We can read that positron emission demands a 1.022 MeV energy difference. I was thinking this was because some kind of internal pair production occurs and the electron is captured whilst the positron is ejected. However I can't find anything that actually says this. Does anybody know of any references that do? Or can anybody give any references that explain why this is definitely not the case?


Suppose we do a simple mass balance on the reactions for beta plus decay and for the electron capture. We'll ignore the kinetic energy of the reaction products to get the minimum energy required.

The neutron is heavier than the proton by $1.293$ MeV, so for beta plus decay the reaction is:

$$ p + 1.804\,\text{MeV} \to n + e^+ + \nu_e $$

where that figure of $1.804$ MeV is the $n-p$ mass difference of $1.293$ MeV plus the positron mass $0.511$ MeV (I'm assuming the neutrino mass is negligable). On the other hand electron capture is:

$$ p + e^- + 0.782\,\text{MeV} \to n + \nu_e $$

where that figure of $0.782$ MeV is the $n-p$ mass difference minus the electron mass.

So both reactions require energy to be supplied, and that energy has to come from a rearrangement of the nucleons. However the beta plus decay requires $1.804$ MeV to go while the electron capture requires only $0.782$ MeV. The difference of $1.022$ MeV is of course just $2m_e$.

And this is all the Wikipedia article means when it says:

electron capture is energetically favored by $2m_e/c^2 = 1.022$ MeV

It just means the electron capture requires $2m_e$ less energy to make it go. It does not imply anything about the detailed mechanism of the reaction.

  • $\begingroup$ The mechanism of the reaction is what I'm really interested in. How does electron capture actually work? You haven't addressed that, but I didn't ask you to, so +1 for answering. $\endgroup$ – John Duffield Jan 1 '17 at 15:23
  • $\begingroup$ @JohnDuffield: that would be an interesting question to answer, though I suspect you would not be impressed with my answer. First it involves QFT, and secondly in a strongly interacting system like a hadron particles don't exist. This sounds like an outrageous claim, but particles are excitations in Fock states and in a strongly interacting system the QFT states are not Fock states, and indeed not even superpositions of Fock states. That's why you'll hear hadrons described as a sea of virtual particles rather than simply three quarks. $\endgroup$ – John Rennie Jan 1 '17 at 16:25
  • $\begingroup$ All points noted John. As you may have deduced, there's quite a lot of particle physics that I'm not impressed with. But thanks for offering something anyway. $\endgroup$ – John Duffield Jan 2 '17 at 13:35

From the wiki link you give:

In a proton, whose charge is +1, there are two up quarks and one down quark. Neutrons, with no charge, have one up quark and two down quarks. Via the weak interaction, quarks can change flavor from down to up, resulting in electron emission. Positron emission happens when an up quark changes into a down quark.

This is a table which shows quark charges:


This is the table of quark decays:


and this is the reaction at fundamental quark level of what is going on:


Reading the diagram time going up the y axis, the up quark turns into the down quark through an off mass shell W+ and a positron plus an nu_e appear. In positron decays of nuclei, a proton turns into a neutron by the diagram on the right.

That is how positron emission can happen if there is available energy. There are more charges than electron and positron charges to keep the charge balance.

Again from your wiki link, for completeness:

For low-energy decays, electron capture is energetically favored by 2m_ec^2 = 1.022 MeV, since the final state has an electron removed rather than a positron added. As the energy of the decay goes up, so does the branching ratio towards positron emission. However, if the energy difference is less than 2mec2, then positron emission cannot occur and electron capture is the sole decay mode.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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