The kinetic energy of the electron in a certain orbit of a hydrogen atom is 3.4 eV. What is its electrostatic potential energy in the next higher orbit? (Take \(E_n=-13.6/n^{2}\ \mathrm{eV}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-3.02\ \mathrm{eV}\)
In any orbit of hydrogen, the kinetic energy is \( K=-E \) and the potential energy is \( U=2E \), where \( E \) is the total energy. Both follow from \( K=e^{2}/8\pi\varepsilon_0r \) and \( U=-e^{2}/4\pi\varepsilon_0r \). First find the orbit from the given kinetic energy.
\( \begin{aligned} E &= -K \\ &= -3.4\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} n^{2} &= \frac{13.6}{3.4} \\ &= 4 \end{aligned} \)
So the electron is in the \( n=2 \) orbit. The next higher orbit is \( n=3 \).
\( \begin{aligned} E_3 &= -\frac{13.6}{3^{2}} \\ &= -1.51\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} U_3 &= 2E_3 \\ &= 2\times(-1.51) \\ &= -3.02\ \mathrm{eV} \end{aligned} \)
NCERT (2026-27):
Electron orbits – “The kinetic energy (K ) and electrostatic potential energy (U) of the electron in hydrogen atom are”
\( \displaystyle K=\frac{1}{2}mv^{2}=\frac{e^{2}}{8\pi\varepsilon_0r} \quad \text{and} \quad \)\( \displaystyle U=-\frac{e^{2}}{4\pi\varepsilon_0r} \)
Electron orbits – “… Thus the total energy E of the electron in a hydrogen atom is”
\( \displaystyle E=K+U=\frac{e^{2}}{8\pi\varepsilon_0r}-\frac{e^{2}}{4\pi\varepsilon_0r} \)
\( \displaystyle =-\frac{e^{2}}{8\pi\varepsilon_0r} \)\( \qquad (12.4) \)