A modern 200 W sodium street lamp emits yellow light of wavelength 0.6μm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is :
A
62×1020
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B
3×1019
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C
1.5×1020
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D
6×1018
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Solution
The correct option is B1.5×1020 Energy converted into light per second E=0.25×200=50J per second
Given : λ=0.6μm
Energy of one photon Ep=hcλ=6.63×10−34×3×1080.6×10−6=33.15×10−20J
Number of photons N=EEp=5033.15×10−20=1.5×1020 photons per second