In a hydrogen atom the effect on potential energy of the electron with increase in radius of orbits will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → potential energy increases
The electrostatic potential energy of the electron (charge \( -e \)) at distance \( r \) from the proton (charge \( +e \)) is
\( \displaystyle U = -\frac{1}{4\pi\varepsilon_0}\,\frac{e^{2}}{r} \)
It is negative, because the force is attractive. As \( r \) increases, \( e^{2}/r \) becomes smaller, so \( U \) becomes less negative, that is, it increases towards zero. In the Bohr model \( U = -27.2/n^{2}\,\mathrm{eV} \): for example \( -27.2\,\mathrm{eV} \) for \( n = 1 \) and \( -6.8\,\mathrm{eV} \) for \( n = 2 \).
Option (1) potential energy remains the same — Incorrect. \( U \) depends on \( r \), so it changes from orbit to orbit.
Option (2) potential energy increases — Correct. A larger orbit gives a less negative, that is higher, potential energy; work must be done against the attraction to move the electron out.
Option (3) potential energy decreases — Incorrect. The magnitude \( |U| \) decreases, but the potential energy itself, being negative, increases.
Option (4) potential energy initially decreases then increases — Incorrect. \( U = -ke^{2}/r \) rises steadily with \( r \); it has no minimum at an intermediate radius.
NCERT (2026-27):
Electron Orbits – “… The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are
\( \displaystyle K = \frac{e^2}{8\pi\varepsilon_0 r} \quad \text{and} \quad U = -\frac{e^2}{4\pi\varepsilon_0 r} \)
(The negative sign in U signifies that the electrostatic force is in the –r direction.)”