If m is the mass of the electron, v its speed and r the radius of a stationary circular orbit of the electron in the hydrogen atom, then the kinetic energy \( K=\frac{1}{2}mv^{2} \) of the electron is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{e^{2}}{8\pi\varepsilon_0 r} \)
In a stationary circular orbit the Coulomb attraction of the nucleus provides the centripetal force on the electron:
\( \displaystyle \frac{mv^{2}}{r}=\frac{1}{4\pi\varepsilon_0}\,\frac{e^{2}}{r^{2}} \)
Multiplying both sides by \( r \):
\( \displaystyle mv^{2}=\frac{e^{2}}{4\pi\varepsilon_0 r} \)
The kinetic energy is half of this:
\( \begin{aligned} K &= \frac{1}{2}mv^{2} \\ &= \frac{e^{2}}{8\pi\varepsilon_0 r} \end{aligned} \)
NCERT (2026-27):
Electron Orbits – “The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are
\( \displaystyle K=\frac{1}{2}mv^{2}=\frac{e^{2}}{8\pi\varepsilon_0 r} \)\( \displaystyle \quad\text{and}\quad U=-\frac{e^{2}}{4\pi\varepsilon_0 r} \)”