Question

A point source causes photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?

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Answer 1

If we use the same light source then by bringing the light source closer to the emitter increases the number of photons falling thereby increasing the intensity.

If no. of photons falling on the plate increases then the no. of photo electrons increases, increasing the photo current.

So photo current α intensity

Now relation between intensity and distance between plate is

$I\propto \frac{1}{4\pi d^2}RememberI=\frac{\left(\frac{E}{t}\right)}{Area}=\frac{P}{4\pi d^2}$

$\text{So saturation current}\alpha IntensityandI\propto \frac{1}{d^2}Photocurrent\to i\propto \frac{1}{d^2}y=\frac{1}{x^2}islikeoption\left(d\right)$