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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$\begingroup$ I'm surprised to see that despite the high number of views nobody mentions that the spin of $^{23}$Na is easily corrected by the Nilsson model. $\endgroup$– Daniel CastroAug 1 at 7:36
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3 Answers
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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 $(-1)^l=+1$.
answered Dec 19, 2013 at 11:20
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2$\begingroup$ 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) $\endgroup$ Dec 19, 2013 at 12:04
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1$\begingroup$ It's aluminum-25 whose spin is actually 5/2 and parity plus, see en.wikipedia.org/wiki/Isotopes_of_aluminum - thanks for your kindness. ... The spin 5/2 isn't that easy to see and the superreliable rules to get the spin in all cases probably doesn't work but in analogy with Hund's rules in chemistry, it's probably favorable to have a larger total J of the protons. You may imagine that the 2 pairs in this case fill the 1/2 and 3/2 pairs of states while the 5/2 is unpaired. If someone knows how to be sure how they pick the states, it is not me. ;-) $\endgroup$ Dec 20, 2013 at 14:59
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2$\begingroup$ You must understand that it's a very complicated system with nonlinear Yang-mills couplings etc. etc. and one calculates the energy level. It's unreasonable to expect that there exists a kindergarten-comprehensible algorithm to reliably answer a nontrivial aspect of the answer. $\endgroup$ Dec 20, 2013 at 14:59
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1$\begingroup$ This seems totally wrong to me. If the nucleus is spherical, then all $j_z$ values are equal in energy. If you have a pair in states with opposite $j_z$ (which makes sense), then the total angular momentum is just the $j$ of the third, unpaired proton, which is 5/2. but in analogy with Hund's rules in chemistry Hund's rule doesn't apply in nuclear physics, because the interaction is attractive. $\endgroup$– user4552Jan 14, 2019 at 20:31
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$\begingroup$ Is it necessary to utter Yang-Mills here, since low-energy phenomenology should be, by construction, devoid of the QFT language, while considering only a handful emergent regions of the space of possibilities? [for all such low-energy effective field theories] Never mind op, this doesn't concern directly to the question you asked, it should be treated just as a remark. $\endgroup$ Mar 9 at 2:19
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As you say neutrons do not contribute since they are all paired. The unpaired proton sits on the $1d_{\frac{5}{2}}$ so based on the extreme limit of the nuclear shell model sodium 23 should have spin=$\frac{5}{2}$ and even parity. Experiments reveal that sodium 23 actually has spin=$\frac{3}{2}$ and even parity. The reason is tricky but simple enough, the extreme limit of the nuclear shell model is just a model so we can not expect it to hold always, is more like a rule of thumb that is helpful because it holds most of the time but it should not surprise you if you find cases in which it does not. You can check by yourself that Thallium 203 is also an exception.
I do not know if you can find other model or improvements of this same model that would account for all the observed spins and parity of nuclei without exception. If you wanna read more on this I suggest Krane page 125.
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Neutrons provide nothing to nuclear spin due to even number, but uncouple proton in 1d5/2.To maximize the Iz (uncouple proton) it should be in 3/2 state because couple protons in 5/2.So I=5/2
