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Looking on Wikipedia's cluster decay article, I see that almost all clusters emitted from a nucleus have an even number of both protons and neutrons (not necessarily the same number of each). The only exceptions listed are:

  • 23F from 231Pa
  • 25Ne from 233U or 235U
  • 29Mg from 235U

If an alpha particle is also considered a cluster, the same is true for it too as it contains two protons and two neutrons.

  • Why are odd numbers of protons or neutrons in cluster decay so rare?
  • Why is the number of neutrons always an equal or greater number of than protons in the cluster?
  • Why couldn't a nucleus emit a single proton?
  • Why doesn't cluster decay involve odd numbers of both protons and neutrons in any of the parent nucleus, the emitted cluster, or the daughter nucleus?
CJ Dennis
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1) Why are odd numbers of protons or neutrons in cluster decay so rare?

Experimentally, even-even nuclei (i.e, a nuclei with even $Z$ and even $A-Z$) are more stable than even-odd or odd-even ones, and both configurations are more stable than the odd-odd one. A stronger p-p (proton-proton) and n-n (neutron-neutron) interaction with regards to p-n interaction inside the nuclei accounts for it.

This is explained in the shell model or in the Fermi gas model by the Pauli exclusion principle. In these models particles are assumed to be all stored below the Fermi level, and effectively free (in a potential well that binds them in the nucleus) since almost any interaction would move them to already occupied energy levels; the only allowed interactions are those that exchange the particles' energies. This strongly discourages p-n interactions, since p and n quantum numbers are different and in general also an energy exchange would violate Pauli's principle. Pauli's principle also guarantees that nuclei with equal Fermi level for both p and n are more stable (this can be seen by a straightforward calculation).

2) Why is the number of neutrons always an equal or greater number of than protons in the cluster?

The potential well that binds neutrons and protons in the nucleus is deeper for neutrons than for protons, due to E.M. repulsion between the latter. Since Fermi levels must be the same, configurations with more neutrons than protons are favored.

I don't have a definitive answer for the last two questions, but I'll try to formulate some hypotheses anyway.

3) Why couldn't a nucleus emit a single proton?

I think this could be explained computing the $Q$-value of nuclear decays and observing that is always smaller for proton emission than for cluster (or $\alpha$) emission, making the former far more difficult to be observed. Note however that fission processes do in general involve protons as side products.

4) Why doesn't cluster decay involve odd numbers of both protons and neutrons in any of the parent nucleus, the emitted cluster, or the daughter nucleus?

The answer for the emitted cluster and the daughter nucleus resides in what is said above. It seems therefore that for an even-even + even-even decay also the parent nucleus should be even-even, preferring to decay in $\beta$ otherwise.

References:
https://web-docs.gsi.de/~wolle/TELEKOLLEG/KERN/LECTURE/Fraser/L10.pdf
http://virgilio.mib.infn.it/~zanotti/FNSN/FNSN_files/FNSN/Nucleus/Fermi_gas_model+Shell_model.pdf

Francesco Arnaudo
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    Proton decay does occur. In general, this answer spends a long time explaining why pairing exists microscopically, which isn't really what the question is asking. –  Jun 20 '19 at 12:08
  • @BenCrowell I see your point, but as far as I know pairing issues are not quantitatively considered in nuclear decay models (such as Gamow theory), and are meant to be understood qualitatively and microscopically/added as corrections to the results given by these models – Francesco Arnaudo Jun 20 '19 at 14:38