Q 12-11-025NEETJEE MainTop questionEasy
In a photoelectric experiment the intensity of the incident light is doubled, its frequency being unchanged. Then
Answer: (A) the saturation current doubles and the stopping potential is unchanged
Twice the intensity means twice as many photons per second, so twice as many photoelectrons: the saturation current doubles. Each photon's energy is unchanged, so $K_{max}$ and the stopping potential stay the same.
Solution by Sreeraj P, M.Sc Physics