The electron in a hydrogen atom moves in a circular orbit of radius \(1.33\times10^{-9}\ \mathrm{m}\). The kinetic 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 1
The correct answer is Option (1) → 0.54 eV
The attraction of the nucleus keeps the electron in its orbit. In such an orbit the kinetic energy is \( K=e^{2}/8\pi\varepsilon_0r \). This is half of \( e^{2}/4\pi\varepsilon_0r \).
Given:
\( r=1.33\times10^{-9}\ \mathrm{m} \)
\( e=1.6\times10^{-19}\ \mathrm{C} \)
\( \begin{aligned} K &= \frac{1}{4\pi\varepsilon_0}\cdot\frac{e^{2}}{2r} \\ &= \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{2\times1.33\times10^{-9}} \\ &= 8.66\times10^{-20}\ \mathrm{J} \end{aligned} \)
Since \( 1\ \mathrm{eV}=1.6\times10^{-19}\ \mathrm{J} \),
\( \begin{aligned} K &= \frac{8.66\times10^{-20}}{1.6\times10^{-19}}\ \mathrm{eV} \\ &= 0.54\ \mathrm{eV} \end{aligned} \)
Therefore, the kinetic energy of the electron in this orbit is about 0.54 eV.
NCERT (2026-27):
Electron orbits – “The kinetic energy (K ) and electrostatic potential energy (U) of the electron in hydrogen atom are”
\( \displaystyle K=\frac{1}{2}mv^{2}=\frac{e^{2}}{8\pi\varepsilon_0r} \)\( \quad \text{and} \quad \displaystyle U=-\frac{e^{2}}{4\pi\varepsilon_0r} \)