The work function of a photoelectric surface is \( 3.3 \) eV. The threshold frequency for the surface is approximately
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8\times10^{14} \) Hz
The work function is the minimum energy needed to free an electron, so a photon at the threshold frequency carries exactly this energy: \( \phi_0 = h\nu_0 \).
\( \phi_0 = 3.3\,\mathrm{eV} = 3.3\times1.6\times10^{-19}\,\mathrm{J} = 5.28\times10^{-19}\,\mathrm{J} \)
\( \displaystyle \nu_0 = \frac{\phi_0}{h} \)
\( \displaystyle \nu_0 = \frac{5.28\times10^{-19}\,\mathrm{J}}{6.63\times10^{-34}\,\mathrm{J\,s}} \)
\( \displaystyle \nu_0 \approx 7.96\times10^{14}\,\mathrm{Hz} \approx 8\times10^{14}\,\mathrm{Hz} \)
Shortcut: A photon of 1 eV has a frequency of about \( 2.42\times10^{14} \) Hz, so \( \nu_0 \approx 3.3\times2.42\times10^{14}\,\mathrm{Hz} \approx 8\times10^{14} \) Hz.
Note: The same question was also asked in CUET (UG) 2023 on 13-06-2023 (Slot 1).
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 – …
