Physics - Modern Physics Question with Solution | TestHub

The electric field of light wave is given as$\overrightarrow{\mathrm{E}\mathrm{ }}={10}^{–3}\cos \left(\frac{2\mathrm{\pi }x}{5\times {10}^{–7}}-2\mathrm{\pi }\times 6\times {10}^{14}\mathrm{ }\mathrm{t}\right)\hat{x}\frac{\mathrm{N}}{\mathrm{C}}$. This light falls on a metal plate of work function$2\mathrm{ }\mathrm{e}\mathrm{V}$. The stopping potential of the photo-electrons is:
Given,$\mathrm{E}$(in$\mathrm{e}\mathrm{V}$)$=\frac{12375}{\mathrm{\lambda }\left(\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{\AA }\right)}$