The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths…

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths $\lambda_1$ and $\lambda_2$ are incident on a metallic surface. If $\lambda_1=3 \lambda_2$ then
  1. $\mathrm{K}_1=\frac{\mathrm{K}_2}{3}$
  2. $\mathrm{K}_1 < \frac{\mathrm{K}_2}{3}$
  3. $\mathrm{K}_1=3 \mathrm{~K}_2$
  4. $3 \mathrm{~K}_1=2 \mathrm{~K}_2$

Solution

Kinetic energy of the photoelectrons $\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi$ $\therefore \quad$ For wavelength $\lambda_1$, $\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_1}-\phi$ $\therefore \quad$ For wavelength $\lambda_1$, $\mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\phi$ Subtracting equation (i) from equation (ii), $\begin{aligned} & \mathrm{K}_2-\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_2}-\phi-\frac{\mathrm{hc}}{\lambda_1}+\phi \\ & \mathrm{K}_2-\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_2}-\frac{\mathrm{hc}}{3 \lambda_2} \\ & \mathrm{~K}_2-\mathrm{K}_1=\frac{2}{3} \frac{\mathrm{hc}}{\lambda_2} \\ & \frac{\mathrm{hc}}{\lambda_2}=\frac{3}{2}\left(\mathrm{~K}_2-\mathrm{K}_1\right) \end{aligned}$ Substituting equation (iii) in equation (ii), $\begin{aligned} & \mathrm{K}_2=\frac{3}{2}\left(\mathrm{~K}_2-\mathrm{K}_1\right)-\phi \\ & 2 \mathrm{~K}_2=3 \mathrm{~K}_2-3 \mathrm{~K}_1-2 \phi \\ & \mathrm{K}_2-3 \mathrm{~K}_1=2 \phi \\ & \mathrm{As}, \phi>0 \\ & \mathrm{~K}_2-3 \mathrm{~K}_1>0 \\ & \mathrm{~K}_1 < \frac{\mathrm{K}_2}{3} \end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 1)

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