The kinetic energy of an electron revolving in a stable orbit around…

The correct answer is Option (2) → -E

In the Bohr model the electrostatic attraction supplies the centripetal force, \( \frac{mv^{2}}{r} = \frac{e^{2}}{4\pi\varepsilon_0r^{2}} \). This gives the kinetic energy, and the potential energy of the electron in the field of the nucleus is known, so the total follows.

    \( \displaystyle K = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\varepsilon_0r} = E \)

    \( \displaystyle U = -\frac{e^{2}}{4\pi\varepsilon_0r} = -2E \)

    \( E_{\mathrm{total}} = K + U = E - 2E = -E \)

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

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

… Thus the total energy E of the electron in a hydrogen atom is

    \( \displaystyle E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r} \qquad (12.4) \)”