When a point source of monochromatic light is | vedclass

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(B) The cut-off voltage (stopping potential) depends only on the frequency of the incident light,not on its intensity. Therefore,it remains $0.6 \ V$.

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The stopping potential will be $0.6 \ V$

Vedclass · Answer

(B) The cut-off voltage (stopping potential) depends only on the frequency of the incident light,not on its intensity. Therefore,it remains $0.6 \ V$.The intensity of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$.When the distance increases from $0.2 \ m$ to $0.6 \ m$,the distance becomes $3$ times $(d' = 3d)$.Thus,the intensity becomes $I' = \frac{I}{3^2} = \frac{I}{9}$.Since the saturation current is directly proportional to the intensity of incident light,the new saturation current will be $I'_{sat} = \frac{18 \ mA}{9} = 2 \ mA$.Comparing this with the given options,option $(b)$ is correct regarding the stopping potential.

When a point source of monochromatic light is at a distance of $0.2 \ m$ from a photoelectric cell,the cut-off voltage and the saturation current are $0.6 \ V$ and $18 \ mA$ respectively. If the same source is placed $0.6 \ m$ away from the photoelectric cell,then:

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