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(A) The intensity $I$ of light at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$,so $I \propto \frac{1}{r^2}$.When the distan

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the emission of electrons per second will be four times

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(A) The intensity $I$ of light at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$,so $I \propto \frac{1}{r^2}$.When the distance is changed from $d$ to $\frac{d}{2}$,the new intensity $I'$ becomes $I' = \frac{1}{(d/2)^2} = 4 	imes \frac{1}{d^2} = 4I$.Since the number of photoelectrons emitted per second is directly proportional to the intensity of incident light,the emission rate becomes $4$ times the original value.According to Einstein's photoelectric equation,$KE_{max} = h
u - \phi$,where $
u$ is the frequency of incident light and $\phi$ is the work function.Since the frequency $
u$ remains unchanged,the maximum kinetic energy $(KE_{max})$ and the stopping potential $(V_s)$ remain unchanged,as $eV_s = KE_{max}$.

The distance between a light source and a photoelectric cell is $d$. If the distance is decreased to $\frac{d}{2}$,then:

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