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Shell Model Spin-parity

  1. Feb 10, 2008 #1
    I am confused how to determine the spin / parity of excited states.

    In my textbook, one of the questions states:


    The ground state of the radioisotope 17-F-9 has spin-parity j_P = (5/2)+ and the first excited state has j_P=(1/2)-. Suggest two possible configurations for the latter state.


    Here is the answer in the back:

    The configuration of the ground state is:

    protons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^2(1d_\frac{5}{2})[/tex]

    To get j_P= (1/2)-, one could promote a p_1/2 proton to the d_5/2 shell giving

    protons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^{-1}(1d_\frac{5}{2})^2[/tex]

    Then by the pairing hypothesis, the two d_5/2 protons could give j_P = 0+ so that the total spin-parity would be determined by the unpaired p_1/2 neutron (j_P=(1/2)-).

    Alternatively, one of the p_3/2 protons could be promoted to the d_5/2 shell, giving

    protons: protons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^{-1}(1p_\frac{1}{2})^2(1d_\frac{5}{2})^2[/tex]

    and the two d_5/2 protons could combine to give j_P = 2+, so that when this combines with the single unpaired j_P = 3/2- proton, the overall spin is j_P = 1/2-


    So here are two things I am confused about:

    Firstly, how can the two d_5/2 protons combine to have j_P = 0+ in the first case and j_P = 2+ in the second case?

    Secondly, how is it that in the second case, the spin-parity ends up being j_P = 1/2-. Is it that the parities of the two are multiplied (ie the parity of the two d_5/2 protons is 1+ and the parity of the unpaired p_3/2 proton is 1-, giving an overall parity of 1-, and then the spin is 2 - 3/2 = 1/2? I don't really get how that works).

    If I can understand this I may be able to even get started on the homework.
  2. jcsd
  3. Feb 11, 2008 #2


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    you have already made this thread in Nuclear- and particle physics section. Double posting is not ok!!!
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