I am calculating the nuclear magnetic moment of aluminium-27 (Z=13). By looking at a nuclear energy diagram it can be seen that $$j=\frac{5}{2}, s=\frac{1}{2}, l=2.$$ Since there are an even number of neutrons and an odd number of protons, we are concerned with just the final odd proton, since the rest of the nucleons will pair up, resulting in a net magnetic moment of zero.
We must now decide on which Schmidt line to use when calculating the g-factor, the two choices being that corresponding to $j=l+\frac{1}{2}$: $$g_J=\Big(1+\frac{1}{2j}\Big)g_l+\frac{1}{2j}g_s$$ or $j=l-\frac{1}{2}$: $$g_J=\frac{1}{j+1}\Big[\Big(j+\frac{3}{2}\Big)g_l-\frac{1}{2}g_s\Big].$$
Once this has been decided, we can simply use $$\mu_J=g_J\mu_Nj.$$
But how can I decide on which one? I am thinking that it would depend on whether the off nucleon is spin up or spin down, but then how would you know that?
Best Answer
Using the shell model (Woods-Saxon with spin-orbit correction potential) you can fill all the neutron shells up to the $1d_{5/2}$ level. You can do the same for the proton shells and it will result in a hole in the $1d_{5/2}$. The total angular moment J of the nucleus corresponds to the total angular moment of the (missing) proton, which is $J^\pi=\frac{5}{2}^+$. The hole is in the stretch level $1d_{5/2}$ which is $j=l+\frac{1}{2}=2+\frac{1}{2}=\frac{5}{2}$. So you should go for the g-factor calculated for $j=l+\frac{1}{2}$: $$g_j=(1+\frac{1}{2j})g_l+\frac{1}{2j}g_s$$