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

  • #1

Summary:

I'm trying to write some nuclear decay questions for my students where they have to use conservation of momentum and conservation of mass-energy to solve problems, but I don't know if it's true that all the released energy is in the form of kinetic energy.

Main Question or Discussion Point

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?
 
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Answers and Replies

  • #2
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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.
 
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