CUET Preparation Today
What is the slope of the graph drawn between frequency of incident light and stopping potential for a given surface? |
$h$ $he^{-1}$ $eh$ $e$ |
$he^{-1}$ |
$ eV_s = h\nu - \phi$ $ V_s = \frac{h}{e} \nu - \frac{\phi}{e}$ slope is he-1 The correct answer is Option (2) → $he^{-1}$ |
