An energy of \( 13.6 \) eV is required to separate a hydrogen atom into a proton and electron. The orbital radius of the electron in hydrogen atom is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5.3\times10^{-11} \) m
In the Bohr model, the energy needed to separate the electron from the proton (the ionisation energy) equals the magnitude of the total energy of the electron in its orbit.
\( \displaystyle E = -\frac{1}{4\pi\varepsilon_{0}}\,\frac{e^{2}}{2r} \)
\( \displaystyle \frac{1}{4\pi\varepsilon_{0}}\,\frac{e^{2}}{2r} = 13.6\ \mathrm{eV} \quad\Rightarrow\quad r = \frac{1}{4\pi\varepsilon_{0}}\,\frac{e^{2}}{2 \times 13.6\ \mathrm{eV}} \)
Converting the energy to joules:
\( 13.6\ \mathrm{eV} = 13.6 \times 1.6\times10^{-19}\ \mathrm{J} \)
\( \displaystyle r = \frac{(9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}) \times (1.6\times10^{-19}\ \mathrm{C})^{2}}{2 \times 13.6 \times 1.6\times10^{-19}\ \mathrm{J}} \)
\( \displaystyle r = \frac{9\times10^{9} \times 1.6\times10^{-19}}{27.2}\ \mathrm{m} \approx 5.3\times10^{-11}\ \mathrm{m} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Atomic Spectra – …
