In a hydrogen atom, the electron revolving in a circular orbit of radius \(r\) has kinetic energy \(K\). In an orbit of radius \(4r\), its kinetic energy is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(K/4\)
The nucleus attracts the electron. This attraction provides the centripetal force that keeps the electron in its orbit. It gives the kinetic energy as \( K=e^{2}/8\pi\varepsilon_0r \), so \( K \) is inversely proportional to \( r \).
\( K\propto \frac{1}{r} \)
\( \begin{aligned} \frac{K^{\prime}}{K} &= \frac{r}{4r} \\ K^{\prime} &= \frac{K}{4} \end{aligned} \)
NCERT (2026-27):
Electron orbits – “The kinetic energy (K ) and electrostatic potential energy (U) of the electron in hydrogen atom are”
\( \displaystyle K=\frac{1}{2}mv^{2}=\frac{e^{2}}{8\pi\varepsilon_0r} \)\( \quad \text{and} \quad \displaystyle U=-\frac{e^{2}}{4\pi\varepsilon_0r} \)