The frequency of the light falling on a photosensitive metal plate is doubled. The maximum kinetic energy of the emitted photoelectrons becomes
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → more than double the earlier value
By Einstein’s photoelectric equation, \( K=h\nu-\phi_0 \). Here \( h\nu \) is the photon energy and \( \phi_0 \) is the work function. The work function stays the same for the same plate. With the frequency doubled:
\( \begin{aligned} K' &= h(2\nu)-\phi_0 \\ &= 2(h\nu-\phi_0)+\phi_0 \\ &= 2K+\phi_0 \end{aligned} \)
Since \( \phi_0\gt0 \), \( K'\gt2K \).
Hence, the maximum kinetic energy becomes more than double the earlier value.
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”
\( K_{\mathrm{max}}=h\nu-\phi_0 \)\( \qquad (11.2) \)