(PHYS722/822 Fall 2018)

- Range of masses 1.673x10
^{-27}kg - several times 10^{30}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 A
^{1/3}-> approximately constant density

- Mattauch mass spectrograph (velocity and momentum filter) - singly ionized atoms
- Comparison to neutral
^{12}C 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(
^{12}C) / 12 = 931.494 MeV/c^{2} - Binding energy B = M(Z,N) - Z*M(
^{1}H) - N*M(n) = D(Z,N) - Z*D(^{1}H) - N*D(n)

- Roughly along line Z=N, N>Z for heavier nuclei
- Absolute limits given by proton or neutron decay ( D(Z,N)
> D(
^{1}H) + 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.

- Nuclei of all sizes have very similar density (about 0.14
nucleons/fm
^{3}); R = R_{0}A^{1/3}(R_{0}= 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
^{235}U, 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)

- Mass formula does NOT depend on I
_{3}= (Z-N)/2 except for proton-neutron mass difference and Coulomb effects. It does depend on (I_{3})^{2}, though. - States in neighboring isobars occur as multiplets with nearly
equal
masses
but different values of I
_{3}up to a maximum |I_{3 max}| = I. - Analogy with spin: Multiplets with 2S+1 members, all members
degenerate
except for effect of external magnetic field; energy only depends on S
^{2}, not S_{z}*VS.*

Multiplets with 2I+1 members, all degenerate except for p-n mass diffference and Coulomb effects. Strong force ONLY depends on I

- Nuclei have spins from 0, 1/2, 1, 3/2, ... to several h-bar.
- Magnetic moments are given by
**m**= g*m_{N}***S**(= nuclear magneton) - g value of 2 is for pointlike Dirac-particle. Proton has g
_{p}= 5.6 and neutron has g_{n}= -3.8 => clear indication of composite structure.

Week 2

- Nuclear density given by r(r) = r
_{inf}/(1+exp((r-R)/a) (r_{inf}= 0.16 N/fm^{3}is the density of inifinte nuclear matter, R = R_{0}A^{1/3}is the nuclear radius, and a is the width of the surface). The average density of medium-sized nuclei is 0.14 N/fm^{3}. - 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 V
_{0}and radius R. Fill 6-dim. phase space V*D^{3}p with nucleons, where each cell of size h^{3}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,
p
_{fermi}= 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 V_{0}= -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 a
_{s}*(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*r_{0}/A * A*R_{0}^{3}/R^{3})^{1/3}(the second term corrects for the actual nuclear radius vs. the constant density approximation R = A^{1/3}*R_{0}). Inserting all constants yields p_{fermi}(protons) = 250 MeV/c*(Z/2A)^{1/3 }* A^{1/3}*R_{0}/R and p_{fermi}(neutrons) = 250 MeV/c*(N/2A)^{1/3 }* A^{1/3}*R_{0}/R. The overall contribution to the total nuclear mass from the kinetic energy due to this "fermi motion" is 3/5*Z*(p_{fermi}(protons))^{2}/2m + 3/5*N*(p_{fermi}(neutrons))^{2}/2m. Expanding this in a Taylor series around the average Z=N=A/2 with I_{3}as the "small" expansion parameter yields 20MeV*A + 44MeV*I_{3}^{2}/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 = A
^{1/3}*R_{0}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, T_{in,ave}(A-1):

A*20MeV = 33MeV + (A-1)*T

- 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)*Y_{lm}(q,f) with R(r)=C*j_{l}(kr) (k is momentum from kinetic energy k^{2}/2m; j_{l}are spherical Bessel functions). Each solution is specified by quantum numbers l = 0,1,2,...; m = -l,...,+l and n (= number of nodes in j_{l}(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^{4}He,^{12}C 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.