nuclear physics - Question on two-neutrino double electron capture

There was a fascinating paper in Nature recently, on the observation of two-neutrino double electron capture in xenon, with a half-life time of $1.8\times 10^{22}$ years.

The process described in the article is $$^{124}\mathrm{Xe} + 2e^- \to {}^{124}\mathrm{Te} + 2 \nu_e.$$

According to Wikipedia, double electron capture can occur only when competing modes are strongly suppressed.

My Question: Why is the single electron capture so strongly suppressed? Why can't we have $$^{124}\mathrm{Xe} + e^- \to {}^{124}\mathrm{I} + \nu_e$$ while the decay mode $$^{125}\mathrm{Xe} + e^- \to {}^{125}\mathrm{I} + \nu_e$$ exists?

Answer

This is explained by Scott Manley in Why a Dark Matter Search Also Observed The Rarest Radioactive Decays at around the 7:20 mark.

The short answer is that the process is electronically forbidden, because the iodine-124 nucleus has a higher binding energy than the xenon-124 nucleus. Using the data from Wikipedia, the masses for the nuclides involved are $$\begin{align} m({}^{124}\mathrm{Xe}) & = 123.905\,893(2) \:\mathrm{u} \\ m({}^{124}\mathrm{I}) & = 123.906\,2099(25) \:\mathrm{u} \\ m({}^{124}\mathrm{Te}) & = 123.902\,8179(16) \:\mathrm{u}. \end{align}$$ This means that the ${}^{124}\mathrm{Xe}\to {}^{124}\mathrm{Te}$ decay is allowed, and releases $$(m({}^{124}\mathrm{Xe}) -m({}^{124}\mathrm{Te}))c^2 = 2.86\:\mathrm{MeV}$$ of energy, whereas that same difference for the decay to iodine yields a negative mass difference, $$(m({}^{124}\mathrm{Xe}) -m({}^{124}\mathrm{I}))c^2 = -0.295\:\mathrm{MeV},$$ which means that the beta decay as it occurs one mass unit up is energetically forbidden. (If you do the same calculation there, you get $(m({}^{125}\mathrm{Xe}) -m({}^{125}\mathrm{I}))c^2 = 1.64\:\mathrm{MeV},$ which is plenty of energy to fuel a beta decay.)

That said, though, this is not enough to rule out an electron-capture mechanism on energetic grounds, since the energy hill from xenon-124 to iodine-124 can be climbed with the annihilation of the electron, $$(m(e^-) + m({}^{124}\mathrm{Xe}) -m({}^{124}\mathrm{I}))c^2 = +0.21\:\mathrm{MeV},$$ so there are definitely substantial details left to explain there, which can hopefully be explained by a nuclear physicist. Still, the difference in the energetics is definitely large enough that the two processes cannot be considered a priori to be roughly equivalent.