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When a neutron is added to the nucleus of Uranium-238, it becomes Uranium-239. Now, the Uranium-239 undergoes a β-decay and becomes Neptunium-239. This Neptunium-239 again undergoes another β-decay and becomes Plutonium-239.

Now the question is, what causes the β-decay in Uranium-239 and Neptunium-239?

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  • $\begingroup$ Are you asking why those elements undergo beta decay rather than alpha decay? $\endgroup$
    – PM 2Ring
    Commented May 19, 2019 at 10:32
  • $\begingroup$ @PM2Ring Nope. I just wanted to understand what causes the β-decay. I understood from the answers that β-decay happens in order to maintain the n/p ratio, that tends to make the atom stable. But, since you mentioned α-decay, now I'm curious to know, in what conditions does α-decay happens? P.S. I know what an α-decay is, but I want to know why that happens. $\endgroup$
    – Vishnu Vivek
    Commented May 19, 2019 at 13:25
  • $\begingroup$ That sounds like a good topic for a fresh question, but check existing questions first so you don't ask a duplicate. But briefly, all heavy nuclei are unstable, and the heavier they are, the higher is the neutron : proton ratio required for (temporary) stability. Alpha emission reduces the mass without disturbing the neutron : proton ratio appreciably. It also doesn't change the parity (odd or even) of the neutron & proton numbers, which is an important factor in stability. See en.wikipedia.org/wiki/Even_and_odd_atomic_nuclei $\endgroup$
    – PM 2Ring
    Commented May 19, 2019 at 13:42
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    $\begingroup$ How is this different from the general question of why beta decay occurs? $\endgroup$
    – user4552
    Commented May 19, 2019 at 14:17
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    $\begingroup$ Of course the general answer is the totalitarian principle, plus energy considerations, plus an allowed interaction vertex from here to there, and energy is the thing that selects the direction which is why the answers are talking about energy. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented May 19, 2019 at 19:33

2 Answers 2

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It is all about stabilising $n/p$ ratio. Atoms tend to stable when their $n/p$ ratio approaches 1. Heavier elements usually have more neutrons than protons. By emitting a $\beta$ particle, their proton count increases, while neutron count decreases.

This is one of the reasons.

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    $\begingroup$ Ratio close to $1$ only for very light nuclides. Not very close to $1$ most heavy ones e.g. $^{238}U$. $\endgroup$
    – badjohn
    Commented May 19, 2019 at 9:49
  • $\begingroup$ @HS Singh Thanks for your answer. But I believe it is n/p ratio rather than p/n. Isn't it? $\endgroup$
    – Vishnu Vivek
    Commented May 19, 2019 at 13:28
  • $\begingroup$ @HS Singh I have made the edits. Thanks. $\endgroup$
    – Vishnu Vivek
    Commented May 22, 2019 at 9:03
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In the nucleus of large atoms, the strong nuclear force must overcome or match the electrostatic repulsion between protons in order to remain stable. Thus if the nucleus is too large due to an abundance of neutrons, the strong force will be unable to keep the nucleus stable for long. This requires a change in the number of neutrons, which explains why beta minus decay occurs.

The jump from uranium to neptunium increases the size of the nucleus by one proton, passing a stable limit. The decay requiring the least energy is beta minus decay, which then occurs:

n→p+e−+ν¯e+(1MeV) (beta minus)

p+(1MeV)→n+e++νe (beta plus)

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