From my understanding, beta-plus decay is only possible when certain energy requirements are met due to the fact that the atom will need to emit a positron particle, which has a rest energy-mass equivalent of 511 KeV. Therefore the difference in the rest energy-mass equivalent of the entire daughter atom vs of the parent atom (or, in other words, its difference in binding energy) has to be at least 511 KeV in order to (in layman terms) ‘produce’ a positron.

However in most physics textbooks they go a step further and mention how the energy requirement is actually double that (1.022 MeV). They explain it’s due to the fact that positron decay converts a proton into a neutron, decreasing the atomic number by 1, meaning the daughter atom would have an excess electron that will also need to be ejected from the atom.

I can't understand how this orbital electron can influence the decay properties of its nucleus. I will illustrate what I don't understand in an imaginary counter-example:

Let's assume the parent nucleus has atomic number Z_1, neutron number N_1, and charge 0 (Z_1 electrons). Let's then assume the theoretical lower-energy daughter state has atomic number Z_2 = Z_1 - 1, neutron number N_2 = N_1 + 1, and charge -1 (Z_2 + 1 electrons). Let's finally assume the transition energy between those two states is lower than 1,022 KeV, say, 612 KeV, and that the binding energy of an electron in that daughter state is 1 KeV.

Why couldn't this nucleus go from the first (Z_1, N_1) state to the (Z_2, N_2) state, use 611 KeV to emit a positron (511 KeV would go towards the "energy-mass" of that positron and 100 KeV would be used as its kinetic energy and the neutrino), and use 1 KeV afterwards to eject one of the valence electrons from the daughter atom, for a total of 612 KeV of transition energy?

  • 1
    $\begingroup$ Re: "I can't understand how this orbital electron can influence the decay properties of its nucleus." - there are a variety of reasons that orbiting electrons impact decay properties. One, of course, is electron capture (a bare Be-7 ion cannot decay by EC with no electrons to capture). There are also low energy beta decays that can only occur with a highly stripped ion because the low energy electron can't escape and has no available energy level to occupy in a neutral atom. Nature is weird... $\endgroup$
    – Jon Custer
    Commented Nov 4, 2022 at 14:24

1 Answer 1


The additional energy requirement is because the daughter atom has an extra electron and therefore extra mass-energy.

The masses we look up in tables of nuclides are the masses of neutral atoms. But when the parent decays to a daughter via inverse beta decay, it ends up with an "extra" electron, which means that it has 511 keV more mass-energy than it would if it were neutral. So for this decay to occur, the mass of the neutral parent has to be greater than the mass of the neutral daughter by (roughly) 1.022 MeV: 0.511 MeV for the ejected positron and 0.511 MeV to account for the extra mass of the daughter (before the electron is ejected.)

By similar logic, in "regular" beta decay the threshold mass-energy difference is pretty negligible. This is because while the neutron decay has to create 0.511 MeV worth of mass-energy in the form of an electron, the daughter is missing an electron and therefore is 0.511 MeV lighter than its neutral mass. So in this case, the two effects cancel out, and beta decay can happen so long as the neutral daughter mass is less than the neutral parent mass.

  • $\begingroup$ hmmm, I have a PhD in nuclear physics and have never considered this. Tell Caltech I want my $600,000 back. $\endgroup$
    – JEB
    Commented Nov 5, 2022 at 5:30

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