question_answer

The electric field of light wave is given as \[\vec{E}={{10}^{-3}}\cos \left( \frac{2\pi x}{5\times {{10}^{-7}}}-2\pi \times 6\times {{10}^{14}}t \right)\]\[\hat{x}\frac{N}{C}.\]This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is : Given, E (in eV) \[=\frac{12375}{\lambda (in{\AA})}\]      [JEE Main 9-4-2019 Morning]

A) 0.48 V

B) 2.0 V

C) 2.48 V

D) 0.72 V

Correct Answer: A

Solution :

[a]

energy of photon

from Einstein's equation

volt

Option [a]