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

The electric field of light wave is given as\(\vec E = {10^{ - 3}}{\rm

1.	The electric field of light wave is given as\(\vec E = {10^{ - 3}}{\rm{cos}}\left( {\frac{{2\pi x}}{{5 \times {{10}^{ - 7}}}} - 2\pi \times 6 \times {{10}^{14}}t} \right)\hat x\frac{N}{C}\) This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is:Given, E (in eV) = \(\frac{{12375}}{{\lambda \left( {{\rm{in\;}}} \right)}}\)
A.	2.0 V
B.	0.72 V
C.	0.48 V
D.	2.48 V
Answer» D. 2.48 V