Photons of energy 4.2 eV fall on the emitter of a photocell. The work function of the emitter is 2.2 eV. The collector is kept at a potential of \(-1.5\ \mathrm{V}\) with respect to the emitter. The maximum kinetic energy of the photoelectrons when they reach the collector is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.5 eV
The fastest electrons leave the emitter with \( K_{\max}=h\nu-\phi_0 \). The collector is negative with respect to the emitter, so the field between the plates pushes the electrons back. Each electron loses 1.5 eV of kinetic energy on its way across.
\( \begin{aligned} K_{\max} &= h\nu-\phi_0 \\ &= 4.2-2.2 \\ &= 2.0\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} K &= K_{\max}-eV \\ &= 2.0-1.5 \\ &= 0.5\ \mathrm{eV} \end{aligned} \)
From the above, \( K=0.5\ \mathrm{eV} \). Hence, the fastest electrons still reach the collector. A retarding potential of 2.0 V would be needed to stop them.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function φ0), the electron is emitted with maximum kinetic energy”
\( K_{\max}=h\nu-\phi_0 \)\( \qquad (11.2) \)
Effect of potential on photoelectric current – “… When the polarity is reversed, the electrons are repelled and only the sufficiently energetic electrons are able to reach the collector A. …”