A light source is at a distance $$d$$ from a photoelectric cell, then the number of photoelectrons emitted from the cell is $$n.$$ If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become
A.
$$\frac{n}{2}$$
B.
$$2\,n$$
C.
$$4\,n$$
D.
$$n$$
Answer :
$$4\,n$$
Solution :
Intensity of light source is given by
$$I \propto \frac{1}{{{d^2}}}$$
where, $$d$$ is the distance of light source from the cell.
So, for two different situations for intensities,
$$\eqalign{
& {\text{or}}\,\,\frac{{{I_1}}}{{{I_2}}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4} \cr
& {\text{or}}\,\,{I_2} = 4\,{I_1} \cr} $$
As number of photoelectrons emitted is directly proportional to intensity, so number of photoelectrons emitted will become 4 times, i.e. $$4\,n.$$
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