I was just watching some videos and came across beta+ radiation (when a positron is emitted). It then occurred to me, how can the following be true, given that a positron and an electron have the same mass:

neutron = proton + electron [eq 1, beta- decay]

proton = neutron + positron [eq 2, beta+ decay]

As this would mean, a neutron = neutron + positron + electron (substitution of eq 2 in eq 1), which seems impossible?

  • $\begingroup$ Great question! Think about the conditions when the decay happens - a free neutron (== outside of a nucleus) can decay, but a free proton cannot. A proton does not have enough energy to decay into a neutron on its own - the energy must come from somewhere else. $\endgroup$
    – Luaan
    Commented May 12, 2020 at 6:33

3 Answers 3


A neutron does not have the same mass as a proton plus an electron, and a proton does not have the same mass as a neutron plus a positron.

  • Proton mass = 938.272 MeV
  • Neutron mass = 939.565 MeV
  • Electron or positron mass = 0.511 MeV


So $m_p+m_e = 938.783 \text{ MeV} \ne m_n$ and $m_n + m_e = 940.075 \text{ MeV} \ne m_p$

Now, your unstated question may be based on thinking that because a neutron can decay into a proton and an electron (and an anti-neutrino) that must mean that it contains a proton and an electron (and an anti-neutrino). That is not the case. When a subatomic particle decays into a different particle the new particles are created, they are not merely separated out of the previous particle. The masses may not balance and the difference is the total KE of the products.

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    $\begingroup$ There's another very important thing to mention - the energy also doesn't balance. A free proton cannot decay into a neutron - you need a proton in a (rather specific) nucleus. The "missing" energy comes from the binding energy of the nucleus - the binding energy for the nucleus with one extra proton happens to be larger than the same nucleus with one extra neutron. The same is true in reverse for neutrons - a free neutron decays readily into a proton (etc.), but a neutron in a stable nucleus cannot - the increase in binding energy is too large. $\endgroup$
    – Luaan
    Commented May 12, 2020 at 6:31

Dale's answer is obviously correct but I'd like to add another way to look at this.

When a neutron decays into a proton and an electron (and an anti-neutrino), while it's true that it doesn't mean that a neutron was a bag of a proton and an electron (and an anti-neutrino), mass conservation should still hold. And it does hold. The mass of the combined system of the proton, the electron (and the anti-neutrino) would be exactly the same as the mass of the neutron. This works out because the mass of a system of particles is not the sum of their masses. Relativity tells us that what gets added together is the four momentum of the constituent particles and you get the mass of the combined system by squaring the total four momentum which is not equal to the sum of the masses of the constituent particles unless there is no relative motion between the constituent particles. So this tells you that a neutron, in fact, can't decay to a state where the proton, the electron, and the anti-neutrino are moving with the same velocity.

  • $\begingroup$ You are right. The mass of the total system is the rest energy $/c^2$. The total system is at rest before and after the decay, hence the mass of the total system is conserved. The problem is that this concept of system mass is not very useful and unusual. $\endgroup$
    – my2cts
    Commented May 10, 2020 at 12:56
  • $\begingroup$ @my2cts I agree but I also think it depends on the context and the other option at hand :P We can as such simply talk about four-momentum conservation and forget about mass conservation as it'd just follow along. But, when people do want to talk about mass anyway, the concept of the system mass is much cleaner than the confusing explanations along the lines of "mass converts to energy". $\endgroup$
    – user87745
    Commented May 10, 2020 at 12:59
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    $\begingroup$ @my2cts I disagree with “not very useful”. Conserved quantities are useful. Invariant quantities are useful. Quantities that are both invariant and conserved are very useful, IMO $\endgroup$
    – Dale
    Commented May 10, 2020 at 16:11
  • $\begingroup$ @Dale Hmm, now that I think about it, the Mandelstam variables, for example, provide a very good instance of an immensely useful application of the "system mass" concept. $\endgroup$
    – user87745
    Commented May 10, 2020 at 16:15
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    $\begingroup$ Ok, that's better. :) But it may not be clear that you mean that the decay products can't have zero total momentum in the neutron's rest frame. $\endgroup$
    – PM 2Ring
    Commented May 11, 2020 at 10:25

A neutron has more mass than a proton and an electron, so it can $\beta^-$ decay, and it does.

A proton does not have more mass than a neutron and a positron (which, as you point out, is the same as the mass of an electron). The laws of arithmetic still hold. So a proton on its own cannot $\beta^+$ decay.

But there are some nuclear isotopes in which changing a proton to a neutron produces a new nucleus with greater binding energy. For example, $^{22}$Na has a mass of 21.994 MeV and can decay to $^{22}$Ne and an electron with 0.003 MeV to spare, as the protons and neutrons in the Neon nucleus arrange themselves more tightly than in the original Sodium. It's in cases like these where the total nuclear energy balance is favourable that positron emission can occur.

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    $\begingroup$ And then there are interesting "borderline" cases like ⁴⁰K, which can do either $\beta^+$ or $\beta^-$ decay. $\endgroup$
    – PM 2Ring
    Commented May 11, 2020 at 10:20

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