When an electron of hydrogen atom is raised from the ground state to an excited state.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The potential energy increases and kinetic energy decreases
In Bohr’s model the electron moves in a circle under the Coulomb attraction. The kinetic and potential energies depend on the orbit radius r, and an excited state has a larger r.
\( \displaystyle K = \frac{e^{2}}{8\pi\varepsilon_0 r} \), \( \displaystyle U = -\frac{e^{2}}{4\pi\varepsilon_0 r} \)
As r increases, K decreases. U is negative, and a larger r makes it less negative, so U increases. The total energy \( E = K + U = -\frac{e^{2}}{8\pi\varepsilon_0 r} \) also rises towards zero, as it must when the electron is raised to a higher level.
NCERT (2026-27):
Electron Orbits – “… The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are
\( \displaystyle K = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad \text{and} \quad U = -\frac{e^2}{4\pi\varepsilon_0 r} \)”