# Deuteron spin and parity

Deuteron is given by j(p)=1(+)

In my textbook it says that the observed parity of + for deuteron means that the orbital angular momentum quantum number, l, is even so is 0 or 2, and s=1.

However, looking back the textbook also says that for odd-odd nuclei, the parity is given by the product of the parities of the two odd shells by p=(-1)^l. If l was 1 in this case, for both cases, wouldn't this give a even parity to the nucleus? Then as j=1, s=0.

I've read somewhere (probably wikipedia) that it's preferable for the neutron and proton to have the same spin, which would mean l=0 or 2 is more favourable, but the book definately makes it sound that it is the parity you can deduce l from.

Any help would be really appreciated!

## Answers and Replies

Bill_K
Science Advisor
What they have in mind for odd-odd nuclei is a situation in which two individual nucleons orbit about a closed shell of other nucleons. In that case, each of the two contributes an orbital wavefunction and the parity receives a factor of -1L for each particle. But in the case of a neutron there is no closed shell, the n and p just orbit each other. So there's only one orbital wavefunction and one factor of -1L.