When a metallic surface is illuminated with radiation of wavelength $\lambda$ the stopping potentials is V. If the same surface is illuminated with radiation of wavelength $2 \lambda$ the stopping potential is $\frac{V}{4}$, the threshold wave length for metallic surface is

$3 \lambda$

The correct answer is Option (4) → $3 \lambda$

By Einstein's photoelectric equation

$\phi=\phi_0+e V_0$

$\frac{h c}{\lambda}=\phi_0+e V$           .....(i)

$\frac{h c}{2 \lambda}=\phi_0+\frac{e V}{4}$          .....(ii)

On putting the value of $e V$ form equation (i) in equation (ii)

$\frac{h c}{2 \lambda} =\phi_0+\frac{1}{4}\left(\frac{h c}{\lambda}-\phi_0\right)$

$\frac{h c}{4 \lambda} =\frac{3 \phi_0}{4}$

$\frac{h c}{4 \lambda} =\frac{3}{4} \frac{\mu c}{\lambda_1} ~~~~\Rightarrow \lambda_1=3 \lambda$