If $K_1\text{ and }{K}_2$ are maximum kinetic energies of photoelectrons emitted when lights of wavelength ${\lambda }_1\text{ and }{\lambda }_2$ respectively incident on a metallic surface. If ${\lambda }_1=3{\lambda }_2$ , then

Kinetic energy of the photoelectrons  $\mathrm{K}=\frac{\mathrm{hc}}{\lambda }-\varphi$

 where   ϕ is the work function of the metal

$$\begin{gathered} \therefore \text{ For wavelength }{\lambda }_1,{\mathrm{K}}_1=\frac{\text{ he }}{{\lambda }_1}-\varphi \\ \text{ For wavelength }{\lambda }_2,{\mathrm{K}}_2=\frac{\mathrm{hc}}{{\lambda }_2}-\varphi \\ \text{ Given: }{\lambda }_1=3{\lambda }_2 \\ \therefore \text{ Equation (1) becomes }{\mathrm{K}}_1=\frac{\mathrm{hc}}{3{\lambda }_2}-\varphi \\ {\mathrm{K}}_2-{\mathrm{K}}_1=\frac{\mathrm{hc}}{{\lambda }_2}-\frac{\mathrm{hc}}{3{\lambda }_2} \\ {\mathrm{K}}_2-{\mathrm{K}}_1=\frac{2}{3}\frac{\mathrm{hc}}{{\lambda }_2}\Rightarrow \frac{\mathrm{hc}}{{\lambda }_2}=\frac{3}{2}\left({\mathrm{K}}_2-{\mathrm{K}}_1\right) \\ {\mathrm{K}}_2=\frac{3}{2}\left({\mathrm{K}}_2-{\mathrm{K}}_1\right)-\varphi \\ \Rightarrow {\mathrm{K}}_2-3{\mathrm{K}}_1=2\varphi \\ \text{ As }\varphi >0\Rightarrow {\mathrm{K}}_2-3{\mathrm{K}}_1>0 \\ \text{ Thus }{\mathrm{K}}_1<\frac{{\mathrm{K}}_2}{3} \end{gathered}$$