If E_p and E_k represents the potential and kinetic energy respective…

The correct answer is Option (1) → \( \displaystyle E_k=-\frac{E_p}{2} \)

In a circular orbit the electrostatic attraction between the nucleus and the electron supplies the centripetal force. This gives the kinetic energy, and it is then compared with the electrostatic potential energy.

    \( \displaystyle \frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}} = \frac{mv^{2}}{r} \Rightarrow mv^{2} = \frac{e^{2}}{4\pi\varepsilon_0 r} \)

    \( \displaystyle E_k = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\varepsilon_0 r},\qquad E_p = -\frac{e^{2}}{4\pi\varepsilon_0 r} \)

Comparing the two, \( E_p = -2E_k \), that is

    \( \displaystyle E_k = -\frac{E_p}{2} \)

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

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