Question

The spin-only magnetic moment of a free ion is $\sqrt{8}\text{B.M}$. The spin angluar momentum of the electron is $x\frac{h}{2\pi }$. Find the value of $x^2$.

Answer 1

Spin only magnetic moment can be given as:
$\sqrt{n(n+2)}=\sqrt{8}B.M.$
Where, $n$ is the number of unpaired electrons
On solving,
$n=2$
$\therefore S=\frac{1}{2}+\frac{1}{2}=1$
Spin angular moment of an electron can be calculated by using the formula,
$L_s=\sqrt{S(S+1)}\frac{h}{2\pi }=\sqrt{2}\frac{h}{2\pi }$
Hence, $x=\sqrt{2}$ and $x^2=2$