A metal surface is illuminated by light of two different wavelengths $248 nm$ and $310 nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$ respectively. If the ratio $u_1:u_2=2:1$ and $hc=1240eV nm$, the work-function of the metal is nearly

Energy corresponding to $284 nm$ wavelength 
                        $=\frac{1240}{248} eV=5 eV$
Energy corresponding to $310 nm$ wavelength
                        $=\frac{1240}{310} eV=4 eV$
                  $$\begin{gathered} \frac{K{E}_1}{K{E}_2}=\frac{{u_1}^2}{{u_2}^2}=\frac{4}{1} \\ =\frac{5 eV-W}{4 eV-W} \end{gathered}$$
$$\begin{gathered} \Rightarrow 16-4W=5-W \\ \Rightarrow 11=3W \\ \Rightarrow W=\frac{11}{3}=3.67 eV \\ \simeq 3.7 eV \end{gathered}$$