The frequency of a photon of energy 2.3 eV emitted due to transition of electron from higher energy level to lower energy level is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 555\times10^{12}\,\mathrm{Hz} \)
The energy of the emitted photon equals the difference between the two energy levels, and it is related to its frequency by Planck's relation \( E = h\nu \).
\( E = 2.3\,\mathrm{eV} = 2.3 \times 1.6\times10^{-19}\,\mathrm{J} = 3.68\times10^{-19}\,\mathrm{J} \)
\( h = 6.63\times10^{-34}\,\mathrm{J\,s} \)
\( \displaystyle \nu = \frac{E}{h} = \frac{3.68\times10^{-19}}{6.63\times10^{-34}}\,\mathrm{Hz} = 5.55\times10^{14}\,\mathrm{Hz} \)
\( \nu = 555\times10^{12}\,\mathrm{Hz} \)
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):
Exercises – …
