A photoelectric surface is illuminated successively by monochromatic light of wavelength $\mathrm{\lambda }\text{ and }\frac{\mathrm{\lambda }}{2}\text{ . }$ 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 of the material is 

(h = Planck's constant, c = speed of light)

 Let ${\varphi }_0$  be the work function of the surface of the material. Then, 

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is

$K_{\underset{1}{\max }}=\frac{hc}{\lambda }-{\varphi }_0$

and that in the second case is

$K_{\underset{2}{\max }}=\frac{hc}{\frac{\lambda }{2}}-{\varphi }_0=\frac{2hc}{\lambda }-{\varphi }_0$

$\text{ But }{K}_{\underset{2}{\max }}=3K_{\underset{1}{\max }}\left(\text{ given }\right)$

$\therefore \frac{2hc}{\lambda }-{\varphi }_0=3\left(\frac{hc}{\lambda }-{\varphi }_0\right)$

$\frac{2hc}{\lambda }-{\varphi }_0=\frac{3hc}{\lambda }-3{\varphi }_0$

$3{\varphi }_0-{\varphi }_0=\frac{3hc}{\lambda }-\frac{2hc}{\lambda }$

$2{\varphi }_0=\frac{hc}{\lambda } \text{ or } {\varphi }_0=\frac{hc}{2\lambda }$