The total energy of the electron in a hydrogen atom is \(-3.4\ \mathrm{eV}\) when it revolves in an orbit of radius \(r\). The energy needed to shift the electron to an orbit of radius \(9r\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3.02 eV
The total energy of the electron in an orbit of radius \( r \) is \( E=-e^{2}/8\pi\varepsilon_0r \). So the energy is inversely proportional to \( r \). A larger orbit has a less negative energy. Energy must therefore be supplied to move the electron outward.
\( \begin{aligned} E^{\prime} &= \frac{E}{9} \\ &= \frac{-3.4}{9} \\ &= -0.38\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} \Delta E &= E^{\prime}-E \\ &= -0.38-(-3.4) \\ &= 3.02\ \mathrm{eV} \end{aligned} \)
Therefore, 3.02 eV must be given to the electron to move it to the orbit of radius \( 9r \).
NCERT (2026-27):
Electron orbits – “… Thus the total energy E of the electron in a hydrogen atom is”
\( \displaystyle E=K+U=\frac{e^{2}}{8\pi\varepsilon_0r}-\frac{e^{2}}{4\pi\varepsilon_0r} \)
\( \displaystyle E=-\frac{e^{2}}{8\pi\varepsilon_0r} \)\( \qquad (12.4) \)