Light of one fixed frequency falls in turn on clean caesium, sodium and copper surfaces. Their work functions are 2.14 eV, 2.75 eV and 4.65 eV. Photoelectrons come out of all three. Which of the following gives the correct order of the stopping potentials \(V_{\mathrm{Cs}}\), \(V_{\mathrm{Na}}\) and \(V_{\mathrm{Cu}}\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(V_{\mathrm{Cs}} \gt V_{\mathrm{Na}} \gt V_{\mathrm{Cu}}\)
For light of frequency \( \nu \), Einstein’s equation gives \( eV_0=h\nu-\phi_0 \). Here \( h\nu \) is the same for all three surfaces. So the metal with the smaller work function gives the larger stopping potential.
\( \displaystyle V_0=\frac{h\nu-\phi_0}{e} \)
The work functions are in the order \( \phi_{\mathrm{Cs}}\ (2.14\ \mathrm{eV}) \lt \phi_{\mathrm{Na}}\ (2.75\ \mathrm{eV}) \lt \phi_{\mathrm{Cu}}\ (4.65\ \mathrm{eV}) \).
Hence, \( V_{\mathrm{Cs}} \gt V_{\mathrm{Na}} \gt V_{\mathrm{Cu}} \).
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… the photoelectric equation … can be written as”
\( eV_0=h\nu-\phi_0;\ \text{for } \nu\ge\nu_0 \)\( \qquad (11.4) \)