# Nuclear Spin of Sodium 23

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.

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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 $(-1)^l=+1$.
Hi LM thanks for the clear explanation. Now suppose if we had two more protons in the $d_{5/2}$, then the spin would have been 1/2. Am i doing it right or not? please provide some info. (PS: I was actually reading your "Why are there spinors?" in your reference frame blog yesterday. So i am doubly happy after seeing your reply,,,Many many thanks) –  bluesquare Dec 19 '13 at 12:04