Answer: 39Ca, 40Ca, 41Ca, 41Sc: Ground & Excited State Spins & Parities

[AlexSm11]

The question:
Use the shell-model diagram provided to justify the spins and parities of the ground-state of 39Ca, 40Ca, 41Ca and 41Sc. Give the spins and parities of the first excited state of 39Ca,41Ca and 41Sc
(Z=20 for CA and Z=21 for Sc)

My problem:
Although I'm fairly certain I've got the spins and parities of the ground states right, I'm ASSUMING (something I know to be dangerous in physics :p) that the first excited state will be the closest higher energy level on the diagram. Is this correct?

Because 2s1/2 state is a lower energy configuration on the diagram than 1d3/2 state.

Thanks in advance

 

[kloptok]

Since both 39-Ca, 41-Ca and 41-Sc have an even number of one nucleon and an odd number of the other one (even #protons and odd #neutrons in Ca, reverse in Sc), the odd nucleon determines the spin and parity of the nucleus. I would, just like you did, assume that it is this last odd nucleon that carries the excitation energy. Then I don't think it is at all generally true that the excitation will be just one energy level higher in the diagram, probably you could have lots of different probabilities for the excitation. BUT, if you don't have any given information about this you should probably assume that the excitation is just to the next level in the shell model diagram. I'm not sure that the levels are the same for neutrons and protons though so there might be a difference between the isotopes.

Now I'm not entirely sure about this reasoning but that's what I would do (and did in my Nuclear Physics course some month ago) in a problem like this. Can you look it up somewhere?