E_1 and E_2 represents potential and kinetic energies of the orbital…

The correct answer is Option (1) → \( \displaystyle E_2 = \frac{-E_1}{2} \)

In Bohr's model the electrostatic attraction supplies the centripetal force for the circular orbit, which fixes the kinetic energy in terms of the radius.

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

    \( \displaystyle E_2 = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\varepsilon_0 r},\qquad E_1 = -\frac{e^{2}}{4\pi\varepsilon_0 r} \)

    \( \displaystyle E_2 = -\frac{E_1}{2} \)

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

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