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Working out magnetic moment and electric quadrupole moment

  1. May 14, 2015 #1
    Could somebody check if I have done this correctly please?

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

    Draw the shell filling for oxygen isotopes and make predictions of their nuclear spin, parity, magnetic moment and electric quadrupole moment.

    2. Relevant equations
    Magnetic moment μ = gj j μN
    Electric quadrupole moment Q ≈ -<r2>[itex]\frac{2j - 1}{2(j + 1)}[/itex]

    3. The attempt at a solution
    So starting off with the isotopes of oxygen: 16O, 17O, 18O


    16O
    This fills the 1p[itex]\frac{1}{2}[/itex] shell.

    spin = 0 because it is even-even
    parity = 1 because even-even
    μ = 0 because even-even
    Q = 0 because it is a closed shell


    18O
    This corresponds to the 1d[itex]\frac{5}{2}[/itex] shell.

    Same values as above except for the electric quadrupole moment:

    Q ≈ -<r2>[itex]\frac{2(0) - 1}{2((0) + 1)}[/itex] = -<r2>[itex]\frac{-1}{2}[/itex] = [itex]\frac{1}{2}[/itex]<r2>


    17O
    This corresponds to a valence neutron in the 1d[itex]\frac{5}{2}[/itex] shell.

    spin = j = [itex]\frac{5}{2}[/itex]
    parity = (-1)l = (-1)2 = 1

    μ = [itex]\frac{5}{2}[/itex] gj μN (I won't work out gj or put the value of μN in)

    Q ≈ -<r2>[itex]\frac{2(\frac{5}{2}) - 1}{2((\frac{5}{2}) + 1)}[/itex] = -[itex]\frac{4}{7}[/itex]<r2>
     
  2. jcsd
  3. May 15, 2015 #2
    I think I may be wrong about the electric quadrupole moments, it's possible that only protons contribute to Q. So because the only thing that changes with oxygen isotopes is neutron number Q will be 0 for all of them?

    EDIT: scratch that neutrons do actually contribute as they attract the protons slightly.
     
    Last edited: May 15, 2015
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