4
$\begingroup$

Question

For nuclides that decay by alpha emission, the Geiger-Nuttall law gives a simple and reasonably accurate estimate of the half-life. Essentially, one can model the alpha particle as a particle in a "box" – the nucleus – and calculate the probability of tunneling out of the box using basic quantum mechanics. The result is that the half-life depends exponentially on the decay energy:

$$\ln \lambda \approx a_0 - a_1 \frac{Z}{\sqrt{E}}$$

where $\lambda$ is the half-life, $Z$ is the atomic number, and $E$ is the decay energy.

Is there any similar simple relationship for beta decay?

Answers that don't work

Beta decay half-lives appear to be very poorly correlated with:

  • Decay energy. Rhenium-187 and lutetium-176 have similar half-lives (about 40 billion years), but lutetium-176 has a decay energy of 1.2 MeV while rhenium-187 has a decay energy of just 2.6 keV.
  • Atomic number or number of neutrons. Cadmium-113 has a half-life of more than 8 quadrillion years, while cadmium-115 has a half-life of just 2 days – even though isotopes 112, 114, and 116 are all stable or have extremely long half-lives.
  • Nuclear shell closure. Even though potassium-40 decays to calcium-40, which has magic numbers of both protons and neutrons, it has a half-life of over a billion years. Meanwhile, potassium-42 has a half-life of only 12 hours.

In short, beta decay half-lives vary widely, and I don't see any obvious rules to help understand why. Is there any way of explaining these results without a very complex calculation?

Review of related Physics.SE questions

The answer to Does the decay energy of a beta decaying isotope become lower as the half life becomes greater (or vice versa)? says that:

That leaves the phase space available to the products as the thing that almost solely determined the lifetime.

In short the phase space available to the interaction depends to a large degree on the total energy and to a much smaller degree on the mass of the recoiling remnant nucleus.

I gave a rather more complete treatment in another answer about the free neutron lifetime.

However, while the theoretical argument that the phase space should depend strongly on the total energy seems plausible, it doesn't seem to match experimental results very well. As I noted above, nuclei with similar half-lives often have completely different decay energies, and vice versa.

The most upvoted answer to Can we predict the half-lives of radioactive isotopes from theory? mentions a

quantity "$ft$" which convolves the half-life of the decay with the electrical interaction between the emitted electron and the positively-charged daughter nucleus.

It is supposed to vary within a narrow range. However, I don't understand this description; also, I looked up some tables of $ft$ values and found that $ft$ actually varies by many orders of magnitude between different nuclei. For example, this textbook chapter (chapter 8 from Modern Nuclear Chemistry by Loveland, Morrissey, and Seaborg) cites $\log(ft)$ values ranging from $-0.27$ to $+7.36$.

$\endgroup$
3
$\begingroup$

This should be a comment, but it is too long.

The alpha particle is a stable nucleus itself, and the decays happen purely from energy consideration, there exist a system with lower energies to which a decay of an alpha gives access.

Beta decay is a two level decay,there should exist a system of nuclei with lower total energy, but also the neutron itself should decay in order to get the beta/electron. It is a two paths decay, the second depending on the weak interaction. That is why a single rule is not possible.

A neutron has a probability to decay within a nucleus, if it is neutron rich, and the neutron's orbital has a probability to be far enough for the residual strong force ( the nuclear force) to be weak and when free a neutron decays. This would depend not only on energies, but also on the specific neutron orbitals. Or it could be a simple energy level accessibility consideration and the neutron is freed and decays.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.