When a metallic surface is illuminated by a light of wavelength $\lambda$, the stopping potential fo

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When a metallic surface is illuminated by a light of wavelength $\lambda$, the stopping potential for the photoelectric current is 3 V , When the same surface is illuminated by light of wavelength $2 \lambda$, the stopping potential is 1 V . The threshold wavelength for this surface is\

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Stopping potential corresponds to maximum kinetic energy of the photoelectron.\ From photoelectric equation\ $\mathrm{h} v=\mathrm{hV}_{0}+\mathrm{KE}_{	ext {max }}$.\ $\mathrm{h} v=\mathrm{h}_{0}+\mathrm{e}$ (stopping potential)\ or $\frac{\mathrm{hc}}{\lambda_{	ext {incident }}}=\frac{\mathrm{hc}}{\lambda_{0}}+\mathrm{eV}_{\mathrm{s}}$\ Given that when $\lambda_{	ext {incident }}=\lambda . \mathrm{V}_{\mathrm{s}}=3 \mathrm{~V}$ and for $$ \begin{aligned} & \lambda_{	ext {incident }}=2 \lambda, \mathrm{~V}_{\mathrm{s}}=1 \mathrm{~V} . \ \Rightarrow & \frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+3 \mathrm{eV} \end{aligned} $$ And $\frac{\mathrm{hc}}{2 \lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+1 \mathrm{eV}$\ On simplifying (i) and (ii) $$ \Rightarrow \lambda_{0}=4 \lambda $$