Question

In a photoelectric experiment, potential is plotted against the current. Then

• A. A and B will have different frequencies
• B. A and B will have same number of photons
• C. A and B will have same intensity
• D. A and B will have different intensities

Answer 1

The correct option is D A and B will have different intensities

We know that,
For incident frequency $\nu ,$
$h\nu =\varphi +K.E_{max}$ and $e{V}_0=K.E_{max}$
$\Rightarrow h\nu =\varphi +e{V}_0$



Since stopping potential $V_0$ is the same for both A and B and $\varphi$ (work function) is a property of the material, we can say that the incident frequency of A and B is the same.

Saturation current (maximum current obtained) depends on the number of electrons that are ejected. And no. of electrons ejected depends on the no. of incident photons. Since saturation current for A and B are different, the number of photons are also different. Hence, A and B have different intensities. Option (d) is correct.