In a hydrogen atom, the electron revolves around the proton in a circular orbit of radius \(5.3\times10^{-11}\ \mathrm{m}\). The magnitude of the electrostatic force between the electron and the proton is (Take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\) and \(e=1.6\times10^{-19}\ \mathrm{C}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(8.2\times10^{-8}\ \mathrm{N}\)
The electron and the proton carry charges of the same size \( e \). They attract each other by Coulomb’s law. This attraction gives the centripetal force that keeps the electron in its orbit.
\( \begin{aligned} F_e &= \frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}} \\ &= \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{(5.3\times10^{-11})^{2}} \\ &= \frac{2.30\times10^{-28}}{2.81\times10^{-21}} \\ &= 8.2\times10^{-8}\ \mathrm{N} \end{aligned} \)
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”
\( F_e=F_c \)
\( \displaystyle \frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}} \)\( \displaystyle =\frac{mv^{2}}{r} \qquad (12.2) \)