 Solution

The qualitative explanation is nuclear pairing: ${}^{40}\mathrm{K}$ is odd-odd ($Z=19,N=21$), while both ${}^{40}\mathrm{Ar}$ and ${}^{40}\mathrm{Ca}$ are even-even. At $A=40$, moving from the odd-odd to either even-even branch lowers the pairing contribution to the atomic mass byTwo neighboring even-even nuclear isobars can therefore both be lower in mass than the intervening odd-odd member. The two possible directions areHowever, the particular coefficients printed in this paper do not quantitatively reproduce both allowed directions. Let $D=m_p+m_e-m_n\simeq -0.7823\mathrm{MeV}$. Using every supplied term, rather than pairing alone, givesThus this parametrization would forbid the calcium beta-minus branch. The physical atomic mass differences are instead approximately $+1.505\mathrm{MeV}$ to argon and $+1.311\mathrm{MeV}$ to calcium, so both physical directions are allowed.^[ref]

The valid interpretation is therefore pairing explains how both directions can be possible, but the supplied smooth mass formula is insufficient to establish both for this isotope. The semi-empirical mass formula has MeV-scale residual errors and omits nuclear shell closure corrections; ${}^{40}\mathrm{Ca}$ has the especially stable $Z=N=20$ shell closures. One must use measured masses or a model with those corrections for a quantitative proof, rather than changing the printed numbers silently.