When a point source of monochromatic light is | vedclass

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(B) The stopping potential (cut-off voltage) depends only on the frequency of the incident light and the work function of the metal surface. It is ind

Vedclass · Accepted Answer

The stopping potential will be $0.6\, V$

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(B) The stopping potential (cut-off voltage) depends only on the frequency of the incident light and the work function of the metal surface. It is independent of the intensity of the light.Since the distance of the source is increased,the intensity of light decreases,but the frequency remains constant. Therefore,the stopping potential remains unchanged at $0.6\, V$.The saturation current is directly proportional to the intensity of the incident light. Intensity $I$ follows the inverse square law,$I \propto 1/d^2$.When the distance increases from $0.2\, m$ to $0.6\, m$ (a factor of $3$),the intensity becomes $(1/3)^2 = 1/9$ of the original value.Thus,the new saturation current will be $18\, mA / 9 = 2\, mA$.Therefore,both options $B$ and $D$ (if corrected) are physically correct,but typically in such problems,the focus is on the invariance of the stopping potential. Given the standard format,$B$ is the primary conceptual answer.

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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