When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5V₀. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V₀. Then the work function of the metallic surface is -

$\frac{hc}{6λ}$

$5eV_0=\frac{hc}{λ}-W$  ...(1)

$eV_0=\frac{hc}{3λ}-W$  ...(2)

Solving equation (1) & (2)

$\frac{hc}{λ}-W=\frac{5hc}{3λ}-5W$

$4W=\frac{2hc}{3λ}⇒W=\frac{hc}{6λ}$