I am having some trouble assigning the odd proton and neutron to the appropriate nuclear shells in the $^8B$ nuclide formed in the $^7\mathrm{Be}(p, \gamma)^8\mathrm{B}$ reaction in the solar plasma.
The $1s_{1/2}$ shell is filled by two $n$ and two $p$. It seems appropriate to place the remaining three protons in the next level $1p_{3/2}$ since that is the lower energy of the two splits at $\mathcal{l}=1$. I would think the odd remaining neutron should go in $1p_{3/2}$ as well, since it is lower energy than $1p_{1/2}$, but I can't obtain the published $J^{\pi}=2^+$ for the resultant nuclear spin with that assignment, i.e., $J = j_1 + j_2 = 3/2 + 3/2 = 6/2 = 3$.
If I kick the odd neutron up to $1p_{1/2}$ then technically I am supposed to subtract the two spins, $J = j_1 - j_2$, since they are in different Schmidt groups. That would give resultant $J= 3/2 - 1/2 = 2/2 = 1$.
If I go ahead and add the two spins as $j_1 + j_2 = 3/2 + 1/2$ then I get the published $J = 2$ and the parity, being the product of two even parity, is the desired positive parity for $J^{\pi} = 2^+$.
Am I correct in this assignment? I haven't been able to find any mention of this nuclide as an exception to the rules as I set out above, perhaps because it is odd-odd and not going to be stable (it $\beta^+$ decays to $^8\mathrm{Be}$ which almost immediately disintegrates to two $\alpha$ particles).
