Lithium-6 isotope has an approximate magnetic momentum of $0.88\ \mu_N$ in its fundamental nuclear state. I'm trying to find its angular momentum and parity.
I found in a standard table: $I=1^+$ and while I get the parity part, I don't understand the rest.
What I have done is filling the shells separately for protons and neutrons and finding that both have their outer particle in the $1 P_{3/2}$ state. Sum of these gives possible total angular momenta to be either $0$, $1$, $2$ or $3$.
So my question is: how do you choose between these to find the real value $I=1$?
Best Answer
Don't forget about the importance of nucleon-nucleon pairing. If you imagine the lithium-6 nucleus as an alpha particle and a deuteron bound in an $s$-wave state with positive parity and no angular momentum, you reproduce the spin and parity. Wikipedia tells me the deuteron has magnetic moment 0.857$\mu_N$, and of course the $\alpha$ has zero; quite close to your value for $^6$Li. You can explain the difference in a hand-waving way by remembering that both the $\alpha$ and deuteron both have some $d$-wave component which will mix in a more complicated way.
To give a more shell-model-friendly explanation, remember that as you fill electron orbitals you have to assign your atom (or ion) a total electronic spin; as you fill nuclear orbitals, you have to keep track of both total spin and total isospin, as permitted by the exclusion principle.