Physics - Modern Physics Question with Solution | TestHub

Kinetic energy of the photoelectrons $$\mathrm{K}=\frac{\mathrm{hc}}{\lambda }-\varphi$$ $\therefore$ For wavelength ${\lambda }_1$, $${\mathrm{K}}_1=\frac{\mathrm{hc}}{{\lambda }_1}-\varphi$$ $\therefore$ For wavelength ${\lambda }_1$, $${\mathrm{K}}_2=\frac{\mathrm{hc}}{{\lambda }_2}-\varphi$$ Subtracting equation (i) from equation (ii), $$\begin{array}{rl}& {\mathrm{K}}_2-{\mathrm{K}}_1=\frac{\mathrm{hc}}{{\lambda }_2}-\varphi -\frac{\mathrm{hc}}{{\lambda }_1}+\varphi \\ & {\mathrm{K}}_2-{\mathrm{K}}_1=\frac{\mathrm{hc}}{{\lambda }_2}-\frac{\mathrm{hc}}{3{\lambda }_2}\\ & {\mathrm{K}}_2-{\mathrm{K}}_1=\frac{2}{3}\frac{\mathrm{hc}}{{\lambda }_2}\\ & \frac{\mathrm{hc}}{{\lambda }_2}=\frac{3}{2}\left({\mathrm{K}}_2-{\mathrm{K}}_1\right)\end{array}$$ Substituting equation (iii) in equation (ii), $$\begin{array}{rl}& {\mathrm{K}}_2=\frac{3}{2}\left({\mathrm{K}}_2-{\mathrm{K}}_1\right)-\varphi \\ & 2{\mathrm{K}}_2=3{\mathrm{K}}_2-3{\mathrm{K}}_1-2\varphi \\ & {\mathrm{K}}_2-3{\mathrm{K}}_1=2\varphi \\ & \mathrm{As},\varphi >0\\ & {\mathrm{K}}_2-3{\mathrm{K}}_1>0\\ & {\mathrm{K}}_1<\frac{{\mathrm{K}}_2}{3}\end{array}$$