A metallic surface is illuminated with radiation of wavelength λ , the stopping potential is V 0 . If…

Let the threshold frequency be ${\lambda }_0$.

By the equation of photoelectric effect, for wavelength $\lambda$,

$e{V}_0=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right) ...\left(i\right)$

For the wavelength $2\lambda$,

$\frac{e{V}_0}{4}=hc\left(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}\right) ...\left(ii\right)$

Dividing (i) by (ii)

$$\begin{gathered} 4=\frac{hc\left({\displaystyle \frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}}\right)}{hc\left({\displaystyle \frac{{\lambda }_0-2\lambda }{2\lambda {\lambda }_0}}\right)} \\ \Rightarrow 4=\frac{2({\lambda }_0-\lambda )}{{\lambda }_0-2\lambda } \\ \Rightarrow 2{\lambda }_0=6\lambda \\ \Rightarrow {\lambda }_0=3\lambda \end{gathered}$$