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Question

Light rays of wavelengths 6000 ˚A and of photon intensity 39.6 watts/m2 incidents on a metal surface. If only 1% of photons incident on the surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be approximately :
[Planck constant h=6.64×1034Js; Velocity of light =3×108ms1]

A
12×1018
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B
10×1018
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C
12×1017
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D
12×1015
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Solution

The correct option is C 12×1017
The intensity of photons I=NhcAλ

Here N is the number of photons and A is the exposed area of the metal.

NA i.e. the number of photons striking the surface of the metal per second per unit area =Iλhc

Substituting the values given in the question, we get Iλhc=12×1019/m2

But only 1 % of photons cause photoemission.

So the number of electrons ejected from the surface of the metal per second per unit area =12×1017/m2

So, the answer is option (C).

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