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Can electron capture proceed with an electron not bound to the nucleus which decays (that is one that is bound to a neighboring atom, bound in a collective state like a conduction band, or free)?

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  • $\begingroup$ An external electron here means an electron that is not in any of the shells of that atom $\endgroup$ – Abishek Shankar Feb 14 '16 at 3:27
  • $\begingroup$ OK. In that case I've edited the text to one that I think better reflects the intent. And I can endorse Bill's answer below. $\endgroup$ – dmckee --- ex-moderator kitten Feb 14 '16 at 16:46
  • $\begingroup$ Yeah that give the question more meaning @dmckee .... Thanks $\endgroup$ – Abishek Shankar Feb 16 '16 at 11:56
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The probability should be very low that this would happen. For EC to happen, there needs to be a probability that the electron could be in the nucleus. An external electron is going to be interacting with the electron cloud of the atom.

The experimental data of x-ray intensities from nuclei following decay by EC indicate a strong drop in capture of n=2 versus n=1 and n=3 versus n=2 electrons. An unbound electron simply has vanishing probability due to shielding effects of the bound electrons.

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Yes. Since they are leptons, they both exhibit the weak nuclear force. However, the force is not to strong considering the protons and neutrons are made out of multiple leptons. They would probably be in a very short, near undetectably short binary orbit and than go back to their parent atoms, or back into space.

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  • $\begingroup$ What "orbit" are you talking about? On atomic scales, quantum effects are fully relevant, and there are no "orbits" for anything. $\endgroup$ – ACuriousMind Feb 13 '16 at 15:03
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    $\begingroup$ Protons and neutrons are not made out of leptons - at least not in the standard model, or any extension I've seen. $\endgroup$ – ragnar Feb 13 '16 at 17:33

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