For a photosensitive surface, work function is \( 3.3\times10^{-19}\,\mathrm{J} \). Taking Planck's constant to be \( 6.6\times10^{-34}\,\mathrm{J\,s} \), the threshold frequency would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 5\times10^{14}\,\mathrm{Hz} \)
At the threshold frequency the photon energy just equals the work function, \( h\nu_{0} = \phi \).
\( \displaystyle \nu_{0} = \frac{\phi}{h} = \frac{3.3\times10^{-19}}{6.6\times10^{-34}} = 5\times10^{14}\,\mathrm{Hz} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Particle Nature of Light: The Photon – …
