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I was reading wikipedia article on beta decay when I came to a section called "Fermi transition". So, the way I understood it is:
"During beta decays, nuclear transitions which involve net zero change in orbital angular momentum of the nucleus is highly favored."
Why is it so? Does it have to do anything with parity conservation? How do I account for such phenomenon?
If the decay products don't have to carry any orbital angular momentum, the angular part of the final-state wavefunction is $s$-wave. If the decay products do have to carry away angular momentum, the angular part is $p$-wave (or higher).
The scaling rule for the overlap between a wavefunction with orbital angular momentum $\ell\hbar$ and a nucleus with radius $r$ is a factor of $(kr)^\ell$, where $\hbar k$ is the momentum scale in the decay. So the larger $\ell$ gets, the smaller the overlap between the initial state and the final state, and the slower the transition. This rule of thumb is good for all decay reactions, not just beta decays.
