A photoelectric surface is illuminated successively by monochromatic light of wavelength \( 2\lambda \) and \( 0.5\lambda \). What will be the work function of the photosensitive material if the maximum kinetic energy of the photo electrons in the second case is 5 times that of the first case? In the given options h is Planck's constant and c is the speed of light.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{hc}{8\lambda} \)
By Einstein's photoelectric equation the maximum kinetic energy of a photoelectron is the photon energy minus the work function \( \phi \): \( K_{\max} = \frac{hc}{\lambda'} - \phi \).
\( \displaystyle K_1 = \frac{hc}{2\lambda} - \phi,\qquad K_2 = \frac{hc}{0.5\lambda} - \phi = \frac{2hc}{\lambda} - \phi \)
Using the given condition \( K_2 = 5K_1 \):
\( \displaystyle \frac{2hc}{\lambda} - \phi = \frac{5hc}{2\lambda} - 5\phi \)
\( \displaystyle 4\phi = \frac{5hc}{2\lambda} - \frac{2hc}{\lambda} = \frac{hc}{2\lambda} \)
\( \displaystyle \phi = \frac{hc}{8\lambda} \)
Check: \( K_1 = \frac{hc}{2\lambda} - \frac{hc}{8\lambda} = \frac{3hc}{8\lambda} \) and \( K_2 = \frac{2hc}{\lambda} - \frac{hc}{8\lambda} = \frac{15hc}{8\lambda} \), so \( K_2/K_1 = 5 \).