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The period of revolution of an electron revolving in nth orbit of H-atom is proportional to
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The correct answer is:
$n^{3}$
The time period of revolution of an electron revolving in $n$th orbit of $\mathrm{H}$-atom is given by
$\begin{array}{ll}
& T_{n}=\frac{2 \pi r_{n}}{v_{n}}=\left(\frac{4 \varepsilon_{0}^{2} h^{2}}{m e^{4}}\right) \frac{n^{3}}{z^{2}} \\
\Rightarrow \quad & T_{n} \propto n^{3}
\end{array}$
$\begin{array}{ll}
& T_{n}=\frac{2 \pi r_{n}}{v_{n}}=\left(\frac{4 \varepsilon_{0}^{2} h^{2}}{m e^{4}}\right) \frac{n^{3}}{z^{2}} \\
\Rightarrow \quad & T_{n} \propto n^{3}
\end{array}$
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