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The magnetic moment of lanthanide ions is determined from which one of the following relation?
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The correct answer is:
$\mu=g \sqrt{J(J+1)}$
In case of lanthanoids, 4 f orbitals lie too deep and hence the magnetic effect of the motion of the electron in its orbital is not quenched out. Here spin contribution $\mathrm{S}$ and orbital contribution $\mathrm{L}$ couple together to give a new quantum number $\mathrm{J}$.
Thus magnetic moment of lanthanoids is given by, $\mu=g \sqrt{J(J+1)}$
where $\mathrm{J}=\mathrm{L}-\mathrm{S}$ when the shell is less than half fill $\mathrm{J}=\mathrm{L}+\mathrm{S}$ when the shell is more than half fill and
$\mathrm{g}=1 \frac{1}{2}+\frac{\mathrm{S}(\mathrm{S}+1)-\mathrm{L}(\mathrm{L}+1)}{2 \mathrm{~J}(\mathrm{~J}+1)}$
Thus magnetic moment of lanthanoids is given by, $\mu=g \sqrt{J(J+1)}$
where $\mathrm{J}=\mathrm{L}-\mathrm{S}$ when the shell is less than half fill $\mathrm{J}=\mathrm{L}+\mathrm{S}$ when the shell is more than half fill and
$\mathrm{g}=1 \frac{1}{2}+\frac{\mathrm{S}(\mathrm{S}+1)-\mathrm{L}(\mathrm{L}+1)}{2 \mathrm{~J}(\mathrm{~J}+1)}$
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