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Using Bohr's model, the orbital period of electron in hydrogen atom in $\mathrm{n}^{\text {th }}$ orbit is $\left(\epsilon_{0}=\right.$ permittivity of free space, $\mathrm{h}=$ Planck's constant, $\mathrm{m}=$ mass of electron,
$\mathrm{e}=$ electronic charge $)$
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$\mathrm{e}=$ electronic charge $)$
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Verified Answer
The correct answer is:
$\frac{4 \epsilon_{0}^{2} \mathrm{n}^{3} \mathrm{~h}^{3}}{\mathrm{me}^{4}}$
(B)
$\begin{array}{l}
\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}} ; \quad \mathrm{r}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2} \epsilon_{0}}{\pi \mathrm{h} \mathrm{e}^{2}} \\
\mathrm{~V}=\frac{\mathrm{e}^{2}}{2 \pi \epsilon_{\mathrm{o}} \mathrm{n}}
\end{array}$
putting the values of $\mathrm{V}$ and $\mathrm{r}$,
$T=\frac{4 \epsilon_{0}{ }^{2} \mathrm{n}^{3} \mathrm{~h}^{3}}{\mathrm{me}^{4}}$
$\begin{array}{l}
\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}} ; \quad \mathrm{r}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2} \epsilon_{0}}{\pi \mathrm{h} \mathrm{e}^{2}} \\
\mathrm{~V}=\frac{\mathrm{e}^{2}}{2 \pi \epsilon_{\mathrm{o}} \mathrm{n}}
\end{array}$
putting the values of $\mathrm{V}$ and $\mathrm{r}$,
$T=\frac{4 \epsilon_{0}{ }^{2} \mathrm{n}^{3} \mathrm{~h}^{3}}{\mathrm{me}^{4}}$
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