Nucleus geometry: sequence of "magic numbers" modelled as nested polyhedrons as filled nucleonic shells.
Question: Is this accurate as drawn, or is a spherical version more correct?
Nature likes spheres. The black lines describe closest links (bonds) between points, points are the proton and neutron centers. So ignoring the actual black lines, and instead see each point as center of a bubble, then does it become spheroid?
The sphere in nature:
least surface area, maximal surface tension in equilibrium of non-crystaline sub-units.
In a body full of non-crystaline sub-units, for each sub-unit (eg. nucleons) to have the maximum number of bonds (contact points), the body must be spherical, any other form will result in fewer contact points in sum.
If the sub-units have weak bonds, the sphere allows the most secure body, simply due to the higher number of contact points per sub-unit. A sphere of weakly bonded sub-units may be stronger than a non-sphere with strongly bonded sub-units.
So regarding the nucleus, if double magic, the body is spherical with maximized contact points.
" ", if not magic, the body will compact towards a sphere-like condition in order to maximize bonds.
The exception occurs when the nucleus is double magic plus one extra nucleon, in which the extra nucleon has the least number of contact points, and so partially protrudes from the sphere. For every additional nucleon added, the sum of contact points is greater, so it tends toward greater sphericity.
This proves that nucleus is never cigar-shaped (except when the sum of nucleons = 2) or disk shaped (except when the sum of nucleons = 3), but instead, the nucleus is always spheroidal except during fusion or fission.
The neutrons act as buffers between the protons allowing this tight spherical fit. If the number of neutrons is less than protons, then the protons can't seat properly into the sphere shell, and will stay on the surface (reducing sphericity of the body) reducing the body density and thus the perceived gravity.