Q 12-12-104JEE MainJEE Main 2020 (7 Jan, Shift 1)Easy
The time period of revolution of an electron in its ground state orbit in a hydrogen atom is $1.6\times10^{-16}$ s. The frequency of revolution of the electron in its first excited state (in s$^{-1}$) is:
Answer: (B) $7.8\times10^{14}$
In the Bohr model $T \propto n^3$, so for $n = 2$:
$$T_2 = 8\times1.6\times10^{-16} = 1.28\times10^{-15}\ \text{s}$$
$$f_2 = \frac{1}{T_2} \approx 7.8\times10^{14}\ \text{s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics