What Is the Likely Spin Parity of an Excited 60Co Nucleus?

[erin85]

Homework Statement

I am asked to determine the most likely I^{[tex]\pi[/tex]} 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^{[tex]\pi[/tex]} for ⁵⁹Co is (7/2)^-, and the ground state for ⁶⁰Co is 5⁺.

Homework 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)^l_p+l_n

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?

 

[Jhero]

Thank you for your question. To determine the most likely I\pi state for an excited 60Co nucleus, we need to consider the spin and parity of the excited state as well as the spin and parity of the ground state of 60Co.

First, we can use the equation you provided, |Ip-In|\leqItot\leq|Ip+In|, to determine the possible values of Itot for an excited odd-odd nucleus. In this case, we have Ip=5/2 and In=3/2. Plugging these values into the equation, we get |5/2-3/2|\leqItot\leq|5/2+3/2|, which simplifies to 1\leqItot\leq4. This means that the possible values for the total spin of the excited state are 1, 2, 3, and 4.

Next, we need to consider the parity of the excited state. Since we know that the ground state of 60Co has a positive parity (5+), the excited state must also have a positive parity in order to conserve parity. This means that the possible values for the total spin and parity of the excited state are 1+, 2+, 3+, and 4+.

To determine which of these states is the most likely, we can look at the energy difference between the excited state and the ground state. The fact that the excited state has an energy of ~7 MeV compared to the ground state suggests that it is a high energy state. This means that the most likely state would have a higher spin and parity compared to the ground state.

Based on the possible values we found earlier, the most likely I\pi state for the excited 60Co nucleus would be 4+. This is because it has the highest possible spin and parity compared to the ground state, and it also has a higher energy which is consistent with a high energy state.

I hope this helps you solve your problem. Good luck with your PhD quals!