A photoelectric surface is illuminated successively by monochromatic light of wavelengths $λ$ and $\frac{λ}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is:
Answer & explanation
Correct answer: option 2
$ K.E (K) = E - \phi = \frac{hc}{\lambda} - \phi $
$ K_1 = \frac{hc}{\lambda} - \phi $
$ K_2 = \frac{2hc}{\lambda} - \phi $
$ K_2 = 3 K_1 $
$\Rightarrow \frac{2hc}{\lambda} - \phi = 3\frac{hc}{\lambda} - 3\phi $
$ \Rightarrow 2\phi = \frac{hc}{\lambda}$
$ \Rightarrow \phi = \frac{hc}{2\lambda}$