# Conceptual doubt regarding nuclear energy levels

I was recently studying about nuclear energy levels and frankly, I thought that I understood the concept pretty well. However, this little problem showed me how wrong I was. The problem is given below:

Plutonium $^{242}_{94}\mathrm{Pu}$ decays into uranium $^{238}_{92}\mathrm{U}$ by alpha decay. The energy of alpha particles takes four distinct values: $4.90\,\rm MeV$, $4.86\,\rm MeV$, $4.76\,\rm MeV$ and $4.60\,\rm MeV$. In all cases, a gamma ray photon is also emitted except when the alpha energy is $4.90\,\rm MeV$. Use this information to suggest a possible nuclear energy level diagram for uranium.

Now, my approach to this problem was to first find the rest energy of the plutonium and then by subtracting the respective alpha particle energies I could show a transition to uranium. However, I am confused as to whether I should consider rest energy of plutonium or its ionisation energy.

Moreover, how exactly does one calculate rest energy? I think that it can be done by finding the atomic mass which in this case is $242\,\rm u$ and then multiplying it by $931.5$.

If my approach is wrong or incomplete, then please suggest some alternatives.

• ionisatiion energy has nothing to do with nuclear levels, it is electron levels and orders of magnitude smaller than the mev energies of nuclear transitions. As in the answer it is simpler to work with energy differences. The rest energy of a nucleus is its mass ; in this link you may get an understaning of how it works en.wikipedia.org/wiki/Nuclear_binding_energy Jan 15 '16 at 11:48

You are told that no photon is emitted when the alpha particle has an energy of 4.9 MeV, so the difference between the plutonium and uranium ground state energies must be $\Delta E =$ 4.9 MeV. If the alpha energy is less than 4.9 MeV and a photon is emitted that means there must be an excited state of the uranium nucleus, and the difference between the alpha energy and 4.9 MeV tells you the energy of the excited state.