(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am asked to determine the most likely I^{[tex]\pi[/tex]}state for an excited^{60}Co nucleus (a^{59}Co nucleus that has just been hit by a neutron). I have determined already from the previous part of the problem that the excited state has energy of ~7 MeV compared to the ground state. Not sure if that is relevant...

From a table of nuclear properties, I know the ground state I^{[tex]\pi[/tex]}for^{59}Co is (7/2)^{-}, and the ground state for^{60}Co is 5^{+}.

2. Relevant equations

Co has 27 protons, so 59Co is an odd-even, and 60Co is an odd-odd. I know if there is more than one unpaired nucleon (odd-odd), we take both of their spins to determine the spin of the entire nucleus, via

|I_{p}-I_{n}|[tex]\leq[/tex]I_{tot}[tex]\leq[/tex]|I_{p}+I_{n}|

And parity = (-1)^{lp+ln}

3. The attempt at a solution

I am just not sure how to figure this out for an excited odd-odd nucleus. This is a PhD quals question, and we aren't given access to a chart showing the order of levels or anything (and we would not be expected to memorize it). I could gather from the two ground states given that the n would have I=3/2... but I don't know what it would have if promoted to another shell/state. Any suggestions?

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# Homework Help: Spin Parity of excited 60Co

Can you offer guidance or do you also need help?

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