The electrostatic potential energy of the electron in a certain orbit of a hydrogen atom is \(-6.80\ \mathrm{eV}\). The minimum energy needed to ionise the atom from this state is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(3.40\ \mathrm{eV}\)
In an orbit of radius \( r \), the kinetic energy is \( K=e^{2}/8\pi\varepsilon_0r \) and the potential energy is \( U=-e^{2}/4\pi\varepsilon_0r \). So \( K \) is half the size of \( U \), and the total energy \( E=K+U \) is half of \( U \). To ionise the atom, the total energy must be raised to zero.
\( \begin{aligned} E &= \frac{U}{2} \\ &= \frac{-6.80}{2} \\ &= -3.40\ \mathrm{eV} \end{aligned} \)
From the above, the electron is bound with an energy of 3.40 eV. Hence, this much energy must be supplied to free it.
NCERT (2026-27):
Electron orbits – “The kinetic energy (K ) and electrostatic potential energy (U) of the electron in hydrogen atom are”
\( \displaystyle K=\frac{1}{2}mv^{2}=\frac{e^{2}}{8\pi\varepsilon_0r} \quad \text{and} \quad \)\( \displaystyle U=-\frac{e^{2}}{4\pi\varepsilon_0r} \)
Electron orbits – “… Thus the total energy E of the electron in a hydrogen atom is”
\( \displaystyle E=K+U=\frac{e^{2}}{8\pi\varepsilon_0r}-\frac{e^{2}}{4\pi\varepsilon_0r} \)
\( \displaystyle =-\frac{e^{2}}{8\pi\varepsilon_0r} \)\( \qquad (12.4) \)