The atomic mass excesses of three tin isotopes ( ) are

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Nuclear Physics · Nuclear constituents, size and binding

The Nucleus Resolved, Ch Spin, parity and the electromagnetic moments of a nucleus

Problem

The atomic mass excesses of three tin isotopes (Z=50Z=50) are Δ(119Sn)=−90.065 MeV\Delta(^{119}\mathrm{Sn}) = -90.065\,\mathrm{MeV}, Δ(120Sn)=−91.099 MeV\Delta(^{120}\mathrm{Sn}) = -91.099\,\mathrm{MeV} and Δ(121Sn)=−89.197 MeV\Delta(^{121}\mathrm{Sn}) = -89.197\,\mathrm{MeV}. ((a)) Use the three-point formula Δn(N)=12[Δ(A−1,Z)+Δ(A+1,Z)−2Δ(A,Z)]\Delta_n(N) = \tfrac12\big[\Delta(A-1,Z) + \Delta(A+1,Z) - 2\Delta(A,Z)\big] to find the neutron pairing gap Δn\Delta_n at N=70N=70 (the isotope 120Sn^{120}\mathrm{Sn}). ((b)) Suppose the last two neutrons in these isotopes occupy the 1h11/21h_{11/2} orbital (j=11/2j=11/2) outside a core with Jπ=0+J^\pi=0^+. List every nuclear spin JJ that the Pauli principle allows for two identical neutrons in that orbital. ((c)) Which single member of the list in part (b) is realised in the ground state of 120Sn^{120}\mathrm{Sn}? Explain why the spectroscopic quadrupole moment of that state vanishes whatever the shape of the nucleus.

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