When a metal surface is illuminated by light of wavelength λ , the stopping potential is 8 V . When the same…

From photoelectric effect equation, we can write $E=\varphi +K_{max}$

and work function can be written as $\varphi =\frac{hc}{{\lambda }_0}$.

Also, maximum kinetic energy $K_{max}=e{V}_0$

For first case:

$8e=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$   ... (i)

For second case:

$2e=\frac{hc}{3\lambda }-\frac{hc}{{\lambda }_0}$    ... (ii)

After using equation (ii) in (i), we get

$4\left(\frac{hc}{3\lambda }-\frac{hc}{{\lambda }_0}\right)=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$

$\Rightarrow {\lambda }_0=9\lambda$