A metallic surface is irradiated by a monochromatic light of frequency $v_{1}$ and stopping potentia

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A metallic surface is irradiated by a monochromatic light of frequency $v_{1}$ and stopping potential is found to be $\mathrm{V}_{1}$. If the light of frequency $v_{2}$ irradiates the surface, the stopping potential will be\

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Maximum kinetic energy $E_{K}=\frac{1}{2} m v^{2}=e V_{s}$ \end{enumerate} Where $\mathrm{V}_{\mathrm{s}}$ is the stopping potential.\ According to Einstein's photoelectric effect $$ \begin{equation*} h v_{1}=\phi+e V_{1} 	ag{i} \end{equation*}$$ $$ \begin{equation*} h v_{2}=\phi+e V_{2} 	ag{ii} \end{equation*} $$ $	herefore \mathrm{h}\left(\mathrm{v}_{1}-\mathrm{v}_{2}ight)=\mathrm{e}\left(\mathrm{V}_{1}-\mathrm{V}_{2}ight)$\ $\frac{\mathrm{h}}{\mathrm{e}}\left(v_{1}-v_{2}ight)=\mathrm{V}_{1}-\mathrm{V}_{2}$ or, $\mathrm{V}_{2}=\mathrm{V}_{1}+\frac{\mathrm{h}}{\mathrm{e}}\left(v_{2}-v_{1}ight)$\