The work function of cesium(Cs) surface is \( 2.14\ \mathrm{eV} \). The wavelength of light incident on the surface, if the photocurrent is brought to zero by a stopping potential of \( 0.50\ \mathrm{V} \) is given by:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sim 471\ \mathrm{nm} \)
By Einstein’s photoelectric equation, the photon energy equals the work function plus the maximum kinetic energy, and the stopping potential measures that kinetic energy: \( K_{\max} = eV_{0} \).
\( \phi_{0} = 2.14\,\mathrm{eV},\ eV_{0} = 0.50\,\mathrm{eV} \)
\( \displaystyle \frac{hc}{\lambda} = \phi_{0}+eV_{0} = 2.14\,\mathrm{eV}+0.50\,\mathrm{eV} = 2.64\,\mathrm{eV} \)
\( \displaystyle \lambda = \frac{hc}{2.64\,\mathrm{eV}} = \frac{(6.63\times10^{-34}\,\mathrm{J\,s})(3\times10^{8}\,\mathrm{m\,s^{-1}})}{2.64\times1.6\times10^{-19}\,\mathrm{J}} \)
\( \lambda \approx 4.71\times10^{-7}\,\mathrm{m} \approx 471\,\mathrm{nm} \)
Shortcut: With \( hc \approx 1240 \) eV nm, \( \lambda \approx 1240/2.64 \approx 470 \) nm, which picks out \( \sim471 \) nm at once.
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 – …
