In radioactivity, the total binding energy of parent nucleus must be greater than the daughter nucleus. In alpha decay, uranium has greater binding energy because it is large but in beta plus decay (positron) proton changes to neutron then binding energy of the daughter nucleus must increase so how can this happen?

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    $\begingroup$ Why must the binding energy of the nucleus increase after $\beta^+$ decay? $\endgroup$ Commented May 16, 2018 at 23:07

1 Answer 1


I don't think it's quite right to say that "binding energy must increase" in a decay. That's a pretty good rule for heavy nuclei, but you're correct that the proton and neutron both have zero binding energy if they are free particles. There are plenty of other decays, like $\pi\to\nu\mu$ or $\mu\to e\nu\nu$, where it doesn't make sense to talk about the binding energy on either side of the reaction.

The real rule is that the rest mass of the particle that decays must be heavier than the combined rest masses of all the particles produced in the decay. That's why a few neutron can decay to a proton but not the other way around. The way that binding energy works in nuclei can change this relationship, however. For example, a potassium-40 nucleus is heavier than an argon-40 nucleus, and also heavier than a calcium-40 nucleus, so potassium-40 undergoes both positive and negative beta decay.

  • $\begingroup$ In other word, i mean that binding energy of nucleus increases as mass number increase so when proton change into neutron, neutron has more mass than proton than will increase its binding energy thats why i am confused $\endgroup$
    – H A B
    Commented May 17, 2018 at 17:44
  • $\begingroup$ But you're asking about beta decays, where the mass number does not change. In that case the you have to keep track of both the binding energy and the proton-neutron mass difference. Professionals use what's called the "mass excess" to make things simpler. I've written about using mass excess to simplify binding energy problems before (e.g. here, here, here). If nobody beats me to it, perhaps later I will add a mass excess discussion to this answer. $\endgroup$
    – rob
    Commented May 17, 2018 at 18:06

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