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A photoelectric cell is illuminated by a small bright source of light placed at 1m. If the same source of light is placed 2m away, the electrons emitted by the cathode

$\begin{array}{1 1} \text{(a) each carries one quarter of its previous momentum.}\\ \text{(b) each carries one quarter of its previous energy.} \\ \text{ (c) are half the previous number.} \\ \text{(d) are one quarter of the previous number. }\end{array}$
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Ans : (d)
On doubling the distance the intensity becomes one fourth i.e. only one fourth of
photons now strike the target in comparison to the previous number. Since
photoelectric effect is a one photon-one electron phenomena, so only one-fourth
photoelectrons are emitted out of the target hence reducing the current to
one fourth the previous value.


answered Dec 27, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1

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