I am actually calculating the nuclear spin of Sodium 23. Here we have 11 protons and 12 neutrons. Now both the nuclei are short of the magic numbers. When I use the shell model for protons and neutrons separately, I found 3 protons in the $1d_{5/2}$ sub-shell and 4 neutrons in the same $1d_{5/2}$ sub-shell. So because of two pairings, neutrons give spin as 0 and because of a pairing in protons, one proton is left out which should give spin as ${1/2}$. But in the book its, $I={3/2}$. Please can anyone explain the fact how the spin of Na nucleus is ${3/2}$.
Thank you in advance.
[Physics] Nuclear Spin of Sodium 23
nuclear-physicsquantum mechanicsquantum-spinspin-models
Best Answer
The number of neutrons is even, so it indeed means that they contribute spin zero and positive parity.
The spin and parity comes from the "last proton" because the number of protons is odd.
The dependence of the energy on the angular momentum is such that the pairs at a high value of $J$ are preferred (lower in energy) due to the special, spin-dependent features of the strong nuclear force (features invisible in the single-nucleon model). That's true despite the fact that the single-particle shells with a lower $J$ could be preferred.
It follows that among the 3 protons in $1d_{5/2}$, the pair really chooses $j_z=\pm 5/2$, the maximum value (in the absolute value). The remaining slots $j_z=\pm 1/2$ and $\pm 3/2$ are available for the last proton. The last proton also prefers the higher value of $J$ so it will sit in the $J=3/2$ state. It's a $d$-shell, i.e. $l=2$, so the parity is $(-1)^l=+1$.