Energy emitted in 0.1sec
=0.1sec×10−3Js−1
=10−4J
If 'n' photons of λ=1000nm are emitted, then 10−4=λnhc
10−4=1000×10−9n×6.63×10−34×3×108
n=5.02×1014=50.2×1013
n=50 (nearest integer)
JEE Main 2021 — Chemistry Physical Chemistry
The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1mW and wavelength of 1000nm, in 0.1 second is x×1013. The value of x is (Nearest integer) (h=6.63×10−34Js,c=3.00×108ms−1):
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