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# Magnetic moment of an electron in the $n^{th}$ orbit of the hydrogen atom is

$\begin {array} {1 1} (a)\;neh/ \pi \: m & \quad (b)\;neh/4 \pi \: m \\ (c)\;meh/ \pi \: n & \quad (d)\;meh/4 \pi \: n \end {array}$

$\large\frac{Magnetic\: moment}{Angular\: momentum}$ $= \large\frac{e}{2m}$
Angular Momentum = $\large\frac{nh}{2 \pi}$
Ans : (b)