The work function of a substance is 2.21 eV. Its threshold frequency will be approximately:
(Given, h = 6.63 x 10–34 Js).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5.33 x 1014 Hz
The threshold frequency is the frequency at which the photon energy just equals the work function.
\( \displaystyle \phi_0 = h\nu_0 \)
Convert the work function from eV to joules.
\( \displaystyle \phi_0 = 2.21 \times 1.6\times10^{-19}\,\mathrm{J} \)
\( \displaystyle \phi_0 = 3.536\times10^{-19}\,\mathrm{J} \)
\( \displaystyle \nu_0 = \frac{\phi_0}{h} = \frac{3.536\times10^{-19}\,\mathrm{J}}{6.63\times10^{-34}\,\mathrm{J\,s}} \)
\( \displaystyle \nu_0 \approx 5.33\times10^{14}\,\mathrm{Hz} \)
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 – …
