In a nuclear decay, is all of the released energy kinetic energy?

I'm trying to make up an example for my students to illustrate that in nuclear decay, mass-energy and momentum are both conserved.

I found this problem: https://physics.stackexchange.com/q...ate-velocity-of-radon-220-nuclear-after-decay

I am trying to modify it so that they have to use both mass-energy and momentum to solve. I wanted them to try calculating the radon's speed using momentum. Then I wanted them to show that if you add up the kinetic energies of the alpha and the radon after the decay, that energy is equal to the mass difference. (In the problem given, this is not true, so I need to modify the numbers).

However, I never took a university course in nuclear decay and realized that perhaps my thinking is wrong. I am assuming that all of the released energy becomes kinetic energy of the products. Is that wrong?

If new particles are created you also have to take their rest energy into account (all beta decays need this). If particles end up in excited states you have to take into account that energy. You can study an alpha decay where the nucleus ends up in the ground state, that makes it easy to calculate the velocity.

This is a very nuanced and practical question. I am currently working on neutron dynamics and the rotational energy of a liquid drop nucleon. So far, only Regge theory proposes angular momentum having some scattering effect in quantum. The reactions usually use the before and after laboratory frames and moving frames to calculate energies and momentum but anything in the neutron or proton as far as angular momentum or angular energy is neglected.

I think when I did these questions we used four frames: the lab before, the lab after, the particle CM frame before, the particles CM frame after. We just calculated the difference in rest mass energy. If I remember in more detail, I'll write how we approached this problem, but it neglects neutron rotation, which I think is non-negligible.