A difference of 2.21 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5.3 \times 10^{14}\,\mathrm{Hz} \)
By Bohr's frequency condition, the energy difference between the two levels is carried away by one photon, \( \Delta E = h\nu \).
\( \Delta E = 2.21\,\mathrm{eV} = 2.21 \times 1.6\times10^{-19}\,\mathrm{J} = 3.54\times10^{-19}\,\mathrm{J} \)
\( h = 6.63\times10^{-34}\,\mathrm{J\,s} \)
\( \displaystyle \nu = \frac{\Delta E}{h} = \frac{3.54\times10^{-19}}{6.63\times10^{-34}} = 5.3\times10^{14}\,\mathrm{Hz} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
