When a photosensitive surface is irradiated by lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$,…

When a photosensitive surface is irradiated by lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$, kinetic energies of emitted photoelectrons are $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ respectively. The work function of the photosensitive surface is
  1. $\frac{\lambda_{2} E_{2}-\lambda_{1} E_{1}}{\lambda_{2}-\lambda_{1}}$
  2. $\frac{\lambda_{1} E_{1}+\lambda_{2} E_{2}}{\lambda_{2}+\lambda_{1}}$
  3. $\frac{\lambda_{1} E_{1}-\lambda_{2} E_{2}}{\lambda_{2}-\lambda_{1}}$
  4. $\frac{\lambda_{2} E_{1}+\lambda_{2} E_{2}}{\lambda_{2}-\lambda_{1}}$

Solution

$\begin{aligned} & \mathrm{E}_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W} \quad \therefore \mathrm{E}_1 \lambda_1=\mathrm{hc}-\mathrm{W} \lambda_1 \\ & \mathrm{E}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W} \quad \therefore \mathrm{hc}=\mathrm{E}_1 \lambda_1+\mathrm{W} \lambda_1 \ldots \ldots \text { (i) } \\ & \therefore \mathrm{E}_2 \lambda_2=\mathrm{hc}-\mathrm{W}_2 \quad \therefore \mathrm{hc}=\mathrm{E}_2 \lambda_2+\mathrm{W} \lambda_2 . \\ & \text { By Eq.(i) and (ii) } \\ & \mathrm{E}_1 \lambda_1+\mathrm{W} \lambda_1=\mathrm{E}_2 \lambda_2+\mathrm{W} \lambda_2 \\ & \therefore \mathrm{E}_1 \lambda_1-\mathrm{E}_2 \lambda_2=\mathrm{W}\left(\lambda_2-\lambda_1\right) \\ & \therefore \mathrm{W}=\frac{\mathrm{E}_1 \lambda_1-\mathrm{E}_2 \lambda_2}{\lambda_2-\lambda_1} \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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