The ratio of total energy to the kinetic energy of an electron in a B…

The correct answer is Option (3) → -1 : 1

In a Bohr orbit the Coulomb attraction provides the centripetal force:

    \( \displaystyle \frac{mv^{2}}{r} = \frac{ke^{2}}{r^{2}} \;\Rightarrow\; K = \frac{1}{2}mv^{2} = \frac{ke^{2}}{2r} \)

The potential energy is \( U = -\frac{ke^{2}}{r} = -2K \), so the total energy is

    \( \displaystyle E = K + U = K - 2K = -K \)

    \( \displaystyle E : K = -1 : 1 \)

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

    \( \displaystyle K = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\varepsilon_{0} r} \quad \text{and} \quad U = -\frac{e^{2}}{4\pi\varepsilon_{0} r} \)

(The negative sign in U signifies that the electrostatic force is in the –r direction.) Thus the total energy E of the electron in a hydrogen atom is

    \( \displaystyle E = K + U = \frac{e^{2}}{8\pi\varepsilon_{0} r} - \frac{e^{2}}{4\pi\varepsilon_{0} r} = -\frac{e^{2}}{8\pi\varepsilon_{0} r} \qquad (12.4) \)

The total energy of the electron is negative.”