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I read in a textbook that the nuclide $^{40}_{20}\ce{Ca}$ is more stable as compared to $^{30}_{13}\ce{Al}$. I checked it online and saw that most sources describe the observation with respect to odd and even number of protons in $\ce{Al}$ and $\ce{Ca}$ respectively.

What is the reason for this? Is it due to proton-proton repulsions overcoming the attractive forces between nucleons?

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closed as off-topic by hBy2Py, M.A.R. ಠ_ಠ, airhuff, ron, Todd Minehardt Apr 10 '17 at 17:40

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  • $\begingroup$ This question on the Physics SE might answer your question:physics.stackexchange.com/questions/149400/…. If I understand your position correctly, you are on the right track with proton-proton repulsion being an issue. $\endgroup$ – Tyberius Apr 10 '17 at 15:51
  • $\begingroup$ Well, that's quite simple: the former is stable, the latter is not. $\endgroup$ – Ivan Neretin Apr 10 '17 at 15:52
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    $\begingroup$ I'm voting to close this question as off-topic because it's a nuclear physics question, not chemistry, and should have been asked on Physics.SE. $\endgroup$ – hBy2Py Apr 10 '17 at 16:04
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In general it is very difficult to estimate the stability of a nucleus based on the number of protons and neutrons it contains, as it is an interplay of different forces (repulsion between protons and the Strong force). However, a crude estimate can be made based on these empirical rules:

  1. There are no stable nuclei with $Z>83$ (Also no stable nuclei for Technetium ($Z=43$) and Promethium ($Z=61$)).
  2. The stability of a nucleus depends on the ratio $$\eta=\frac{A-Z}{Z}$$ there are no stable isotopes with $\eta<1$, except for $^1$H and $^3$He. Stable nuclei with relatively small atomic number have $\eta \gtrsim 1$, while nuclei with larger $Z$ need more neutrons to stabilize the nuclear charge, for instance $^{206}_{\,\,82}$Pb has $\eta=\tfrac{124}{82}\approx1.51$.
  3. Most stable nuclei have an even number of protons and neutrons, there are only a few stable nuclei that have both an odd number of protons and neutrons.
  4. Nuclei that have a number of protons or neutrons that equals the so-called magic numbers 2, 8, 20, 58, 50, 82, 126 are especially stable. Nuclei that have both a magic number of neutrons and protons (such as $^4_2$He, $^{16}_{\,\,8}$O, $^{40}_{20}$Ca, $^{48}_{20}$Ca, and $^{208}_{\,\,82}$Pb) are exceptionally stable.

In your example, $^{40}_{20}$Ca belongs to this special category.

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