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How does beta emission by a nuclei increases the general stability of it? I surely am not a specialist or anything, but this just don't make sense to me, since the objective of radiation is to release energy, but I can't figure out exactly how it's done.

The loss of energy from the electron getting expelled, in addition to its kinetic energy, roughly gives half a MeV (source). Besides, as the strong force acts similarly on protons and neutrons, I guess it won't change considerably in piratical effects. Thus, in my conception, at least, the eletrostatic repulsion from the proton added should overlap the energy who got rid in the process.

Hence, I am confused. I wish to understand how it plays a role in nucleus stabilization.

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While this is nuclear physics, the relative number and relative stability of isotopes is of some relevance to chemistry. Further, I spend more time than I'd like perhaps at the intersection of materials science, chemistry, and nuclear physics, so...

There are several good on-line resources to get information on nuclear energy levels and the relative stability of nuclei. These include: The Nuclear Data Evaluation Project at Triangle Universities Nuclear Laboratory (TUNL), and the Evaluated Nuclear Data Structure File data base at the National Nuclear Data Center hosted at Brookhaven National Laboratory. From these you can get more information than you might want with regard to what is known about nuclear energy levels and stability of the elements.

One cut at looking at stability is the so-called isobar diagram. This compares the relative nuclear energy levels of all nuclei with the same $A$, where $A = P + N$, that is, the sum of the number of protons and neutrons in the nucleus. Since $\beta \pm$ converts between protons and neutrons, keeping $A$ constant, this is the best vantage point for the purpose at hand.

Borrowing from TUNL one finds for $A=14$: enter image description here

So, we see 4 different nuclei shown on the isobar diagram. The lowest energy (most tightly bound) nuclei is $^{14}$N. To the left is $^{14}$C, which you likely know is unstable and $\beta -$ decays to $^{14}$N, used for carbon dating. To the right is $^{14}$O and you will note that it $\beta +$ decays to $^{14}$N. Neither $^{14}$C or $^{14}$O has as much binding energy as $^{14}$N due to the fact that protons and neutrons are not actually identical in the nucleus. So, the ground state of $^{14}$C or $^{14}$O is about 2.4 MeV higher than $^{14}$N.

High and off to the left you can barely see $^{14}$B, 24.71 MeV above $^{14}$N. It desperately would like to be $^{14}$N, but can't get there directly through the available $\alpha$, $\beta$ or $\gamma$ decay channels. So, it happily $\beta -$ decays to $^{14}$C with a half-life of 16 ms, which will then ultimately decay down to $^{14}$N.

Now, your misconception is that you don't know nuclear physics (and I don't mean that negatively - you applied a fairly simple model, did not get the expected answer, so you asked a question here - GREAT!). But, in the nucleus you have Coulomb repulsion, the strong force, and the weak force, and it all has to be treated quantum mechanically. Taken all together one can determine the binding energy (total or per nucleon) of nuclei, the excited nuclear energy levels, and possible decay paths. It is great fun, but is a different world to explore.

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