# Spin Parity of excited 60Co

1. Sep 10, 2008

### erin85

1. The problem statement, all variables and given/known data
I am asked to determine the most likely I$$\pi$$ state for an excited 60Co nucleus (a 59Co 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$$\pi$$ for 59Co is (7/2)-, and the ground state for 60Co 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

|Ip-In|$$\leq$$Itot$$\leq$$|Ip+In|

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?