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Single particle shell model for nucleus

  1. Apr 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Predict the ground state spin and parity for [itex]^{7}_{3}Li[/itex], [itex]^{15}_{7}N[/itex]. [itex]^{31}_{15}P[/itex].

    2. Relevant equations

    Fill up shells for neutrons and protons separately.
    Shells in increasing energy are: [itex]1S_{1/2}[/itex], [itex]1P_{3/2}[/itex], [itex]1P_{1/2}[/itex], [itex]1S_{1/2}[/itex], [itex]1D_{5/2}[/itex], [itex]1D_{3/2}[/itex]
    and have occupancies: 2, 4, 2, 6, 4, 8, 6

    A missing proton is a hole with parity -1.
    An unpaired neutron has parity +1.

    3. The attempt at a solution

    [itex]^{7}_{3}Li[/itex]: We have odd protons so only interested in this. [itex]1S_{1/2}[/itex] fills up. Then only one proton out of four sits in [itex]1P_{3/2}[/itex]. It has j = 3/2 and parity -1.

    [itex]^{15}_{7}N[/itex]: We are interested in the 7 protons. We will be left with one hole in [itex]1P_{1/2}[/itex] so it has j = 1/2 and parity -1.

    [itex]^{31}_{15}P[/itex]: We are interested in the 15 protons. We fill up 2, 4, 2, 6 and then are left with 1 out of 4 in [itex]1D_{3/2}[/itex]. This unpaired proton should have spin 3/2 and parity -1 if I use my same method.
    However the answer given is that the unpaired proton sits in the [itex]2S_{1/2}[/itex] so has spin
    1/2 and parity +1. How am I to know that jumping up to 2S is more favourable than carrying on through the 1X's? Why is the parity now +1. Does it flip when you jump from 1X to 2X?
     
  2. jcsd
  3. Apr 3, 2014 #2

    BvU

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    Can you check your list ? there are 6 shells with ##1S_{1/2}## appearing twice (I take it the second should be ##2S_{1/2}## ?) and there are 7 occupancies, with unlikely values. I would expect something like 2,4,2,2,6,4 but that still brings you to ##1D_{5/2}## with P, so it looks as if the ##2S_{1/2}## sits between ##1D_{5/2}## and ##1D_{3/2}## , which isn't impossible. See http://jol.liljenzin.se/KAPITEL/CH11NY3.PDF 310-313
     
  4. Apr 3, 2014 #3
    Sorry for sloppy post, and also you are right about the ordering. Let me write it out correctly.

     
  5. Apr 3, 2014 #4
    [itex]^{31}_{15}P[/itex]: We are interested in the 15 protons. We fill up 2, 4, 2, 6 and then are left with 1 out of 2 in [itex]2S_{1/2}[/itex]. The hole should have spin 1/2 as given in solution.

    Regarding the parity, does it flip when we jump up an energy level. (Going from [itex]1S_{1/2}[/itex] to [itex]2S_{1/2}[/itex] is jumping up an energy level right? Increasing the quantum number n.)
     
  6. Apr 4, 2014 #5

    BvU

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    I don't really know much about parity and I don't understand the "a missing proton is a hole with parity -1".

    As far as I know parity depends on angular momentum L as (-1)L and the primary quantum number n doesn't appear. S states are L=0 states, so definitely even parity: there is no ##\theta,\, \phi## dependence.
     
  7. Apr 4, 2014 #6
    I found this on wikipedia http://en.wikipedia.org/wiki/Nuclear_shell_model#Other_properties_of_nuclei.

    "All protons in the same level (n) have the same parity (either +1 or −1), and since the parity of a pair of particles is the product of their parities, an even number of protons from the same level (n) will have +1 parity"

    Although it doesn't state it explicitly this suggests that the parity does switch as you move up n.
     
  8. Apr 5, 2014 #7

    BvU

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    I am not buying it. The whole S, P, D thing comes from splitting the r dependency in the wavefunction from the ##\theta, \phi## dependence. n has to do with the r (##|\vec r|##) dependence -- even parity, period. Proton itself has intrinsic party +1 (definition). Odd parity can come from l and m and it turns out p = (-1)l.

    Wikipedia isn't always right. I'm not always right either, so if someone could set this straight, I'd be grateful.
     
  9. Apr 5, 2014 #8
    Yes, the missing proton is a hole with parity -1 is the problem statement here I think. I read this out of context and it doesn't help.

    I've learned one should just think in terms of unpaired protons in terms of holes. So the parity is simply +1 as it is in an l = 0 shell and the unpaired proton has +1 parity.

    I get it now. Thanks!
     
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