In a hydrogen atom, the electron revolves in an orbit of radius \(4.29\times10^{-9}\ \mathrm{m}\). The electrostatic potential energy of the electron in this orbit is about (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 3
The correct answer is Option (3) → \(-0.34\ \mathrm{eV}\)
The potential energy of the electron at a distance \( r \) from the proton is \( U=-e^{2}/4\pi\varepsilon_0r \). The minus sign comes from the attraction between the two charges.
\( \begin{aligned} U &= -\frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r} \\ &= -\frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{4.29\times10^{-9}} \\ &= -5.37\times10^{-20}\ \mathrm{J} \end{aligned} \)
Divide by \( 1.6\times10^{-19} \) to change joules into electron volts.
\( \begin{aligned} U &= -\frac{5.37\times10^{-20}}{1.6\times10^{-19}}\ \mathrm{eV} \\ &= -0.34\ \mathrm{eV} \end{aligned} \)
From the above, \( U=-0.34\ \mathrm{eV} \). Hence, the potential energy of the electron in this orbit is about \( -0.34\ \mathrm{eV} \).
NCERT (2026-27):
Electron orbits – “The kinetic energy (K ) and electrostatic potential energy (U) of the electron in hydrogen atom are”
\( \displaystyle K=\frac{1}{2}mv^{2}=\frac{e^{2}}{8\pi\varepsilon_0r} \quad \text{and} \quad \)\( \displaystyle U=-\frac{e^{2}}{4\pi\varepsilon_0r} \)