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Question

# A 600 W mercury lamp emits a monochromatic radiation of wavelength 331.3 nm. How many photons are emitted from the lamp per second? h=6.626×10−34 Js;velocity of light =3×108 ms−1

A
1021
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B
1022
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C
1020
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D
1019
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Solution

## The correct option is A 1021According to Planck's quantum theory, E=nhcλ Where, E = energy of the radiation =600 W=600 J/s n = number of photons emitted per second h = Planck's constant =6.626×10−34 J s c = speed of light =3×108 m/s λ = wavelength of the radiation =331.3 nm=331.3×10−9m On subtituting the values in the equation and solving for n, n=Eλh.c=600×331.3×10−96.626×10−34×3×108 On solving n=1×1021 photons/second

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