In the ground state of a hydrogen atom, the electron moves in an orbit of radius \(5.3\times10^{-11}\ \mathrm{m}\) with a speed of \(2.2\times10^{6}\ \mathrm{m\,s^{-1}}\). The time taken by the electron to complete one revolution is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.5\times10^{-16}\ \mathrm{s}\)
In one revolution the electron covers the circumference \( 2\pi r \) of its orbit. Its speed stays the same, so the time period is this distance divided by the speed.
\( \begin{aligned} T &= \frac{2\pi r}{v} \\ &= \frac{2\times3.14\times5.3\times10^{-11}}{2.2\times10^{6}} \\ &= 1.5\times10^{-16}\ \mathrm{s} \end{aligned} \)
This period is the reciprocal of the frequency of revolution \( \nu=v/2\pi r \), which is about \( 6.6\times10^{15}\ \mathrm{Hz} \).
NCERT (2026-27):
Example 12.4 – “… Thus, the frequency of the electron moving around the proton is”
\( \displaystyle \nu=\frac{v}{2\pi r} \)\( \displaystyle =\frac{2.2\times10^{6}\ \mathrm{m\,s^{-1}}}{2\pi(5.3\times10^{-11}\ \mathrm{m})}\approx6.6\times10^{15}\ \mathrm{Hz} \)