Magnetic moment due to the motion of the electron in $n$th energy state of hydrogen atom is proportional to …

1. $n^{-2}$
2. n
3. $n^2$
4. $n^3$

Solution

Magnetic moment of a moving electron, which shown in the figure below, Current, $i=\frac{e}{T}=\frac{e \omega}{2 \pi}\left(\because \omega=\frac{2 \pi}{T}\right)$ As, magnetic moment, $ M=i A=\frac{e \omega}{2 \pi} \pi R^2 $
$\Rightarrow \quad M=\frac{e v}{2 R} R^2 \quad\left(\because \omega=\frac{v}{R}\right)$ $\Rightarrow \quad M=\frac{e v R}{2}=\frac{e L}{2 m}$ where, $\quad L=m v r$ and $L=\frac{n h}{2 \pi}$ So, $\quad M=\frac{e n h}{2 \pi m} \Rightarrow M \propto n$ $\because e, h, m$ are constant of electron, Hence, the correct option is (b)

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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