Isotope that decays when ionized

Question

Some time ago, I read about a certain isotope that is stable when neutral but decays with electron emission (beta) when being completely ionized, but I can't find which one it was.

Which isotope decays when fully ionized?

Answer 1

I guess you have read about an isotope decaying via electron capture (which is kind of an inverse $\beta$ decay). There are quite many radio-isotopes decaying in this way, for example: $${}^{59}\text{Ni} + e^- \to {}^{59}\text{Co} + \nu$$ $${}^{40}\text{K} + e^- \to {}^{40}\text{Ar} + \nu$$

In this decay the atomic nucleus captures one of the surrounding electrons (usually one of the innermost $K$ shell). Of course this process can only happen if the atom or ion has at least one electron. It cannot occur if the ion is completely ionized, i.e. it has no electrons at all.

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Another quite rare phenomenon is bound-state $\beta^-$ decay. Here the created anti-neutrino takes almost all of the decay energy, and the created electron gets very little energy so that it fails to escape the atom, and instead integrates into the atomic orbital.

You probably have read about the completely ionized rhenium ion which decays quickly (with half-life $32.9$ years) by bound-state $\beta^-$ decay $${}^{187}\text{Re}^{75+} \to {}^{187}\text{Os}^{76+} + e^- + \bar{\nu}$$

whereas the neutral rhenium atom is almost stable (with half-life $42$ billion years) $${}^{187}\text{Re} \to {}^{187}\text{Os}^+ + e^- + \bar{\nu}$$

This particular rhenium isotope ${}^{187}$Re has a very small $\beta^-$ decay energy (only $3$ keV). This energy (or more precisely: the energy part delivered to the electron, not to the antineutrino) is not enough for the electron to escape the neutral atom. And it cannot find a place in the shell because all electron orbits of the atom are already occupied. But when the ion is fully ionized (i.e. all electrons stripped off), then the energy is enough for the electron to reach a low electronic orbital of the ion.

See also the original article by Bosch et al. (1996)
"Observation of Bound-State $\beta^-$ Decay of Fully Ionized ${}^{187}$Re: ${}^{187}$Re-${}^{187}$Os Cosmochronometry".

Answer 2

I’m nearly certain you are thinking of beryllium-7, but that you have remembered the condition backwards.

Neutral $\rm^7Be$ can decay to $\rm^7Li$ by electron capture, with energy about $\rm 860\,keV$. Positron-emission decays are always disfavored relative to electron-capture decays, because the final state with an electron missing is lower in mass than the final state with the positron added. Since the total $\rm^7Be$ decay energy is less than the mass difference $2m_\mathrm e = \rm 1022\,keV$, the positron-emission mode is completely forbidden. Beryllium-7 is not found in beryllium ores on Earth, but completely-ionized $\rm^7Be$ is a stable component of cosmic rays.

A read through all the NNDC dataset finds a number of other nuclei with electron-capture $Q$-values below $\rm1\,MeV$, starting with $\rm^{41}Ca$. However, cosmic ray populations are heavily skewed towards the low-mass end of the chart of isotopes; I’ve only ever heard people talk about beryllium-7 having this property.

For the condition you describe, where an ionized parent nucleus can decay but the neutral parent atom is stable, the decay energy would have to be less than the electron binding energy for the daughter atom. If that were the case, the ionized nucleus could decay to a bound state of the daughter and the beta electron, with the antineutrino carrying away the energy. But the neutral atom would be “Pauli-blocked” from decaying, with its bound electrons already occupying the possible final states for the $\beta^-$.

I believe there are no $\beta^-$ emitters with energies this low. If such a decay existed, it’d be an interesting place to try and measure the mass of the electron antineutrino, by doing precision mass spectrometry on the parent ion and the daughter ion, to be compared with recoil measurements on the daughter following the decay. Instead, that experimental energy has gone into the KATRIN experiment, which analyzes the $\beta^-$ decay of tritium to helium-3, with endpoint energy $\rm17\,keV$.

However, as revealed in the comments on another answer, my belief was incorrect. Neutral dysprosium-163 is stable against $\beta^-$ decay, with $Q$-value $\rm-2.6\,keV$; the linked paper observes the beta-decay of the bare nucleus. The next candidate would be $\rm^{148}Eu$, with $Q$-value $\rm-27\,keV$.