From evolutionwiki:

"Potassium 40 decays into argon 40 through a process known as electron capture. In electron capture, an electron from the innermost electron shell "falls" into the nucleus, causing a proton to convert into a neutron."

What's the implication of an electron "falling" into the nucleus, what happens to the electron? Is it the electron that subsequently will be emitted? Why does it not decay by emitting an alpha particle?


2 Answers 2


The decay of potassium-40 to argon-40 is either a $\beta^+$ decay in which what is emitted is not an electron but a positron $$ {}^{40}{\rm K} \to {}^{40}{\rm Ar} + e^+ + \nu_e $$ or, more frequently (if we have whole atoms), an electron capture that you mentioned in which no charged leptons are emitted at the end! About 11% of the potassium-10 decays proceed in this way. $$ {}^{40}{\rm K} + e^- \to {}^{40}{\rm Ar} + \nu_e $$ The remaining 89% of the decays of potassium-40 go to calcium-40 (the beta-plus decay is a small fraction of a percent). Note that the two reactions above differ by moving the positron from the right hand side to the left hand side so that the sign has to be changed.

Potassium-40 has 19 protons and 21 neutrons. Argon-40 has 18 protons and 22 neutrons. So if we focus on the "minimum part" of the nuclei, the reactions above may be reduced either to $$ p \to n + e^+ + \nu_e$$ or $$ p+e^- \to n + \nu_e $$ which are the standard reactions switching protons and neutrons. In particular, the second reaction displayed right above this line is the more "microscopic" description of the electron capture you're primarily interested it.

These reactions preserve the electric charge, baryon number, and lepton number. They also have to preserve energy. A free proton couldn't decay to the neutron and other two particles because it's lighter. Even a proton and a low-velocity electron wouldn't have enough mass/energy to produce the neutron (plus the neutrino) as in the second reaction.

But when the protons and neutrons are parts of whole nuclei, the energies of the initial and final nuclei are affected by the nuclear interactions. In particular, the argon-40 nucleus (and especially atom) is highly bound which means lighter and the reactions where argon-40 appears as a product are therefore "more possible".

To summarize, the electron capture (=falling of the electron) simply means that the proton has a nonzero probability to meet with one of the electrons – probably in the inner shells – and merge into a new particle, a neutron, plus a neutrino. This process can't occur in the vacuum due to the energy conservation but in the context of the nucleus, the interactions with other neutrons and protons make the final state with the new neutron favorable.

On the contrary, alpha decays are more rare. Among 24 isotopes of potassium, only potassium-36 may alpha-decay. Carbon-14 doesn't alpha-decay, either. Among the isotopes of carbon, only carbon-9 alpha-decays. Both of these alpha-decays must be preceded by a beta-decay. Usually just heavy enough nuclei (with too small an excess of neutrons) alpha-decay.

  • $\begingroup$ "an electron capture that you mentioned in which no leptons are emitted at the end!" Maybe better to say "no charged leptons" because the neutrino is present in the final state. $\endgroup$ Commented Jun 26, 2016 at 19:38
  • $\begingroup$ Yup, your correction is right. $\endgroup$ Commented Jun 26, 2016 at 20:38
  • $\begingroup$ From the quark level Feynmann diagram ($u\to d+e^+ +\nu_e$), does $\beta^+$ decay process mean that initial u quark (in this case) is heavier than d quark? (I don't think so.) If not, how to draw the diagram for this process? I think the binding energy has to become some virtual particles (but what's this virtual particle?) and it hits u quark and then this virtual u quark decays? $\endgroup$
    – luyuwuli
    Commented Oct 4, 2016 at 13:57
  • $\begingroup$ Dear @luyuwuli - the up-quark is indeed lighter in "isolation" (except that it can't be isolated) than the down-quark but only the total energy/mass is conserved. There are other contributions to it, from interactions of the quarks with each other and so on, and those change between the initial and final protons/neutrons/nuclei. When you look, you see that the rest mass of the initial nucleus exceeds the sum of the final rest masses so the energy conservation law doesn't ban this process. $\endgroup$ Commented Oct 7, 2016 at 8:47
  • 1
    $\begingroup$ @LubošMotl Sure, treating the whole process as one quark's flavor change is oversimplified. I think since we agree on the fundamental process , it's time to end the discussion. Thanks for your patience and explanation. $\endgroup$
    – luyuwuli
    Commented Nov 21, 2016 at 6:20

Electron capture

Electron capture (K-electron capture, also K-capture, or L-electron capture, L-capture) is a process in which the proton-rich nucleus of an electrically neutral atom absorbs an inner atomic electron, usually from the K or L electron shell. This process thereby changes a nuclear proton to a neutron and simultaneously causes the emission of an electron neutrino.

p + e− → n + ν_e

K,L,M are the principle quantum number(n) of the orbitals of the electrons.

The orbitals of electrons, give the probability of the electron to be found at (x,y,z). For l=0 ( the angular momentum quantum number) the electrons have a probability of passing through the nucleus. If the nucleus is unstable with respect to another, i.e. there exists the extra energy needed for creating a neutron and an electron neutrino from a proton and and electron, one has electron capture and a lower by one Z nucleus appears.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.