When a point source of monochromatic light is | vedclass

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(B) The stopping potential $(V_S)$ depends only on the frequency of the incident light and the work function of the metal surface. It is independent o

Vedclass · Accepted Answer

The stopping potential will be $0.6 \ V$.

Vedclass · Answer

(B) The stopping potential $(V_S)$ depends only on the frequency of the incident light and the work function of the metal surface. It is independent of the intensity of light or the distance of the source from the cell.Therefore,the stopping potential remains $0.6 \ V$.The saturation current $(I_S)$ is directly proportional to the intensity of the incident light. Since the source is a point source,the intensity $I$ follows the inverse square law: $I \propto \frac{1}{r^2}$.Given $r_1 = 0.2 \ m$ and $r_2 = 0.6 \ m$,the ratio of intensities is $\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}ight)^2 = \left(\frac{0.2}{0.6}ight)^2 = \left(\frac{1}{3}ight)^2 = \frac{1}{9}$.Since $I_S \propto I$,the new saturation current $I_{S2} = I_{S1} 	imes \frac{1}{9} = 18 \ mA 	imes \frac{1}{9} = 2 \ mA$.Comparing the options,the correct statements are that the stopping potential remains $0.6 \ V$ (Option $B$) and the saturation current becomes $2 \ mA$ (which is not listed,but $B$ is a correct statement).

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

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