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Question

5% energy of a monochromatic light source is converted into light of wavelength 5600 A0. If the power of the source be 100 W, then how many photons of light will be emitted by it per second?

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Solution

λ=5600A

Energy of one photon is E=hν=h×cλE=6.63×1034×3×108/(5600×1010)=3.616×1019J

A100watt bulb requires 100joule of energy per second.

Energy radiated by the bulb as visible light per second =5% of100=5Joule

Number of photons emitted per second=Energy radiated per sec/ energy of one photon

=5/(3.616×1019)

=1.38×1019


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