If the surface of a metal is successfully exposed to radiation of ${\lambda }_1=350\mathrm{nm}\text{ and }{\lambda }_2=450\mathrm{nm}$ the maximum velocity of photoelectrons will differ by a factor 2. The work function of this metal is  

Let A be the work function of metal 

$\begin{array}{r}\frac{hc}{{\lambda }_1}=A+\frac{m{v}_1^2}{2}\\ \frac{hc}{{\lambda }_2}=A+\frac{m{v}_2^2}{2}\end{array}$

$hc\left[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right]=\frac{m}{2}\left(v_1^2-v_2^2\right)=\frac{m}{2}\left[4v_2^2-v_2^2\right]=\frac{3m{v}_2^2}{2}$

$\begin{array}{l}\frac{{\mathrm{mv}}_2^2}{2}=\frac{\mathrm{hc}}{3}\left[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right]\\ A=\frac{hc}{{\lambda }_2}-\frac{{\mathrm{mv}}_2^2}{2}=\frac{hc}{{\lambda }_2}-\frac{hc}{3}\left[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right]=\frac{4hc}{3{\lambda }_2}-\frac{hc}{3{\lambda }_1}\\ =-\frac{hc}{3}\left[\frac{1}{{\lambda }_1}-\frac{4}{{\lambda }_2}\right]\\ =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3}\left[\frac{4}{450\times {10}^{-9}}-\frac{1}{350\times {10}^{-9}}\right]\\ \therefore A=3.93\times {10}^{-9}\mathrm{J}.\end{array}$