Why do some isotopes have different quadrupole/octupole moments, when these moments of charge density should be independent of mass?
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The charge density stems entirely from the charged particles in the nucleus, but the higher moments depend largely (in a sense, "entirely") on their arrangement. Placing 8 protons at the vertices of a cube produces notably different higher moments than, say, placing all along a line segment, or in an octagon. (This should be apparent from the different symmetries of the system).
As you add more mass, you introduce more neutrons that will affect the spatial organization of the nucleons. This can induce or break symmetry of the protons' arrangement, modifying their moments.
Alex Meiburg
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In atomic physics, we have Hund's rules which tell us whether the next electron added to an atom will fill an $s,p,d,f$ orbital. One consequence of the Wigner-Eckart theorem is that the angular momentum of an object constrains its multipolarity. A spinless object may carry only monopole moment; a spin-half object may carry monopole and dipole moments, but no higher; a spin-one object may carry monopole, dipole, and quadrupole moment; etc. A full shell, whether it's an $s$ shell or an $f$ shell, has spherical symmetry. So for atoms, it's the multipolarity of the partially-filled electron orbitals that determines the multipolarity of the atom.
In the shell model, we construct "nucleon orbitals" within the nucleus analogous to the electron orbitals outside the nucleus. They are likewise grouped by their angular momentum, and nuclei with full nucleon shells are especially tightly bound in the same way that the noble gases are especially chemically inert. Away from the magic nuclei, it is the multipole moments of nucleons in partially filled shells which gives net quadrupole, octupole, etc. moment to a stable nucleus.
rob
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