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(B) The intensity $I$ of light from a point source at a distance $d$ is given by $I = \frac{P}{4\pi d^2}$,where $P$ is the power of the source.Thus,$I

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Become four times

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(B) The intensity $I$ of light from a point source at a distance $d$ is given by $I = \frac{P}{4\pi d^2}$,where $P$ is the power of the source.Thus,$I \propto \frac{1}{d^2}$.When the distance is reduced to $d' = d/2$,the new intensity $I'$ becomes $I' = \frac{P}{4\pi (d/2)^2} = \frac{P}{4\pi (d^2/4)} = 4 	imes \frac{P}{4\pi d^2} = 4I$.The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light $(N \propto I)$.Since the intensity becomes $4$ times,the number of electrons emitted per second will also become $4$ times the original value.

$A$ photoelectric cell is illuminated by a point source of light kept at a distance $d$. If the distance is reduced to $d/2$,then the number of electrons emitted per second will be:

Similar Questions

Which graph represents the relationship between the intensity of incident light on a metal plate and the resulting photoelectric current?

$A$ point source of light is used in an experiment of photoelectric effect. If the distance between the source and the photoelectric surface is doubled, which of the following may result?

Consider the two following statements $A$ and $B$ and identify the correct choice given in the answers:

$A$ photocell is illuminated by a small bright light source placed at a distance of $1 \ m$. When the same light source is placed at a distance of $\frac{1}{2} \ m$,the number of electrons emitted by the photocathode will be:

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Reason: The work function of the metal is its characteristic property.

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