Why is helium-3 stable? Why is helium-3 stable? Besides hydrogen, helium-3 is the only isotope that has a neutron-to-proton ratio less than 1. Why is it not radioactive?
 A: To what would $^3$He decay (strongly)? The options are:
$$^3{\rm He}\rightarrow D+p$$
$$^3{\rm He}\rightarrow 2p+n$$
neither of which makes sense based on mass. A weak decay to $^3$Li is out of the question (https://en.wikipedia.org/wiki/Isotopes_of_lithium#Lithium-3).
The fact that tri-neutrons and tri-protons do not exist, not even as resonances, tells you $^3$He is isospin 1/2, as its mirror nucleus, $^3$H.
The strong force is approximately iso-scalar, meaning $^3$He and $^3$H have the same nuclear wave functions, with neutron and proton swapped. (The latter decays to the former via the weak interaction because it is energetically possible).
So, the answer to the question, "Why is helium-3 stable" comes down to binding energy. Why the binding energy is what it is comes down to the nuclear wave function. The nuclear wave function of helium-3 and the triton is, and has been for decades, an active area of research. For example, the Triton collaboration at J-Labs Hall-A: https://www.jlab.org/news/releases/physicists-study-mirror-nuclei-precision-theory-test.
I have been to many seminars on the topic (but not recently). One topic that has never come up is the semi-empirical mass formula, it is simply not used at $A=3$, as it treats the nucleus (approximately) as liquid drop with volume terms, surface terms, and so on.
The largest component of the wave function has the two protons in a spin-singlet $S$-wave, with the neutron also in an $S$-wave (hence: no electric quadruple moment). The nucleus provides the simplest place to study three-nucleon forces. It is small enough to allow recent quark model investigations (e.g. https://www.sciencedirect.com/science/article/pii/S2211379717308094).
Beyond that, the question cannot be answered here. The subject is far too large.
