The ground state energy of the electron in the hydrogen atom is −13.6…

The correct answer is Option (3) → \( 2.2\times10^{6}\,\mathrm{m/s} \)

For the electron in a hydrogen orbit, the kinetic energy is \( K=e^{2}/8\pi\varepsilon_0r \) and the total energy is \( E=-e^{2}/8\pi\varepsilon_0r \). So \( K=-E \), and in the ground state the kinetic energy is 13.6 eV.

    \( \begin{aligned} K &= 13.6\,\mathrm{eV} \\ &= 13.6\times1.6\times10^{-19}\,\mathrm{J} \\ &\approx 2.18\times10^{-18}\,\mathrm{J} \end{aligned} \)

From \( K=\frac{1}{2}m_ev^{2} \), the speed is:

    \( \begin{aligned} v &= \sqrt{\frac{2K}{m_e}} \\ &= \sqrt{\frac{2\times2.18\times10^{-18}\,\mathrm{J}}{9.1\times10^{-31}\,\mathrm{kg}}} \\ &= \sqrt{4.8\times10^{12}\,\mathrm{m^{2}/s^{2}}} \\ &\approx 2.2\times10^{6}\,\mathrm{m/s} \end{aligned} \)

NCERT (2026-27):
Electron Orbits – “The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are

    \( \displaystyle K=\frac{1}{2}mv^{2}=\frac{e^{2}}{8\pi\varepsilon_0 r} \)\( \displaystyle \quad\text{and}\quad U=-\frac{e^{2}}{4\pi\varepsilon_0 r} \)

(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

    \( \displaystyle E=K+U=\frac{e^{2}}{8\pi\varepsilon_0 r}-\frac{e^{2}}{4\pi\varepsilon_0 r}= \)\( \displaystyle -\frac{e^{2}}{8\pi\varepsilon_0 r} \)\( \qquad (12.4) \)”