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  • When radiation of wavelength is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength <i

When radiation of wavelength \lambda is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3\lambda, the stopping potential is V/4 . If the threshold wavelength for the metallic surface is n\lambda then value of n will be____________
Option: 1 9
Option: 2 6
Option: 3 3
Option: 4 12

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best_answer

\frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\mathrm{eV} \quad \ldots \ldots(\mathrm{i}) \\ \frac{\mathrm{hc}}{3 \lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\frac{\mathrm{e} \cdot \mathrm{V}}{4} \quad \ldots \ldots(\mathrm{ii}) \\ (multiply \ \ equation (ii) \ by \ 4)\\ we \ get \ \frac{4 \mathrm{hc}}{3 \lambda}=\frac{4 \mathrm{hc}}{\lambda_{0}}+\mathrm{eV} \quad \ldots (iii) \\ Now \ From \ (i) \ \& \ (iii) \ we \ get \\ \frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}=\frac{4 \mathrm{hc}}{3 \lambda}-\frac{4 \mathrm{hc}}{\lambda_{0}} \\ -\frac{h c}{3 \lambda}=-\frac{3 h c}{\lambda_{0}} \\ \Rightarrow 9 \lambda=\lambda_{0}\\ \mathrm{n}=9

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