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14/01/2022 3:32 pm
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The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x × 1013. The value of x is _____. (Nearest integer)
(h = 6.63 × 10-34 Js, c = 3.00 × 108 ms-1)
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14/01/2022 3:37 pm
Energy emitted in 0.1 sec.
= 0.1 sec x 10-3 J/s
= 10-4 J
If 'n' photons of λ = 1000 nm are emitted,
then; 10-4 = n x \(\frac{hc}{\lambda}\)
⇒ 10-4 = \(\frac{n \times 6.63 \times 10^{-34} \times 3 \times 10^8}{1000 \times 10^{-9}}\)
⇒ n = 5.02 x 1014 = 50.2 x 1013
= 50 (Nearest integer)