In a hydrogen atom, the electron revolves in a circular orbit of radius \(5.3\times10^{-11}\ \mathrm{m}\) with a speed of \(2.2\times10^{6}\ \mathrm{m\,s^{-1}}\). The centripetal acceleration of the electron is about
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(9.1\times10^{22}\ \mathrm{m\,s^{-2}}\)
An electron moving in a circle is always accelerated towards the centre. The attraction of the nucleus supplies the centripetal force \( mv^{2}/r \). So the centripetal acceleration is \( a=v^{2}/r \).
\( \begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(2.2\times10^{6})^{2}}{5.3\times10^{-11}} \\ &= 9.1\times10^{22}\ \mathrm{m\,s^{-2}} \end{aligned} \)
Hence, the centripetal acceleration of the electron is about \( 9.1\times10^{22}\ \mathrm{m\,s^{-2}} \).
NCERT (2026-27):
Electron orbits – “… The electrostatic force of attraction, Fe between the revolving electrons and the nucleus provides the requisite centripetal force (Fc) to keep them in their orbits. Thus, for a dynamically stable orbit in a hydrogen atom Fe = Fc”
\( \displaystyle \frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}}= \)\( \displaystyle \frac{mv^{2}}{r} \)\( \qquad (12.2) \)