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According to Bohr's theory, the moment of momentum of an electron revolving in $4^{\text {th }}$ orbit of hydrogen atom is:
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Verified Answer
The correct answer is:
$2 \frac{h}{\pi}$
Moment of momentum is $\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{P}}$
$\begin{aligned}
& \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \mathrm{m} \overrightarrow{\mathrm{v}} \\
& \mathrm{L}=\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}=\frac{4 \mathrm{~h}}{2 \pi}=\frac{2 \mathrm{~h}}{\pi}
\end{aligned}$
$\begin{aligned}
& \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \mathrm{m} \overrightarrow{\mathrm{v}} \\
& \mathrm{L}=\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}=\frac{4 \mathrm{~h}}{2 \pi}=\frac{2 \mathrm{~h}}{\pi}
\end{aligned}$
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