According to Bohr’s theory, the ratio of the kinetic energy to the total energy of the electron in a hydrogen atom at ground state is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1:-1\)
In Bohr’s model, the electrostatic attraction provides the centripetal force, which makes the potential energy twice the kinetic energy in size and negative.
\( \displaystyle \frac{mv^{2}}{r} = \frac{ke^{2}}{r^{2}} \;\Rightarrow\; K = \frac{1}{2}mv^{2} = \frac{ke^{2}}{2r} \)
\( \displaystyle U = -\frac{ke^{2}}{r} = -2K \)
\( \displaystyle E = K + U = -K \)
\( \displaystyle \frac{K}{E} = \frac{K}{-K} = 1 : -1 \)
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
\( \displaystyle E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r} \qquad (12.4) \)”