# On what basis can we say that one nuclide is more stable than another one? [closed]

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?

• 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. Apr 10, 2017 at 15:51
• Well, that's quite simple: the former is stable, the latter is not. Apr 10, 2017 at 15:52
• 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. Apr 10, 2017 at 16:04

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 $\eta=\tfrac{124}{82}\approx1.51$.
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.