Why is it that we mostly speak of alpha and beta decay (when looking at emitted particles with rest mass)? Why don't we speak of decays that emit something like $2n3p$, $3n2p$ and so on? Do they occur at all? Are they highly improbable?

I know that $\text{He}$ has very high binding energy per nucleon, is this relevant for my question?

  • $\begingroup$ Protons and neutrons can also be emitted. $\endgroup$
    – Mithoron
    Commented Feb 17, 2016 at 1:07
  • $\begingroup$ In addition to single nuleons we now have evidence that some nuclei near the drip line decay by two-proton emission. Spontaneous fission could be thought of an extreme case of cluster decay. $\endgroup$
    – rob
    Commented Feb 21, 2016 at 23:24

1 Answer 1


These were historically the first emissions that were detected and most of the decay modes are of the type.

Beta decay is as a result of a neutron in the nucleus decaying into a proton. It is energetically favourable for the nucleus to make this transformation.

The Helium nucleus is very stable. As you quite rightly pointed out is has a high binding energy per nucleon. It is to do with energy levels in the nucleus which can be compared to the energy levels of orbiting electrons in an atom. Just like electron shells being filled gives very stable atoms - Helium, Neon, Argon etc, so it is true of neutrons and protons. When the first shell is full you form a very stable nucleus composed of two neutrons and two protons which is the Helium nucleus.

So the examples you gave, 2n3p, 3n2p, etc are nowhere near as stable as a Helium nucleus and are not emitted.

The gamma radiation is just a way for a nucleus to get rid of some surplus energy from an excited nucleus just like em radiation being given off when electrons drop down energy levels when an atom is excited. gamma's are much more energetic than the photons given off from electron transitions in an atom because the spacing between the energy levels in a nucleus is very much greater that of electron energy levels in an atom.

To compound the shells not being full, the 2n3p nucleus is made more unstable because of the extra Coulomb repulsive force as compared with that in the 2n2p nucleus.
So I assume that the probability of the 2n3p surviving to get out of the nucleus is very small.
Although there is not this extra Coulomb repulsive force within a 3n2p nucleus it is again unstable compared with the 2n2p nucleus because of an unclosed shell and so does not survive to get out.
So because of the very high stability of 2n2p that is the nucleus which escapes.

  • 1
    $\begingroup$ Thanks, I know about the nature of betas and gammas as well :) So, it has to do with Pauli's principle. Can you also explain how "stableness" relates to "probability of emission"? E.g. why does it not happen that 2n3p is emitted and then quickly decays to 2n2p+p? $\endgroup$
    – Jasper
    Commented Feb 16, 2016 at 23:15
  • 1
    $\begingroup$ The statement that only two protons and two neutrons can occupy an energy level is not correct. That is approximately true for atomic systems where spin orbit coupling is weak and L and S are separately good quantum numbers. In that case, the two levels correspond to approximately degenerate spin-up and spin-down states. However, in nuclei, there is a strong spin orbit coupling in the mean field approximation, and so L and S are no longer good quantum numbers and J=L+S is the number that labels orbits, carrying a degeneracy 2J+1. That's why 8Be isn't a closed shell. $\endgroup$
    – ragnar
    Commented Feb 17, 2016 at 6:19
  • $\begingroup$ @ragnar Thank you for your comment which I have acted on. Does the fact that to get out pf the parent nucleus the particle must tunnel affect the probability of escape? Perhaps a less stable particle has a lower energy with the parent nucleus and hence has to tunnel further? $\endgroup$
    – Farcher
    Commented Feb 17, 2016 at 8:32

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