The kinetic energy of an electron in ground level in hydrogen atom is…

The correct answer is Option (1) → - 2K, - K

In the Bohr model the Coulomb attraction provides the centripetal force, which fixes the kinetic energy in terms of the potential energy.

    \( \displaystyle \frac{mv^{2}}{r} = \frac{e^{2}}{4\pi\varepsilon_0 r^{2}} \Rightarrow K = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\varepsilon_0 r} \)

    \( \displaystyle U = -\frac{e^{2}}{4\pi\varepsilon_0 r} = -2K \)

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

So the potential energy is −2K and the total energy is −K.

Note: The same question was also asked in CUET (UG) 2024 on 16-05-2024 (Slot 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.”