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Spin of Sodium 23

  1. Dec 19, 2013 #1
    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 1d5/2 sub-shell and 4 neutrons in the same 1d5/2 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.
  2. jcsd
  3. Dec 19, 2013 #2


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    Hi bluesquare,

    Looking at wikipedia (http://en.wikipedia.org/wiki/File:Shells.png), would the shell model not lead to a spin of 5/2?
    However, I remember that deformation of the nuclei can re-order the level.
    Could that not be an explanation why the sheel model does not lead to the experimental spin?

  4. Dec 19, 2013 #3
    Yeah I know that we have deformities but I am unable to find it anywhere. Can you site any reference. Meanwhile i tried to calculate the spin of Ti -47 and from the shell model i got the spin as 1f7/2.. But when i checked in it is actually 5/2..So i am like ....

    thank you for the reply though.
  5. Dec 19, 2013 #4


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    "For 23Na, the observed spin must be explained by coupling of equivalent nucleons or by deformation." -- Burcham
  6. Dec 19, 2013 #5

    Vanadium 50

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    The approach you are using predicts that the only allowed nuclear spins are 0, 1/2 and 1 - you allow exactly one unpaired nucleon per shell. This is not borne out by data, so the shell model must be oversimplified. The oversimplification is that the shells in fact do depend on the other nuclei, and that moves the energy levels around, so you have the shell order change slightly.
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