In a certain experiment, the photoelectric cut-off voltage is 3.5 V. What is the maximum kinetic energy of the emitted photoelectrons? (Take \( e=1.6\times10^{-19}\,\mathrm{C} \).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5.6\times10^{-19}\,\mathrm{J} \)
The cut-off (stopping) potential \( V_0 \) is just enough to stop even the most energetic photoelectrons, so their maximum kinetic energy is \( K_{\mathrm{max}}=eV_0 \).
\( \begin{aligned} K_{\mathrm{max}} &= eV_0 \\ &= 1.6\times10^{-19}\,\mathrm{C}\times3.5\,\mathrm{V} \\ &= 5.6\times10^{-19}\,\mathrm{J} \end{aligned} \)
NCERT (2026-27):
Effect of Potential on Photoelectric Current – “… Photoelectric current is zero when the stopping potential is sufficient to repel even the most energetic photoelectrons, with the maximum kinetic energy (Kmax), so that”
\( K_{\mathrm{max}}=e\,V_0 \)\( \qquad (11.1) \)