Properties of Nuclei Z protons and N neutrons held together with a short-ranged force  gives binding energy P and n made from quarks. Most of the mass due to the strong interactions binding them together. P and n are about 1 Fermi in size and the strong force doesn’t compress. Size ~ range of strong force  all nuclei have the same density and higher A nuclei are bigger (unlike atoms). Volume=aA radius = bA1/3

Protons vs Neutron neutron slightly heavier than proton and so it decays. No reason “why” just observation quark content: n = udd and p = uud (plus g, qqbar) Mass up and down quarks 5-10 Mev three generations of quarks. Only top quark ever observed as “bare” quark. Somehow up quark seems to be slightly lighter than down quark

Nuclei Force Strong force binds together nucleons Strong force nominally carried by gluons. But internucleon force carried by pions (quark-antiquark bound states) as effective range too large for gluons Each p/n surrounded by virtual pions. Strong force identical p-p, p-n, n-n (except for symmetry/Pauli exclusion effects) Range of 1 F due to pion mass n p n p p

Nuclear Sizes and Densities Use e + A  e + A scattering completely EM pe = 1000 MeV/c  wavelength = 1.2 F now JLAB, in 60s/70s SLAC up to 20 GeV( mapped out quarks) Measurement of angular dependence of cross section gives charge distribution (Fourier transform) Can also scatter neutral particles (n, KL) in strong interactions to give n,p distributions Find density ~same for all but the lowest A nucleii

Nuclear Densities can write density as an energy density Note Quark-Gluon Plasma occurs if

P383 Model of Nuclei “billiard ball” or “liquid drop” Adjacent nucleons have force between them but not “permanent” (like a liquid). Gives total attractive energy proportional to A (the volume) – a surface term (liquid drop) Repulsive electromagnetic force between protons grows as Z2 Gives semi-empirical mass formula whose terms can be found by fitting observed masses Pauli exclusion as spin ½  two (interacting) Fermi gases which can be used to model energy and momentum density of states Potential well is mostly spherically symmetric so quantum states with J/L/S have good quantum numbers. The radial part is different than H but partially solvable  shell model of valence states and nuclear spins

Semiempirical Mass Formula M(Z,A)=f0 + f1 +f2 + f3 + f4 + f5 f0 = mp Z + mn (A-Z) mass of constituents f1 = -a1A A ~ volume  binding energy/nucleon f2 = +a2A2/3 surface area. If on surface, fewer neighbors and less binding energy f3 = +a3Z2/A1/3 Coulomb repulsion ~ 1/r f4 = +a4(Z-A/2)2/2 ad hoc term. Fermi gas gives equal filling of n, p levels f5 = -f(A) Z, N both even = 0 Z even, N odd or Z odd, N even = +f(A) Z., N both odd f(A) = a5A-.5 want to pair terms (up+down) so nuclear spin = 0 Binding energy from term f1-f5. Find the constants (ai’s) by fitting the measured nuclei masses

Semiempirical Mass Formula volume surface Coulomb Eb= DE/A Total N/Z asymmetry A the larger the binding energy Eb, the greater the stability. Iron is the most stable can fit for terms good for making quick calculations; understanding a small region of the nuclides.

www.meta-synthesis.com/webbook/33_segre/segre.html number of protons most stable (valley) Number of neutrons

Semiempirical Mass the “f5” term is a paring term. For nuclei near U there is about a 0.7 MeV difference between having both n and p paired up (even A), odd A (and so one unpaired), and another 0.7 MeV for neither n or p having paired spin (even A) so ~5.9 MeV from binding of extra n plus 0.7 MeV from magnetic coupling easier for neutron capture to cause a fission in U235. U236 likelier to be in an excited state.

Fermi Gas Model p,n spin ½ form two Fermi gases of indistinguishable particles  p n through beta decays (like neutron stars) and p/n ratio due to matching Fermi energy In finite 3D well with radius of nucleus. Familiar: Fermi energy from density and N/A=0.6 Slightly lower proton density but shifted due to electromagnetic repulsion

Fermi Gas Model II V = depth of well = F(A) ~ 50 MeV Fermi energy same for all nuclei as density = constant Binding energy B = energy to remove p/n from top of well ~ 7-10 MeV V = EF + B Start filling up states in Fermi sea (separate for p/n) Scattering inhibited 1 + 2  1’ + 2’ as states 1’ and 2’ must be in unfilled states  nucleons are quasifree B V vs (ignore Coulomb) n p n p

Nuclei If ignore Coulomb repulsion, as n p through beta decay, lowest energy will have N=Z (gives (N-Z) term in mass formula) proton shifted higher due to Coulomb repulsion. Both p,n fill to top with pn coupled by Weak interactions so both at ~same level (Fermi energy for p impacted by n) n p

Nuclei: Fermi motion. Mostly skip if p,n were motionless, then the energy thresholds for some neutrino interactions are: but Fermi momentum allows reactions to occur at lower neutrino energy. dN/dp p

Nuclei:Pauli Suppression But also have filled energy levels and need to give enough energy to p/n so that there is an unfilled state available. Simplest to say “above” Fermi Energy at low energy transfers (<40 MeV) only some p/n will be able to change states. Those at “top” of well. Gives different cross section off free protons than off of bound protons. Suppression at low energy transfers if target is Carbon, Oxygen, Iron... In SN1987, most observed events were from antineutrinos (or off electrons) even though (I think) 1000 times more neutrinos. Detectors were water and scintillator (CH)

Nuclear Shell Model Potential between nucleons can be studied by studying bound states (pn, ppn, pnn, ppnn) or by scattering cross sections: np  np pp  pp nD  nD pD  pD If had potential could solve Schrod. Eq. Don’t know precise form but can make general approximation 3d Finite Well with little r-dependence (except at edge of well) Almost spherically symmetric (fusion can be modeled as deformations but we’ll skip) N-N interactions are limited (at high A) due to Pauli exclusion. p + n  p’ + n’ only if state is available

Infinite Radial Well Already Did Write down equation Solve for L=0 For L>0, angular momentum term goes to infinity at r=0. Reduces effective wavelength, giving higher energy Go to finite well. Wave function extends a bit outside well giving longer effective wavelength and lower energy (ala 1D square wells) In nuclei, potential goes to infinity at r=0 (even with L=0) as that would be equivalent to nucleon “inside” other nucleon

Angular part If V(r) then can separate variables y(r,q,f) = R(r)Y( q,f) have spherical harmonics for angular wave function Angular momentum then quantized like in Hydrogen (except that L>0 for n=1, etc) Energy doesn’t depend on m Energy increases with increasing n (same l) Energy increases with increasing l (same n) will have 1s, 1p, 1d....2s, 2p, 2d... with energy 1p > 1s; 2s > 1s but “unknown” if 1p vs 2s If both n,l vary then use experimental observation to determine lower energy Energy will also depend on strong magnetic coupling between nucleons Fill up states separately for p,n

L,S,J Coupling: Atoms vs Nuclei ATOMS: If 2 or more electrons, Hund’s rules: Maximize total S for lowest E (S=1 if two) Maximize total L for lowest E (L=2 if 2 P) Energy split by total J (J=3,2,1 for S=1,L=2) NUCLEI: large self-coupling. Plus if 2 p (or 2 n) then will anti-align giving a state with J=0, S=0, L=0 leftover “odd” p (or n) will have two possible J = L + ½ or J = L – ½ higher J has lower energy if there are both an odd P and an odd n (which is very rare in stable) then add up Jn + Jp Atom called LS coupling nuclei called jj Note that magnetic moments add differently as different g-factor for p,n

Spin Coupling in Nuclei All nucleons in valence shell have same J Strong pairing causes Jz antiparallel (3 and -3) spin wavefunction = antisymmetric space wavefunction = symmetric This causes the N-N to be closer together and increases the attractive force between them e-e in atoms opposite as repulsive force Even N, even Z nuclei. Total J=S=L=0 as all n,p paired off Even N, odd Z or odd N, even Z. nuclear spin and parity determined by unpaired nucleon Odd N, odd Z. add together unpaired n,p Explains ad hoc pairing term in mass formula

Energy Levels in Nuclei Levels in ascending order (both p,n) State n L degeneracy(2j+1) sum 1S1/2 1 0 2 2*** 1P3/2 1 1 4 6 1P1/2 1 1 2 8*** 1D5/2 1 2 6 14 2S1/2 2 0 2 16 1D3/2 1 2 4 20*** 1F7/2 1 3 8 28*** 2P3/2 2 1 4 32 1F5/2 1 3 6 38 2P1/2 2 1 2 40 1G9/2 1 4 10 50*** *** “magic” number is where there is a large energy gap between a filled shell and the next level. More tightly bound nuclei. (all filled subshells are slightly “magic”)

Magic Numbers Large energy gaps between some filled shells and next (unfilled) shell give larger dE/A and more made during nucleosynthesis in stars # protons #neutrons 2 He 2 He-4 6 C 6 C-12 8 O 8 O-16 20 Ca 20 28 Ni 28 Cr-52(24,28) 50 Sn 50 Ni-78 82 Pb 82 126 136 Ni-78 (2005) doubly magic. While it is unstable, it is the most neutron rich. Usually more isotopes if p or n are magic. Sn has 20 isotopes, 10 of which are stable

Nuclear Spin and Parity If Even-Even then all n,p pair up and i=0 and P=even O16 (8p,8n) If Even-Odd or Odd-even, leftover p,n determine spin and parity N15 (7p,8n) unpaired p in 1p1/2 i=1/2 parity is odd O17 (8p,9n) unpaired n in 1d5/2 i=5/2 parity is even as d  l=2 P=(-1)l Odd-Odd need to combine spin of p and spin of n indeterminate, use experiment N14 (7p,7n) unpaired in 1p1/2 i=0 or 1 see i=1. P = odd*odd = even

Magic Numbers protons 8 = O 20 = Ca 28 = Ni 50 = Sn 82 = Pb

Cosmic Abundance of Elements Nature prefers elements with an even number of protons and an even number of neutrons as those nuclei have larger binding energy Carbon, Oxygen, Silicon, Iron are abundant All elements heavier than lead are unstable, radioactive, decay to lead