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While studying binding energy per nucleon curve, I came across the fact that the curve has peaks at nuclei with magic number of protons or neutrons. But one thing that is very confusing here. I have studied that nuclei with magic number of protons or neutrons are stable nuclei and nuclei with proton and neutron number both magic number i.e. "doubly magic" nuclei are highly stable. Now, as we know, the magic numbers are 2, 8, 20, 28, 50, 82, 126. But the binding energy per nucleon curve do not have any peaks after Fe-56. Why is that? For nuclei such as N=50 and Z=50 should also have peaks in the curve. I do not understand this thing. If anyone could shed some light on this, it will be a great help.

P.S. : I have gone through this related question's answers, but it does not really answer my question. Link: Peaks in binding energy per nucleon

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  • $\begingroup$ Are you sure there are no local maxima? They may not be as large as the ones for low mass number, because there are more nucleons, but I think the BE curve is not totally smooth. $\endgroup$
    – ProfRob
    Commented Nov 8, 2021 at 8:21
  • $\begingroup$ Yes I am pretty sure. Actually the bump in binding energy that we see for the lighter ones with magic number nuclei are very high compared to heavier ones and the increase is also not so prominent for the heavier ones. $\endgroup$
    – Neutralino
    Commented Nov 8, 2021 at 8:39

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As so often with nuclear physics, this is a complicated issue whose quantitative treatment needs either difficult calculations or handwaving, but let's try to give an overview.

A doubly magic nucleus doesn't always have a high binding energy per nucleon. It needn't even be stable. Helium-$10$ is a famous example: it's not even particle-bound. You'll probably object that's nowhere near the optimum $N$ vs. $Z$ for $A=10$, viz. $\frac{N}{Z}=1+\frac{a_C}{2a_A}A^{2/3}$ or however you'd rearrange it. But this famous result, which makes neutrons substantially outnumber protons in large stable nuclei, also makes it hard for large doubly magic nuclei to perform well.

For example, you mentioned the next magic atomic number after iron is $Z=50$, but no stable isotope of tin is doubly magic. Or if you consider $N=50$, neither $Z=28$ (nickel) nor $Z=82$ (lead) is anywhere near stable. (While lead is famous for a doubly magic isotope, nickel ironically prefers to be an isotone of iron-$56$, which isn't even singly magic.)

At even higher magic numbers, just having a stable nucleus - never mind one that's doubly magic, with $N=Z$ or otherwise - is increasingly difficult, because large nuclei in general can easily fail. For starters, nothing transuranic will occur in nature. Admittedly we're nowhere near $Z=92$ yet, but the fact two pre-uranium elements don't exist naturally either illustrates just how many $(Z,\,N)$ combinations can fail. This shows "work out $N/Z$ for your chosen $A$, and hence determine $N$ and $Z$" sometimes only tells us the least unstable isobar.

We may as well compare iron-$56$, for which neither $Z$ nor $N$ is magic, to nickel-$56$, for which both are with $Z=N$ even, i.e. a multiple of helium-$4$. While stellar nucleosynthesis largely consists in building up such nuclei, there comes a point where $N$ should be quite a bit more than $Z$. Therefore, binding energy per nucleon peaks not at such a nucleus, but the decay endpoint of one. It's no surprise nickel-$56$ becomes cobalt-$56$ by positron decay, which then becomes iron-$56$ by a second position decay. It's also no surprise, if you consider each iron isotope's possible origin, that iron-$56$ is not only stable but almost all of the element.

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  • $\begingroup$ "For starters, nothing transuranic will occur in nature." - Can you please explain why this is true? Apart from the half life explanation. $\endgroup$
    – Neutralino
    Commented Nov 8, 2021 at 19:38
  • $\begingroup$ @SAGARMODAK Again, it's a huge topic, but think how hard it is for a star or anything else in nature to form a nuclei of $93$ or more protons, and in particular of at least $240$ nucleons or so. First, you need a fusion pathway where nothing is too rare and short-lived to be involved in the next step, and then their $N/Z$ need to be unusually high for their $A$ (with all the usual stability implications) just to match the desired endpoint. $\endgroup$
    – J.G.
    Commented Nov 8, 2021 at 19:45
  • $\begingroup$ is there any good texts or articles that you can recommend which explains these things e.g. the absence of transuranic elements in nature? $\endgroup$
    – Neutralino
    Commented Nov 8, 2021 at 19:49

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