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Looking at the the binding energy per nucleon chart: enter image description here

I observe peaks for N=4,8,12,16,20,24 while I expected to observe peaks for 2, 8, 20, 28, 50, 82, and 126 because I have heard that in correspondence of these numbers there is the completion of a nuclear orbital, so I expect a local maximum of the stability an so a peak of binding energy. Is my reasoning wrong?

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Your peaks at $A=4n$ are the so-called "alpha-particle nuclei," which have even proton number, even neutron number, and $N=Z$. In a very hand-wavy way you can say that alpha-particle nuclei are especially tightly bound because the alpha particle is especially tightly bound.

Inside of a nucleus you have separate orbitals for neutrons and protons. The magic numbers you're thinking of, where a nucleon orbital closes, do correspond to especially tightly bound nuclei, but to see it you need to look at a full table of isotopes. For instance, tin (with $Z=50$) has ten or eleven stable isotopes, while indium and antimony ($Z=49,51$) have only two stable isotopes each. There are also six different elements with stable isotopes having $N=50$, which kind of stands out on the chart. There are some "doubly-magic" nuclei with closed shells in both $N$ and $Z$, such as calcium-40, calcium-48, which are both stable, and tin-100, tin-132, both near the drip lines.

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