# Ground states in the shell model for odd-even nuclei

I understand that even-even nuclei (Z and N number) have zero spin because of pairing.

Even-odd nuclei have the spin of the odd nucleon, and parity is given by $(-1)^L$ - so my question is, how do we work out the state which this odd nucleon is in?

As an example: $^3_2 \text{He}$ the odd nucleon, is a neutron which is the state $1\text{s}_\frac{1}{2}$ ($l=0,$ $s=1/2$) so the answer for the ground state spin-parity is $\frac{1}{2}^+$... Or with $^9_4 \text{Be}$ the most energetic neutron is the $1\text{p}_{\frac{3}{2}}$ giving $\frac{3}{2}^-$ but how is this worked out? I assume it is something to do with the nucleons occupying all the lowest energy states and finding the highest energy spare nucleon - How will I know which quantum numbers are the lowest energy?

Edit: for example, do I need to refer to this table? http://en.wikipedia.org/wiki/File:Shells.png

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