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(B, D) The stopping potential (cut-off voltage) depends only on the frequency (or wavelength) of the incident light and is independent of the intensit

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the stopping potential will be $0.6\, V$

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(B, D) The stopping potential (cut-off voltage) depends only on the frequency (or wavelength) of the incident light and is independent of the intensity or the distance of the source from the photoelectric cell. Therefore,the stopping potential remains $0.6\, V$.The saturation current is directly proportional to the intensity of light. Since the source is a point source,the intensity $I$ follows the inverse square law: $I \propto \frac{1}{d^2}$.Given the initial distance $d_1 = 0.2\, m$ and initial current $I_1 = 18.0\, mA$,and the new distance $d_2 = 0.6\, m$,the new current $I_2$ is calculated as:$I_2 = I_1 	imes \left(\frac{d_1}{d_2}ight)^2 = 18.0 	imes \left(\frac{0.2}{0.6}ight)^2 = 18.0 	imes \left(\frac{1}{3}ight)^2 = 18.0 	imes \frac{1}{9} = 2.0\, mA$.Thus,both the stopping potential remains $0.6\, V$ and the saturation current becomes $2.0\, mA$.

When a monochromatic point source of light is at a distance of $0.2\, m$ from a photoelectric cell,the cut-off voltage and the saturation current are respectively $0.6\, V$ and $18.0\, mA$. If the same source is placed $0.6\, m$ away from the photoelectric cell,then

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