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^{9}Be and

^{9}B are both 3/2

^{-}. Assuming in both cases that the spin and parity are characteristic only of the odd nucleon, show how it is possible to obtain the observed spin-parity of

^{10}B(3

^{+}). What other spin-parity combinations could also appear? (These are observed as excited states of

^{10}B.)

Here's what I think I know. The parity is (-1)

^{l}. So for

^{9}Be and

^{9}B, l = 1. It makes sense that these two nuclei have half-integer spin because A = 9 is odd. It makes sense that

^{10}B has integer spin because A = 10 is even. The fact they're telling me to consider the spin and parity is charactersitic of the odd nucleon is supposed to be hinting at something, but I'm not sure what. I realize

^{9}Be has an extra neutron and one less proton than

^{9}B, but I don't know if I'm supposed to be getting any useful information from the fact that the unpaired particle is a proton in one case and a neutron in the other. If anyone can fill in any of these blanks for me, I would appreciate it.

Aside from this question, and in general, I'm confused about how nucleons pair off and have their spin cancel out with other nucleons. From what I've heard in class so far even numbers of nucleons should just pair off and leave nuclei with either 0 or 1/2 spin. :(