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Predicting spin and parity of excited states from shell mode

  1. May 14, 2016 #1
    1. The problem statement, all variables and given/known data


    Consider the following example from a previous exam. We are to predict the spin and parity for F(A=17,Z=9), Florine, in the ground state and the first two excited states using the shell model.

    Ground state:
    Neutrons: (1s 1/2)^2 (1p 3/2)^4 (1p 1/2)^2

    Protons: (1s 1/2)^2 (1p 3/2)^4 (1p 1/2)^2 (1d 5/2)^1

    Thus J^P = (5/2)^+

    First excited state: Promote one proton from the (1p 1/2) level
    Neutrons: (1s 1/2)^2 (1p 3/2)^4 (1p 1/2)^2

    Protons: (1s 1/2)^2 (1p 3/2)^4 (1p 1/2)^1 (1d 5/2)^2

    Does the spin and parity come from (a) the hole created or (b) the new filled level. If (a) then J^P = (1/2)^-. If (b) then J^P = (0)^+.



    Second excited state: Promote one neutron from the (1p 1/2) level
    Neutrons: (1s 1/2)^2 (1p 3/2)^4 (1p 1/2)^1 (1d 5/2)^1

    Protons: (1s 1/2)^2 (1p 3/2)^4 (1p 1/2)^2 (1d 5/2)^1

    Does the spin and parity come from (a) the hole created or (b) the new filled level. If (a) then the possible J values range from |1/2 - 5/2| to (1/2 + 5/2) in steps of one, so J=2 or 3. The parity multiplies so we have a minus from the l=1 neutron and a positive from the l=2 proton thus parity is minus and J^P = (2)^- or (3)^-. If (b) then J ranges from |5/2 - 5/2| to (5/2 + 5/2) in steps of one so J = 0,1,2,3,4 or 5. The parities both come from the same l level (d) so is positive. Thus J^P = (0)^+, (1)^+, (2)^+, (3)^+, (4)^+ or (5)^+.

    So how do we get the spin's and parities from the excited states? Is it determined by the hole created or the new excited level?

    Also how do I know which would be the first excited level and which would be the second? And is there a way to figure out which levels get filled up first when calculating the ground state, or should I just remember the order of the levels?

    We are not expected to know the Nordheim rules.
     
  2. jcsd
  3. May 19, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. May 21, 2016 #3
    I have come to the conclusion that any unpaired nucleons will contribute to the spin/parity, not just the ones from the hole created or the new level filled. And if it so happens that an excited state yields three or more unpaired nucleons, then the shell model is no longer reliable in predicting the spin and parity of the nucleus.
     
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