The kinetic energy of an electron in the ground state of hydrogen atom is 13.6 eV. The potential energy (P.E) and total energy (E) of electron in this state is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( P.E = -27.2\,\mathrm{eV},\ E = -13.6\,\mathrm{eV} \)
In the Bohr model the Coulomb attraction supplies the centripetal force, which gives a fixed link between the energies of the electron in any orbit.
\( \displaystyle \frac{m v^{2}}{r} = \frac{k e^{2}}{r^{2}} \;\Rightarrow\; K = \frac{1}{2}m v^{2} = \frac{k e^{2}}{2r} \)
\( \displaystyle U = -\frac{k e^{2}}{r} = -2K \)
\( \displaystyle E = K + U = -K \)
With \( K = 13.6\,\mathrm{eV} \):
\( U = -2 \times 13.6\,\mathrm{eV} = -27.2\,\mathrm{eV} \)
\( E = -13.6\,\mathrm{eV} \)
The total energy is negative because the electron is bound to the nucleus.
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. This implies the fact that the electron is bound to the nucleus. If E were positive, an electron will not follow a closed orbit around the nucleus.”