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
Concept. Intensity sets the number of photons; frequency sets each photon's energy.
Step 1. Double intensity. Twice the photoelectrons: saturation current doubles; $K_{\max}$ and $V_0$ unchanged.
Answer. The saturation current doubles and the stopping potential is unchanged (Option A).
Solution by Sreeraj P, M.Sc Physics