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Question

3.0mW of 400nm light is incident on a photoelectric cell, if 0.1% of the photons are contributing in ejection of electrons, then the current in the cell is

A
0.48μA
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B
Resistance value not given
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C
Zero
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D
0.96μA
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Solution

The correct option is D 0.96μA
For wavelength 400 nm , energy of each photon be E=hcλ
If n be the number of photons. total energy = nE.
Given nE=3mW
n= 3E=6.04×1015
0.1% of the photons are contributing in ejection of electrons =6.04×1012
Current , I= dQdt=6.04×1012e
e is charge of coulomb.
e=1.6×1019 Coulomb
I=0.96μA

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