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The Wikipedia article on alpha decay stated:

The strength of the attractive nuclear force keeping a nucleus together is thus proportional to the number of nucleons, but the total disruptive electromagnetic force trying to break the nucleus apart is roughly proportional to the square of its atomic number.

How can this linear/quadratic scaling be justified? To me it is anything but obvious.

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The reason for this effect is that the attractive nuclear potential is short ranged, while electromagnetism is long ranged. Each nucleon in a nucleus interacts almost exclusively with its nearest neighbors, and in a large nucleus, the constituent nucleons are tightly packed, with each nucleon having the same number of nearest neighbors. The (negative) potential energy associated with the nuclear force is thus proportional to the total number of nucleons, as each nucleon experiences just the stabilizing potential of all its neighbors.

The electromagnetic repulsion between protons is much weaker than the nuclear attraction at short ($\sim 1$ fm) ranges, but it falls off much more slowly with distance than the (exponentially decaying) strong interaction. So there is a substantial positive potential between every single pair of protons in a nucleus. For atomic number $A$, the number of pairs is $A(A-1)/2$, so the total (positive) electromagnetic potential energy scales like $A^{2}$.

As result, large nuclei are eventually destabilized by the electrostatic repulsion of their constituent protons.

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  • $\begingroup$ clear and concise. $\endgroup$ – niels nielsen Apr 22 at 0:01

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