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I was reading about atomic orbitals and came across the fact that electrons in s-orbitals have a non-zero probability density at the nucleus. That got me wondering: -

Can electrons in s-orbitals actually contribute to the stability of a nucleus in any meaningful way?

To explore this, I imagined a hypothetical scenario using a deuterium (D / $^2\text{H}$) atom (1 proton, 1 neutron, 1 electron). Suppose the s-orbital electron happens to be inside the nucleus and undergoes electron capture: -

  1. The proton captures the electron and becomes a neutron, emitting a neutrino.
  2. Then the other remaining neutron absorbs that neutrino and becomes a proton instead, thereby releasing an electron back into the orbital.

This hypothetical loop made me wonder; even if such a process is incredibly rare, could there be any theoretical or practical effect on nuclear stability due to the presence of an s-orbital electron?

Then I thought of tritium (T / $^3\text{H}$), which has a half-life of about 12.33 years. If s-electrons played any role in stabilizing the nucleus, wouldn't the half-life of ionized tritium (i.e. $T^+$, with no electrons) be different from that of negatively charged tritium (i.e. $T^−$, with 2 electrons in the 1s orbital)?

I haven't been able to find any experimental data confirming whether tritium ions with different electron counts (neutral, positively charged, or negatively charged) have identical or different half-lives. Does anyone have a source on this? Or could someone help clarify whether electrons in s-orbitals could, even theoretically, play any role—however tiny—in nuclear stability?

Swarnim Khosla
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1 Answers1

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The section https://en.wikipedia.org/wiki/Radioactive_decay#Changing_rates in Wikipedia discusses several examples of the presence or absence of electrons’ affecting nuclear decay rates. See also https://en.wikipedia.org/wiki/Electron_capture.

tparker
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