When a metallic surface is illuminated with radiation of wavelength...

Given,

Stopping potential is $V$ when the metal is illuminated with $\lambda$.

Stopping potential is $\dfrac{V}{4}$ when the metal is illuminated with $2\lambda$.

In photo electric effects

$e{V}_{0} = hν - W$

$e{V}_{0} = \dfrac{hc}{\lambda} - W$

$eV = \dfrac{hc}{\lambda} - W$ ...(i)

$e\dfrac{V}{4} = \dfrac{hc}{2\lambda} - W$ ....(ii)

From (i) and (ii)

$\dfrac{hc}{\lambda} - W = 4\left(\dfrac{hc}{2\lambda} - W\right)$

$\dfrac{hc}{\lambda} - W = \dfrac{2hc}{\lambda} - 4W$

$3W = \dfrac{hc}{\lambda} ⇒ W = \dfrac{hc}{3\lambda}$

\dfrac{hc}{{\lambda}_\max} = \dfrac{hc}{3\lambda} ⇒ {\lambda}_\max = threshold wavelength $= 3\lambda$