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Magnetic moment due to the motion of the electron in $n$th energy state of hydrogen atom is proportional to ..........
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The correct answer is:
n
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).
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).
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