radiation - $alpha$ decay to more than one nuclear state

I do not understand how $\alpha$ decay can be a probabilistic process such that there are multiple products from the decay. For example:

$^{241}\mathrm{Cm}$ decays to the excited states of $^{237}\mathrm{Pu}$, *

0.202MeV, spin: 5/2 with probability 13%,

0.125MeV, spin 3.2 with probability 17%

and 0.145MeV, spin 1/2 with probability 70%

Also there is no decay to the $^{237}\mathrm{Pu}$ ground state.

I dont not understand how the decay has this choice?

* class notes

Answer

The simplest way to think of it is to note that the nucleons in a heavy nucleus are arranged in a shell structure not unlike the electrons in an atom, and when you want to form that alpha particle you have to grab two protons and two neutrons to do it.

They do not have to be the highest energy ones.

If you pull some of them out of low lying shells you leave a hole in the remnant nucleus which means that it is in an excited state. This process is not unlike what happens in the production of characteristic atomic x-rays.

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Now, there are some limits. In particular energy conservation set an absolute limit on how excited the remnant can be.

There are also correlations between the nucleons in the parent nucleus that contribute to the selection process (imprecisely some pairs of nucleons are "near" each other more often than others).