 7 2 pe2 p E
dpE dpe dEe 2 c
If m  0 then p  E / c and
d GF2 pe2 p2 GF2 pe2 E2 GF2 pe2 ( E - Ee )2



3 7
3 7 3
dpe 2 c 2 c
2 3 7 c3
Spectra shifted for b+ w.r.t. b-!
Possible changes of nuclear spins are not taken into account
If the change is > 1, the decay is suppressed
Kurie plots and the neutrino
mass
• Studying b-spectrum around the end point can be used to measure me
• Kurie plots are the most obvious
2 2
d  F ( Z , Ee )GF pe  E - Ee 

dpe
2 3 7 c3
H ( Ee ) 
2
Qb
d
1
 E - Ee
dpe pe K ( Z , pe )
F(Z, Ee) and constants are here
3H
3
spectrum and the neutrino mass
H  3 He  e -   e
3H
most suitable isotope
• Low Qb, Qb  E0 = 18.6 keV
• Simple atom
World’s best result (Mainz, Troitsk)
me < 2.2 eV/c2
Future experiment: KATRIN
Sensitivity: me ~ 0.2 eV/c2
Probably the lowest possible limit for this technique
Katrin detector transportation
from Deggendorf to Karlsruhe (400km away) but had to make a detour of…9000 km