A point of source of 6 watts emits monochromatic light of wavelength 5000˙A. The number of photons striking normally per second per unit area of the surface distant 5 m from the source will be
A
4.82
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B
4.82×10−4
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C
4.82×10−6
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D
4.82×1016
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Solution
The correct option is A4.82×1016 The photons will be emitted symmetrically in all directions.
Power of Source = 6W
Energy=nhν=nhcλ where n is no.of photons emitted per second.
Soving the above equation we get n.
The no.of photons falling on 1m2 area is equal to n4πd2 where d is the distance of the point.