When radiation of wavelength λ is incident on a metallic surface, the stopping potential of ejected…

$V_{\mathit{S}}=hv-\varphi$

$4.8=\frac{hc}{\lambda }-\varphi ...\left(\mathrm{i}\right)$

$1.6=\frac{hc}{2\lambda }-\varphi ...\left(\mathrm{ii}\right)$

Using above equation $\left(\mathrm{i}\right)$ $-\left(\mathrm{ii}\right)$

$3.2=\frac{hc}{\lambda }-\frac{hc}{2\lambda }$

$3.2=\frac{\mathrm{hc}}{2\lambda } ...\left(\mathrm{iii}\right)$ 

$\left[\lambda =\frac{hc}{6.4}\right]$

Put in equation $\left(\mathrm{ii}\right)$

$\varphi =1.6$

$\frac{hc}{{\lambda }_{\mathrm{th}}}=1.6$

${\lambda }_{\mathrm{th}}=\frac{hc}{1.6}$

$=\left(\frac{hc}{6.4}\right)\times 4=4\lambda$