Lecture Notes for Graduate Nuclear
Physics
(PHYS722/822 Fall 2018)
Week 1
Global Properties of Nuclei
- Range of masses 1.673x10-27
kg - several times 1030 kg (neutron stars). Nuclei proper
have
masses of about A*1.673x10-27
kg with A = 1...300. A = Z + N (number of protons plus number of
neutrons).
Heaviest nuclide found so far: Z = 118, N = 176, A = 294 ("Oganesson")
- Charge Q = Ze
- Spin = 0, 1/2, 1, 3/2, ...10 hbar in ground state
- Size roughly proportional to A1/3 -> approximately constant density
Mass measurements and definitions
- Mattauch mass spectrograph (velocity and momentum filter) -
singly
ionized
atoms
- Comparison to neutral 12C atom -> all masses given
for
neutral
atoms
- Mass excess: D(Z,N) = M(Z,N) -
(Z+N)*u
with u (= 1 a.m.u. = atomic mass unit) = m(12C) / 12 =
931.494
MeV/c2
- Binding energy B = M(Z,N) - Z*M(1H) - N*M(n) = D(Z,N)
- Z*D(1H) - N*D(n)
Range of stable nuclei
- Roughly along line Z=N, N>Z for heavier nuclei
- Absolute limits given by proton or neutron decay ( D(Z,N)
> D(1H) + D(Z-1,N)
or D(Z,N) > D(n)
+ D(Z,N-1) )
- Minimum mass near Z=N because of Pauli principle. Other nuclei
are
unstable
because of beta-decay or electron capture.
- Upper limit for A determined by Coulomb repulsion. Heavier nuclei
are
unstable
against alpha-decay or fission.
Liquid Drop model
- Nuclei of all sizes have very similar density (about 0.14
nucleons/fm3);
R = R0A1/3 (R0 = 1.2 fm)
- Nucleons inside nuclei only interact with their nearest neighbors
- Liquid Drop model is a classical approximation of many-nucleon
quantum
system, which guides us in our understanding of collective behaviour of
nuclei
- => shape of nucleon-nucleon interaction with repulsive core
and
"short"
range (a few fm) attraction
- => Nuclei can be described as drops of an incompressible
liquid
(with "van
der Waals" short-range forces); nucleons move "freely" throughout nuclear volume
- => Weizsaecker mass formula (see Povh et al.); have to add
Coulomb
repulsion,
asymmetry term (Pauli principle) and even/odd staggering term (pairing
force) "by hand"
- Explains valley of stability (single vs. double parabola for
m(Z-N) at
fixed A) against p, n, and beta+/- decay
- Explains limit of stability and actual process of fission
(including
alpha-decay)
- Actual process of fission: elongation following heating
(through n
capture
on 235U, for example) with subsequent breakup
- Nuclear reactions (either "direct" through "liquid exchange" or
via
"compound
nuclei" involving hot intermediate system followed by "nucleon
evaporation")
- Vibrations (quadrupole shape oscillations, Giant Dipole
Isovector
Resonance
- protons oscillate against neutrons)
- Rotations up to 60-80 hbar (rigid rotator: Eex = L2/2I; surface
shape
rotation;
elongation increasing with rotational velocity)
Isospin
- Mass formula does NOT depend on I3 = (Z-N)/2 except
for
proton-neutron
mass difference and Coulomb effects. It does depend on (I3)2,
though.
- States in neighboring isobars occur as multiplets with nearly
equal
masses
but different values of I3 up to a maximum |I3 max
| = I.
- Analogy with spin: Multiplets with 2S+1 members, all members
degenerate
except for effect of external magnetic field; energy only depends on S2,
not Sz VS.
Multiplets with 2I+1 members, all degenerate except for p-n mass
diffference
and Coulomb effects. Strong force ONLY depends on I2, not I3.
Nuclear forces are Charge Symmetric (unchanged
under
reversal
of sign of I3 = interchange of p's with n's) and even Charge
Independent (unchanged under any rotation in isospin
space
- all members of a given isospin multiplet behave the same).
Spin and magnetic moment
- Nuclei have spins from 0, 1/2, 1, 3/2, ... to several h-bar.
- Magnetic moments are given by m
= g*mN*S
(= nuclear magneton)
- g value of 2 is for pointlike Dirac-particle. Proton has gp
= 5.6 and neutron has gn = -3.8 => clear indication of
composite
structure.
Week 2
Fermi Gas model
- Nuclear density given by r(r) = rinf/(1+exp((r-R)/a)
(rinf = 0.16 N/fm3 is
the density of inifinte nuclear matter, R = R0A1/3
is the nuclear radius, and a is the width of the surface). The average
density of medium-sized nuclei is 0.14 N/fm3 .
- Potential energy for a single nucleon inside the nucleus follows
roughly
the same form because of short-range nature of nuclear force.
Self-consistent
solution of Schroedinger equation with potential determined by nuclear
density = Hartree-Fock method.
- Simple model: replace "realistic" potential by square well with
depth V0
and radius R. Fill 6-dim. phase space V*D3p
with nucleons, where each cell of size h3 can be occupied by
only 2 nucleons of the same kind (p or n), with spin up and down (Pauli
principle).
- First order result: all nuclei have the same maximum nucleon
momentum,
pfermi = 250 MeV/c. All states up to that maximum are
filled.
The root-mean-square average momentum is 200 MeV/c, and the average
kinetic
energy of all nucleons is 20 MeV. Since the binding energy is about 8
MeV/nucleon,
we conclude V0 = -28 MeV.
- The maximum kinetic energy is 33 MeV, which would lead to a
positive
total
energy (unbound nucleon). However, there is an additional "energy
barrier"
because the remaining nucleus would have to shrink against the "Fermi
gas
pressure", which comes out to 13 MeV for all nuclei. So the most
energetic
nucleon is still bound by 8 MeV.
- Fermi gas model explains why nucleons are "free" although their
mean
free
pathlength in nuclei should be less than 2 fm: there are no "open
states"
available into which they can scatter (Pauli principle).
- Fermi gas model also explains the asymmetry term as*(Z-N)2/4A
in the mass formula, at least its form.
- Refinements: Finite temperature (>0) => some nucleons
occupy even
higher
momentum states; some intra-nuclear scattering can occur. Fermi-gas
model
can explain level density for first few excited levels.
- Fermi momentum (=maximum momentum) for protons is proportional to
(Z/Volume)1/3
= (Z*r0/A * A*R03/R3)1/3
(the second term corrects for the actual nuclear radius vs. the
constant
density approximation R = A1/3*R0). Inserting all
constants yields pfermi(protons) = 250 MeV/c*(Z/2A)1/3 *
A1/3*R0/R and pfermi(neutrons) = 250
MeV/c*(N/2A)1/3 *
A1/3*R0/R. The overall contribution to the total
nuclear mass from the kinetic energy due to this "fermi motion" is
3/5*Z*(pfermi(protons))2/2m
+ 3/5*N*(pfermi(neutrons))2/2m. Expanding this in
a Taylor series around the average Z=N=A/2 with I3 as the
"small"
expansion parameter yields 20MeV*A + 44MeV*I32/A,
which is the same form (but not quite size) as the asymmetry term in
the
Weizsaecker mass formula.
- More detail on the "Fermi gas pressure": Assuming the constant
density
approximation R = A1/3*R0 holds for both parent
(A)
and daughter (A-1) nuclei, and Z=N=A/2 for both as well, we have a
Fermi
momentum of 250 MeV/c and therefore an average kinetic energy
of
20 MeV for both. For the parent, the total kinetic energy =
A*20MeV
can be written as sum of the kinetic energy 33MeV of the least bound
nucleon
plus the (A-1) times the initial average kinetic energy of all
remaining
nucleons, Tin,ave(A-1):
A*20MeV = 33MeV + (A-1)*Tin,ave(A-1) => (A-1)*Tin,ave(A-1)
= A*20MeV - 33MeV = (A-1)*20MeV - 13MeV. After the most energetic
nucleon
has been removed, these same A-1 remaining nucleons have to increase
their
kinetic energy (their momenta increase due to the smaller size of the
remaining
daughter nucleus) until their average energy is again 20MeV and their
total
energy is therefore (A-1)*20MeV. By comparison, it is clear that a
total
of 13 MeV are needed to "lift" all of these remaining nucleons up to
their
new kinetic energies. This energy has to be subtracted from the
single-particle
energy of the most energetic nucleon (+5 MeV) to get the true
separation
energy, which is therefore -8MeV (the same as the average binding
energy).
Biggest nucleus described by Fermi Gas: Neutron Star. Stability up to 2.x solar masses.
Shell model
- Try to solve Schroedinger equation for a single nucleon in
Nuclear
potential
- Exact solutions for simple approximations:
- Spherical Well potential with infinitely high walls: y(r)
= R(r)*Ylm(q,f) with R(r)=C*jl(kr)
(k is momentum from kinetic energy k2/2m; jl are
spherical Bessel functions). Each solution is specified by quantum
numbers
l = 0,1,2,...; m = -l,...,+l and n (= number of nodes in jl(kr)
between r=0 and r=R). Nomenclature for l: 0,1,2,3,4,... = s,p,d,f,g,...
Energy eigenvalues are order as 1s, 1p, 1d, 2s, 1f, 2p, 1g, 2d, 3s,...
yielding "magic numbers" (filled shells) for total proton or neutron
numbers
of 2,8,18,20,34,40,58,...
- Harmonic Oscillator potential has solutions with eigenvalues E
=
(N+3/2)hbarw,
where each N is composed again of radial n (nodes) and l. Similar
levels
as spherical well (but some are degenerate, e.g. 1d and 2s) leading to
fewer "magic numbers": 2,8,20, 40,...
- Expect full solution somewhere "in between" (using
self-consistent
shape
of potential - Wood Saxon, see Lecture 3).
- However, observed "magic numbers" are at 2,8,20,28,50,82,126
which
disagrees
for the larger ones. (Magic numbers are Z and N for those nuclei which
are particularly stable = low mass relative to their neighbors, have
very
high excitation energies for first excited state, are much more
abundant
in nature, etc.). Solution: Strong l.s coupling for nuclear
force leads to splitting of l-levels according to overall j quantum
number
(j = l+s).
- Other evidence for l.s coupling: Left-right asymmetry
for
scattering
of spin-up (polarized) protons from J=0 nuclei like 4He, 12C
etc. Origin of this coupling is basically relativistic (any covariant
force
acting on a Dirac particle leads to a l.s coupling term in
the
Dirac equation); example: magnetic moment interacting with magnetic
field
of moving charge.
- Consequences: magic numbers are predicted correctly; all nuclei
with Z
and N magic (doubly magic - closed shells) have spin-parity 0+. Single
particle picture: nuclei with just one nucleon in addition to closed
shells
(or one short of a closed shell -> "hole") have spin, parity and
magnetic
moment completely determined by that one "valence" nucleon/hole. See
textbooks
for consequences, e.g. for nuclear magnetic moment.
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