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$^3\rm He$ has a lower (nuclear) mass than tritium which is why the latter decays into the former. This is not explained by the semi-empirical mass formula, which would predict a lower binding energy (and therefore higher mass) for $^3\rm He$ because of the $Z(Z-1)$-term.

Is there any theoretical explanation of the mass difference?

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    $\begingroup$ Isn't tritium already heavier than helium-3 without even considering the binding energy? I guess the answer is that the binding energy just isn't enough to make up for the heavier mass of the neutron. $\endgroup$
    – AfterShave
    Sep 22, 2022 at 6:30

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The $Z(Z-1)$ term is only relevant for large nuclei with high Z (quadratic scaling). By the way, the whole model (except for maybe the pairing effect) is intended to be used for large nuclei where the individual nucleon vanishes in the "sea" of other nucleons.

The main effect in Tritium vs 3He comes from the fact that the neutron is heavier than the proton.

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    $\begingroup$ The proton/neutron difference is about $1.3$ MeV, but the binding energy difference is also quite large, being $8.5$ MeV for tritium vs only $7.8$ MeV for helium-$3$. $\endgroup$
    – J.G.
    Sep 22, 2022 at 9:09
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The SEMF assumes the validity of the "liquid drop" model of the nucleus which itself presupposes that the nucleus is a tiny (but perhaps lumpy) sphere. The smallest nuclei are decidedly not spherical and so it should not be a surprise that they do not follow the predictions of the SEMF.

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