Electromagnetic radiations of two different frequencies, whose photons have energies \( 6\,\mathrm{eV} \) and \( 9\,\mathrm{eV} \), respectively, successively illuminate a metal of work function \( 2\,\mathrm{eV} \). The ratio of maximum kinetic energy of the photoelectrons emitted due to illumination by the two radiations will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4:7 \)
By Einstein’s photoelectric equation, the maximum kinetic energy is the photon energy minus the work function: \( K_{\mathrm{max}} = h\nu - \phi_0 \).
\( K_1 = 6 - 2 = 4\,\mathrm{eV} \)
\( K_2 = 9 - 2 = 7\,\mathrm{eV} \)
\( K_1:K_2 = 4:7 \)