question_answer

The electric field of light wave is given as \[\overrightarrow{E}={{10}^{3}}\cos \] \[\left( \frac{2\pi x}{5\times {{10}^{-7}}}-2\pi \times 6\times {{10}^{14}}t \right)\hat{x}\frac{N}{C}\] This light falls on a metal plate of work function 2eV. The stopping potential (in V) of the photo-electrons is: Given, \[E\,(\text{in}eV)=\frac{12375}{\lambda \,(\text{in}\,\overset{\text{o}}{\mathop{\text{A}}}\,)}\]

Answer:

0.48