A caesium photocell,with a steady potential d | vedclass

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(A) The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light.For a point source,the intensity $

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Are one quarter as numerous

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(A) The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light.For a point source,the intensity $I$ follows the inverse square law,$I \propto \frac{1}{d^2}$,where $d$ is the distance from the source.Let $N_1$ be the number of photoelectrons emitted at distance $d_1 = 50 \ cm$ and $N_2$ be the number at $d_2 = 1 \ m = 100 \ cm$.Since $N \propto I$,we have $\frac{N_1}{N_2} = \frac{I_1}{I_2} = \left( \frac{d_2}{d_1} \right)^2$.Substituting the values: $\frac{N_1}{N_2} = \left( \frac{100}{50} \right)^2 = (2)^2 = 4$.Therefore,$N_2 = \frac{N_1}{4}$.Thus,the number of photoelectrons becomes one-quarter of the original number.

$A$ caesium photocell,with a steady potential difference of $60 \ V$ across it,is illuminated by a bright point source of light $50 \ cm$ away. When the same light source is placed $1 \ m$ away,the photoelectrons emitted from the cell:

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