Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, illuminate a metal of work function 0.5 eV. The ratio of the maximum KE of the emitted electrons will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1:4 \)
By Einstein’s photoelectric equation, the maximum kinetic energy is the photon energy minus the work function, \( K_{\max} = E - \phi \). Both photons exceed the work function (0.5 eV), so electrons are emitted in each case.
\( K_1 = 1 - 0.5 = 0.5\,\mathrm{eV} \)
\( K_2 = 2.5 - 0.5 = 2.0\,\mathrm{eV} \)
\( \displaystyle \frac{K_1}{K_2} = \frac{0.5}{2.0} = \frac{1}{4} \)