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Nuclear spin of Fluorine 19

  1. Feb 4, 2015 #1
    I was trying to calculate the ##^{19}_{9}F_{10}## nuclear spin using the nuclear shell method. From what i know the nuclear spin for a nucleous with odd ##A## is the total angular momentum ##J## of the stand-alone nucleon:
    P: ##(1d_{5 \backslash 2})^1## (stand-alone nucleon)
    N: ##(1d_{5 \backslash 2})^2##
    which in this case is a proton with ##J=\frac{5}{2}## so it leads to a nuclear spin of ##I=\frac{5}{2}##. But this isn't what i found in lecture, from the pdg (http://ie.lbl.gov/toi/nuclide.asp?iZA=90019) ##^{19}_{9}F_{10}## result to have ##I=\frac{1}{2}##.
    I missed something?
     
  2. jcsd
  3. Feb 4, 2015 #2

    ChrisVer

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    Your filling is incorrect.
    You have 9 protons and 10 neutrons. Well no need to care about the neutrons, since they will pile up into pairs together (even number). From the protons you will have 1 extra remaining unpaired. You will fill until [itex]1d_{5/2} [itex]. So your nucleus spin will be determined by the unpaired proton, the proton has spin 1/2 so 1/2 is the spin of your nucleus...
    Sometimes things are mixed with J, S in nuclear physics, calling both of them "spins". The total angular momentum is indeed 5/2, the spin is 1/2.
     
    Last edited: Feb 4, 2015
  4. Feb 4, 2015 #3
    My apologies but i'm not understanfding what you mean, for example take the ##^{43}_{21}Sc_{22}##. It have 22 neutron (so we don't need to care about), and 21 proton which are filled until ##1f_{7 \backslash 2}##. There's only one proton unpaired so nuclear spin ##I## should be ##\frac{1}{2}## (equal to the spin of the proton) but is ##I=\frac{7}{2}## (always from pdg http://ie.lbl.gov/toi/nuclide.asp?iZA=210043)
     
  5. Feb 4, 2015 #4

    Vanadium 50

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    I don't understand what either of you is saying. 19F can be modeled as an 18O plus a proton. 18O is 0+, so 19F is 1/2+.
     
  6. Sep 20, 2015 #5
    have you failed to recognize the difference between net spin and total magnetic moment?
     
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