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Surely I must have a misconception?

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Short answer

Protons are the only stable free baryons.

In a bound state, stability of nucleons (in sense of their mutual conversion via weak force) is determined by the mean nucleon binding energy.

Positron emitters are proton rich nuclei with decreased mean nucleon binding energy, partly due enlarged proton repulsion, partly due less advantages nuclear binding, often with odd proton and neutron numbers.

The extra energy of less stable nuclei with higher than optimal proton/neutron ratio for a given nucleon number is released by conversion of a proton to a neutron, emitting positron and electron neutrino.

The other possible beta decay mode of proton rich nuclei is an electron capture.

Some interesting details about relative nucleus stability

The mean binding energy is determined by proton/neutron ratio, that is initially optimal as 1:1. For heavy nuclei a higher neutron/proton ratio is needed.

But the energy is also determined by various nucleon parity rules. E.g generally, nuclei with even count of protons a/o neutrons are more stable than those with odd counts.

As consequence, there is the rule elements with the odd proton number have at the most 2 stable isotopes.

$\require{mhchem} \ce{Tc}$ and $\ce{Pm}$ have for these reasons bad luck as each of their stable isotope candidates have more stable nuclei among their element neighbors. Therefore, both elements are radioactive in spite of being light elements.

Plus, there are so called magic numbers $2, 8, 20, 28, 50, 82, 126$  of protons a/o neutrons, making the nuclei (relatively) extra stable.

E.g $\ce{^{40}_{20}Ca}$ is dual magic number nuclei

The numbers are the close analogy of respective electron numbers filling atomic orbitals ($2, 8, 18, 32, ...$)

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  • $\begingroup$ And this is not proton decay but rather inverse neutron decay. $\endgroup$ – my2cts Jun 23 at 11:45
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    $\begingroup$ @my2cts Sure. I have not said it is a proton decay. The sentence should read as The other possible beta decay mode of "proton rich kernels“ is an electron capture. $\endgroup$ – Poutnik Jun 23 at 11:47
  • $\begingroup$ My comment is intended as a additional clarification of your wholly correct answer. $\endgroup$ – my2cts Jun 23 at 12:13
  • $\begingroup$ I see. OK. Thanks. $\endgroup$ – Poutnik Jun 23 at 12:14
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    $\begingroup$ Most of this answer isn't really relevant. The question doesn't ask for details of how to find out which nuclei are more stable and which are less stable. Positron emitters are proton rich nuclei with decreased mean nucleon binding energy, due enlarged proton repulsion. This is not quite right. The existence of a line of stability doesn't involve electrical repulsion. The electrical repulsion of the protons simply changes the shape of the line of stability for heavy nuclei. $\endgroup$ – Ben Crowell Jun 23 at 16:01
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It is very important to understand that (as you see in the comments), there are to cases:

  1. Free protons (isolated in a vacuum) are stable.

  2. Protons bound inside a nucleus may be unstable.

Positron emission and proton decay are different things:

  1. proton decay is a hypothetical form of particle decay, that has never been observed, the proton is said to be stable, because the half life is at least $1.67×10^{34} $ years

  2. positron emission (a type of radioactive decay) is not proton decay, because with positron emission, the proton interacts with other particles within the atom (according to the SM, protons are stable, because baryon number (quark) is conserved, and protons will not decay into other particles on their own)

Positron emission is a type of beta decay, where a proton inside a nucleus is converted into a neutron (while releasing a positron and an electron neutrino), to increase stability of the nucleus. It happens typically in proton-rich nuclei.

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    $\begingroup$ Protons bound inside an atom are not stable. I think you mean protons bound inside a nucleus, and that they may be unstable. positron emission, according to the SM, protons are stable, because baryon number (quark) is conserved, and protons will not decay into other particles on their own. This sentence doesn't make sense. How does "positron emission" connect gramatically to the rest of the sentence? It happens typically in large, (proton-rich) nuclei. It happens mainly in proton-rich nuclei (or also in odd-odd nuclei near stability), but they don't have to be large. $\endgroup$ – Ben Crowell Jun 23 at 15:59
  • $\begingroup$ E.g. $^{40}\mathrm{K}$ (19p,21n) undergoes positron emission as well, aside of electron emission and electron capture. $\endgroup$ – Poutnik Jun 23 at 18:08
  • $\begingroup$ @BenCrowell thank you I edited $\endgroup$ – Árpád Szendrei Jun 23 at 18:58

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