I recently read about a beta decay nucleus (rhenium 187) which physicists got to decay faster by stripping the nucleus of all its electrons.

Original half life: 42 X 10^9 years

"Changed" half life: 33 years!

Why does this happen, and is this only a characteristic of rhenium 187, or to all beta decaying nuclei?

Edit: link to paper: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.77.5190

  • $\begingroup$ @Ben Crowell -- I disagree. It’s not implausible. Since the decay process is $n\to pe\bar{\nu }$, electrons in S-orbitals could block final states via the exclusion principle and should slow down the decay. Removing them would clear the way. $\endgroup$ Commented Jun 19, 2018 at 11:17
  • $\begingroup$ @BertBarrois: Aha, that makes sense, because of the unusually low beta decay energy. $\endgroup$
    – user4552
    Commented Jun 19, 2018 at 14:25
  • $\begingroup$ How do you strip a nucleus of ALL of its electrons? $\endgroup$
    – Jiminion
    Commented Oct 16, 2018 at 13:37
  • $\begingroup$ @Jiminion violently. The general way to strip nuclei is to launch a beam of them at high speeds through foil. It's messy and only a fraction will lose all their electrons. Those are the ones that you keep. $\endgroup$ Commented Apr 24, 2023 at 1:09

1 Answer 1


Unfortunately I don't have access to the paywalled PRL paper, but I think Bert Barrois's comment sounds right, so I'll go ahead and post a possible explanation as an answer. Maybe others who have access to the article, or who can find other information, can provide better answers.

The Q value of this decay is unusually low at 2.5 keV (which is presumably the reason for the very long half-life). More typical beta decay energies would be in the MeV range. The energy of the K shell in osmium is 74 keV, and the less tightly bound orbitals would go down in energy from there to whatever the ionization energy of osmium is, presumably on the order of 1 eV.

So if 187Re tries to emit a beta with an energy in the range from 0 to 2.5 keV, the beta is emitted with an energy that is sort of in the middle of the range of energy levels occupied by the electron cloud. Although this is just a hand-waving argument, it does seem plausible, as Bert Barrois suggests, that this frustrates the decay due to the exclusion principle.

Why does this happen, and is this only characteristic to rhenium 187, or all beta decaying nuclei?

Assuming this explanation is right, then it's a mechanism that would apply only in very unusual cases. In most cases, beta decay energies are on the order of MeV, which is at least one order of magnitude greater than the K-shell energies, even in heavy elements.

  • 1
    $\begingroup$ Having access to the paywalled PRL paper, you're essentially correct. The process being talked about is bound-state beta decay, where instead of producing a free electron, the nucleus produces an electron in one of the unoccupied orbitals, which is obviously only possible for an ionized nucleus. $\endgroup$ Commented Jun 19, 2018 at 14:42
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    $\begingroup$ There are a handful of nuclides like this, where the energy is small enough that the electron cloud affects the final state significantly. There's a light nucleus (maybe beryllium-7?) which is unstable against electron capture but doesn't have enough energy to create a positron. This isotope is not found as a neutral atom on Earth but is a stable ion in cosmic rays. $\endgroup$
    – rob
    Commented Jun 19, 2018 at 15:04

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