When the wavelength of radiation falling on a metal is changed from

(2)

Using equation, $=\frac{h c}{\lambda}-\phi$

$K E_{\max }=\frac{h c}{\lambda}-\phi=\frac{h c}{500}-\phi$          ....(1)

Again, $3 K E_{\max }=\frac{h c}{200}-\phi$                                    ....(2)

Dividing equation (2) by (1),

$\frac{3 K E_{\max }}{K E_{\max }}=\frac{3}{1}=\frac{\frac{h c}{200}-\phi}{\frac{h c}{500}-\phi}$

Putting the value of $h c=1237.5$ and solving we get, work function, $\phi=0.61 \mathrm{eV}$.