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Electron capture is a kind of decay by which a nuclear transformation takes place. Below is an example of it.

$$ _{29} ^{64} \text{Cu} + e^- \to\ {}_{28}^{64} \text{Ni} + {\nu}_e$$

Of course, with the nucleus being positively charged and electron being negatively charged, there is an attraction between them, but electrons generally don't fall into the nucleus. I am unable to understand exactly how is the nucleus able to capture electron; using which interaction or force? The uncertainty principle won't allow an electron in the nucleus, though it is being captured. Exactly how?

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Electron capture occurs via a weak nuclear reaction. The electron itself ceases to exist; instead a neutron and neutrino appear. It is a reversed reaction of the neutron weak decay, if you like.

A bound electron wave function is not zero in the nucleus. There is a finite probability to find an electron within the nucleus volume. So they can interact and they do interact.

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    $\begingroup$ Just to clarify, there's also a proton from the old nucleus involved that ceases to exist in the reaction, right? $\endgroup$ – b_jonas Dec 27 '12 at 16:39
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    $\begingroup$ @b_jonas: Yes, it disappears too. $\endgroup$ – Vladimir Kalitvianski Dec 27 '12 at 16:44
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    $\begingroup$ The reason electron capture can happen is because S states have an orbital that overlaps the nucleus, i.e. there is a probability for a proton to be superposed on an electron and the weak interaction can take over, if the energetics allows it, i.e. enough energy to allow a proton instead of a neutron . (64 neuclei before and after) $\endgroup$ – anna v Apr 3 '17 at 11:51
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    $\begingroup$ Indeed, electrons in $s$ wave orbitals have a contact density at the nucleus. This is also responsible for NMR (nuclear magnetic resonance) shifts and other hyperfine effects. $\endgroup$ – Pieter May 7 '18 at 11:11

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