Working out magnetic moment and electric quadrupole moment

  • Thread starter Flucky
  • Start date
  • #1
Flucky
95
1
Could somebody check if I have done this correctly please?

1. Homework Statement

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

Homework Equations


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

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>
 

Answers and Replies

  • #2
Flucky
95
1
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.
 
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