The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6 \times 10^{- 16}$ s. The frequency of revolution of the electron in its first excited state $(i n s^{- 1})$ is

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$7.8 \times 10^{14}$

$6.2 \times 10^{15}$

$5.6 \times 10^{12}$

$1.6 \times 10^{14}$

answer is A.

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Detailed Solution

$\begin{array}{l} W e k n o w T i m e P e r i o d, T \propto \frac{r}{v} \propto \frac{n^{2}}{z} \times \frac{n}{z} \propto \frac{n^{3}}{z^{2}} \\ \Rightarrow \frac{T_{1}}{T_{2}} = \frac{n_{1}^{3}}{n_{2}^{3}} = \frac{1}{8} \\ \Rightarrow T_{2} = 8 T_{1} = 8 \times 1.6 \times 10^{- 16} = 12.8 \times 10^{- 16} \\ N o w, f r e q u e n c y, f_{2} = \frac{1}{T_{2}} = \frac{1}{12.8 \times 10^{- 16}} \approx 7.8 \times 10^{14} \end{array}$