The kinetic energy of an electron in a given orbit in a hydrogen atom…

The correct answer is Option (3) → \( \displaystyle\frac{e^{2}}{8\pi\epsilon_{o}r} \)

In a circular orbit the electrostatic attraction provides the centripetal force on the electron.

    \( \displaystyle \frac{e^{2}}{4\pi\epsilon_{o}r^{2}} = \frac{mv^{2}}{r} \;\Rightarrow\; mv^{2} = \frac{e^{2}}{4\pi\epsilon_{o}r} \)

The kinetic energy is half of \( mv^{2} \):

    \( \displaystyle K = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\epsilon_{o}r} \)

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27):
Electron Orbits – “… The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are

    \( \displaystyle K = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad \text{and} \quad U = -\frac{e^2}{4\pi\varepsilon_0 r} \)”