A and B are two metals with threshold frequencies \( 1.8\times10^{14}\,\mathrm{Hz} \) and \( 2.2\times10^{14}\,\mathrm{Hz} \) respectively. Two identical photons of energy 0.825 eV each are incident on them, one on each metal. Photoelectrons are emitted from (Take \( h=6.6\times10^{-34}\,\mathrm{J\,s} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A alone
A metal emits photoelectrons only if the frequency of the incident light is above its threshold frequency, whatever the intensity of the light. The frequency of each photon follows from \( E=h\nu \).
\( \begin{aligned} E &= 0.825\,\mathrm{eV} \\ &= 0.825\times1.6\times10^{-19}\,\mathrm{J} \\ &= 1.32\times10^{-19}\,\mathrm{J} \end{aligned} \)
\( \begin{aligned} \nu &= \frac{E}{h} \\ &= \frac{1.32\times10^{-19}\,\mathrm{J}}{6.6\times10^{-34}\,\mathrm{J\,s}} \\ &= 2.0\times10^{14}\,\mathrm{Hz} \end{aligned} \)
From the above, \( \nu \) is above the threshold frequency of A (\( 1.8\times10^{14}\,\mathrm{Hz} \)) and below that of B (\( 2.2\times10^{14}\,\mathrm{Hz} \)).
Option (1) B alone — Incorrect. The photon frequency \( 2.0\times10^{14}\,\mathrm{Hz} \) is below the threshold frequency of B, so B emits no photoelectrons.
Option (2) A alone — Correct. The photon frequency is above the threshold frequency of A. So each photon has more energy than the work function of A and can eject an electron from A.
Option (3) neither A nor B — Incorrect. The threshold frequency of A, \( 1.8\times10^{14}\,\mathrm{Hz} \), is lower than the photon frequency, so A does emit photoelectrons.
Option (4) both A and B — Incorrect. The threshold frequency of B, \( 2.2\times10^{14}\,\mathrm{Hz} \), is higher than the photon frequency, and below the threshold there is no emission at all.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Since Kmax must be non-negative, … photoelectric emission is possible only if hν > φ0 or ν > ν0, where”
\( \displaystyle \nu_0=\frac{\phi_0}{h} \qquad (11.3) \)
“… Thus, there exists a threshold frequency ν0 (= φ0/h) for the metal surface, below which no photoelectric emission is possible, no matter how intense the incident radiation may be or how long it falls on the surface.”