Physics › Dual Nature of Matter and Radiation › Photoelectric Effect
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
$\frac{\lambda_{2} E_{2}-\lambda_{1} E_{1}}{\lambda_{2}-\lambda_{1}}$ $\frac{\lambda_{1} E_{1}+\lambda_{2} E_{2}}{\lambda_{2}+\lambda_{1}}$ $\frac{\lambda_{1} E_{1}-\lambda_{2} E_{2}}{\lambda_{2}-\lambda_{1}}$ $\frac{\lambda_{2} E_{1}+\lambda_{2} E_{2}}{\lambda_{2}-\lambda_{1}}$
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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