When wavelength of incident radiation on the metal surface is reduced from ' $\lambda_1$ ' to ' $\lambda_2$…

When wavelength of incident radiation on the metal surface is reduced from ' $\lambda_1$ ' to ' $\lambda_2$ ' the kinetic energy of emitted photoelectrons is tripled. The work function of metal $[\mathrm{h}=$ Plank 's constant, c = velocity of light]
  1. $\frac{\mathrm{hc}}{2}\left[\frac{3 \lambda_1-\lambda_2}{\lambda_1 \lambda_2}\right]$
  2. $\frac{\mathrm{hc}}{2}\left[\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right]$
  3. $\operatorname{hc}\left[\frac{3 \lambda_1-\lambda_2}{\lambda_1 \lambda_2}\right]$
  4. $\operatorname{hc}\left[\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right]$

Solution

Let $\mathrm{K}$ represent kinetic energy $\begin{aligned} & \therefore \mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 \\ & \text { and } \mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \\ & \mathrm{~K}_2=3 \mathrm{~K}_1 \\ & \therefore \frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0=3 \frac{\mathrm{hc}}{\lambda_1}-3 \mathrm{~W}_0 \\ & \therefore 2 \mathrm{~W}_0=\frac{3 \mathrm{hc}}{\lambda_1}-\frac{\mathrm{hc}}{\lambda_2} \\ & \therefore \mathrm{W}_0=\frac{\mathrm{hc}}{2}\left(\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right) \end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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