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Nucleus contains nucleons viz; protons and neutrons. Nucleus is not just pile of balls (neutrons and protons) as is generally depicted in many cases. And we don't know much about nucleus. Obviously linear variation of size with atomic mass number is experimented fact. What are theoretical explanations for this?

I mean nucleus density is not uniform, there is possibility of probability shells for nucleons, I don't think the nucleons just stacked up like balls. My question is not why is this variation; but why this linear variation?

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    $\begingroup$ this is a funny statement as it does not seem to vary linearly !! have a look here catalog.flatworldknowledge.com/bookhub/… . the whole thing depends on quantum mechanics. $\endgroup$
    – anna v
    Commented Dec 2, 2017 at 16:07
  • $\begingroup$ 80+ years of nuclear physics have taught a lot about the nucleus. Amongst other things, the size grows like $A^{1/3}$, as discussed here: en.m.wikipedia.org/wiki/Atomic_nucleus#Nuclear_models $\endgroup$
    – ZeroTheHero
    Commented Dec 2, 2017 at 16:20
  • $\begingroup$ What makes you think it's a linear variation? If that is published somewhere, don't believe it; it's wrong. $\endgroup$
    – Bill N
    Commented Dec 3, 2017 at 4:48
  • $\begingroup$ Isn't it linear? $\endgroup$
    – Rana
    Commented Dec 3, 2017 at 7:15

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The volume of a nucleus varies approximately as the atomic number $A$, and for that (large) subset of nuclei that are spherical or ellipsiodal, the median axis varies approximately as $A^{1/3}$.

These results can be had from data and also arrived at as results in shell models.

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  • $\begingroup$ +1, but: -- These results can be had from data and also arrived at as results in shell models. This is not really true in the vast majority of calculations using some version of the nuclear shell model. Normally the size is put in by hand, unless you're doing something like Hartree-Fock, and even then, the parameters of the nuclear force model are tuned up in order to reproduce liquid drop stuff like $A^{1/3}$. $\endgroup$
    – user4552
    Commented Mar 1, 2019 at 13:35
  • $\begingroup$ @Ben Fair enough. I suspect that I was thinking about Weleka's book/class when I wrote that. As I recall he builds a simplified model of the nuclear force in chapter one of his text and later plugs that simple model into a Hartree-Fock to show that he can construct the basic measured trends. Certainly, the results are not correct in detail until you tweak the potential model, but getting the $r \propto A^{1/3}$ result is tractable. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Mar 1, 2019 at 15:57

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