If the frequency of light in a photoelectric experiment is doubled, the stopping potential will :
Answer & explanation
Correct answer: option 3
Let Initial frequency is f. Initial Stopping potential is $ V_s = \frac{hf - \phi}{e}$
When frequency is doubled then Stopping Potential is
$ V_s' = \frac{2hf - \phi}{e} = \frac{2(hf - \phi) + \phi}{e} $
$\Rightarrow \text{Stopping Potential is } V_s' = 2V_s + \frac{\phi}{e}> 2V_s$