Lecture Notes for Graduate Nuclear
(PHYS722/822 Fall 2018)
Global Properties of Nuclei
- Range of masses 1.673x10-27
kg - several times 1030 kg (neutron stars). Nuclei proper
masses of about A*1.673x10-27
kg with A = 1...300. A = Z + N (number of protons plus number of
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) -
- Comparison to neutral 12C atom -> all masses given
- Mass excess: D(Z,N) = M(Z,N) -
with u (= 1 a.m.u. = atomic mass unit) = m(12C) / 12 =
- 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
because of beta-decay or electron capture.
- Upper limit for A determined by Coulomb repulsion. Heavier nuclei
against alpha-decay or fission.
Liquid Drop model
- Nuclei of all sizes have very similar density (about 0.14
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
system, which guides us in our understanding of collective behaviour of
- => shape of nucleon-nucleon interaction with repulsive core
range (a few fm) attraction
- => Nuclei can be described as drops of an incompressible
der Waals" short-range forces); nucleons move "freely" throughout nuclear volume
- => Weizsaecker mass formula (see Povh et al.); have to add
asymmetry term (Pauli principle) and even/odd staggering term (pairing
force) "by hand"
- Explains valley of stability (single vs. double parabola for
fixed A) against p, n, and beta+/- decay
- Explains limit of stability and actual process of fission
- Actual process of fission: elongation following heating
on 235U, for example) with subsequent breakup
- Nuclear reactions (either "direct" through "liquid exchange" or
nuclei" involving hot intermediate system followed by "nucleon
- Vibrations (quadrupole shape oscillations, Giant Dipole
- protons oscillate against neutrons)
- Rotations up to 60-80 hbar (rigid rotator: Eex = L2/2I; surface
elongation increasing with rotational velocity)
Nuclear forces are Charge Symmetric (unchanged
of sign of I3 = interchange of p's with n's) and even Charge
Independent (unchanged under any rotation in isospin
- all members of a given isospin multiplet behave the same).
- Mass formula does NOT depend on I3 = (Z-N)/2 except
mass difference and Coulomb effects. It does depend on (I3)2,
- States in neighboring isobars occur as multiplets with nearly
but different values of I3 up to a maximum |I3 max
| = I.
- Analogy with spin: Multiplets with 2S+1 members, all members
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
and Coulomb effects. Strong force ONLY depends on I2, not I3.
Spin and magnetic moment
- Nuclei have spins from 0, 1/2, 1, 3/2, ... to several h-bar.
- Magnetic moments are given by m
(= 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
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
the same form because of short-range nature of nuclear force.
solution of Schroedinger equation with potential determined by nuclear
density = Hartree-Fock method.
- Simple model: replace "realistic" potential by square well with
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
- First order result: all nuclei have the same maximum nucleon
pfermi = 250 MeV/c. All states up to that maximum are
The root-mean-square average momentum is 200 MeV/c, and the average
energy of all nucleons is 20 MeV. Since the binding energy is about 8
we conclude V0 = -28 MeV.
- The maximum kinetic energy is 33 MeV, which would lead to a
energy (unbound nucleon). However, there is an additional "energy
because the remaining nucleus would have to shrink against the "Fermi
pressure", which comes out to 13 MeV for all nuclei. So the most
nucleon is still bound by 8 MeV.
- Fermi gas model explains why nucleons are "free" although their
pathlength in nuclei should be less than 2 fm: there are no "open
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
momentum states; some intra-nuclear scattering can occur. Fermi-gas
can explain level density for first few excited levels.
Biggest nucleus described by Fermi Gas: Neutron Star. Stability up to 2.x solar masses.
- Fermi momentum (=maximum momentum) for protons is proportional to
= (Z*r0/A * A*R03/R3)1/3
(the second term corrects for the actual nuclear radius vs. the
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
A1/3*R0/R. The overall contribution to the total
nuclear mass from the kinetic energy due to this "fermi motion" is
+ 3/5*N*(pfermi(neutrons))2/2m. Expanding this in
a Taylor series around the average Z=N=A/2 with I3 as the
expansion parameter yields 20MeV*A + 44MeV*I32/A,
which is the same form (but not quite size) as the asymmetry term in
Weizsaecker mass formula.
- More detail on the "Fermi gas pressure": Assuming the constant
approximation R = A1/3*R0 holds for both parent
and daughter (A-1) nuclei, and Z=N=A/2 for both as well, we have a
momentum of 250 MeV/c and therefore an average kinetic energy
20 MeV for both. For the parent, the total kinetic energy =
can be written as sum of the kinetic energy 33MeV of the least bound
plus the (A-1) times the initial average kinetic energy of all
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
has been removed, these same A-1 remaining nucleons have to increase
kinetic energy (their momenta increase due to the smaller size of the
daughter nucleus) until their average energy is again 20MeV and their
energy is therefore (A-1)*20MeV. By comparison, it is clear that a
of 13 MeV are needed to "lift" all of these remaining nucleons up to
new kinetic energies. This energy has to be subtracted from the
energy of the most energetic nucleon (+5 MeV) to get the true
energy, which is therefore -8MeV (the same as the average binding
- Try to solve Schroedinger equation for a single nucleon in
- 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
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
- Harmonic Oscillator potential has solutions with eigenvalues E
where each N is composed again of radial n (nodes) and l. Similar
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
of potential - Wood Saxon, see Lecture 3).
- However, observed "magic numbers" are at 2,8,20,28,50,82,126
for the larger ones. (Magic numbers are Z and N for those nuclei which
are particularly stable = low mass relative to their neighbors, have
high excitation energies for first excited state, are much more
in nature, etc.). Solution: Strong l.s coupling for nuclear
force leads to splitting of l-levels according to overall j quantum
(j = l+s).
- Other evidence for l.s coupling: Left-right asymmetry
of spin-up (polarized) protons from J=0 nuclei like 4He, 12C
etc. Origin of this coupling is basically relativistic (any covariant
acting on a Dirac particle leads to a l.s coupling term in
Dirac equation); example: magnetic moment interacting with magnetic
of moving charge.
- Consequences: magic numbers are predicted correctly; all nuclei
and N magic (doubly magic - closed shells) have spin-parity 0+. Single
particle picture: nuclei with just one nucleon in addition to closed
(or one short of a closed shell -> "hole") have spin, parity and
moment completely determined by that one "valence" nucleon/hole. See
for consequences, e.g. for nuclear magnetic moment.
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