- #1
passingthru
- 19
- 0
Forgive me if this has already been discussed. I've been searching and haven't found it. When summing the spin of an atom in the Russell-Saunders coupling scheme, some texts include the spin of an odd nucleon, and some don't. For example, if an atomic energy state is described as ${3}^P_{2}$, this means that the total spin is one, and since $L=1$, $J=S+L=2$. If this is a hydrogen atom, apparently, the spin of the proton is included. Many discussions of this never even bring up the nucleus. When should the spin of an odd nucleon be included?