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Q.

A monochromatic infrared range finder of power 1mW emits photons with wavelength 1000 nm in 0.1 seconds. The number of photons emitted in 0.1 seconds is
(Given h =6.626×10^-34) Js, c=3×10⁸ ms^-1, Avogadro number =6.022×10²³)

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a

5 x 10¹⁴

b

5 x 10¹¹

c

30 x 10³⁴

d

30 x 10³⁷

answer is A.

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Detailed Solution

detailed_solution_thumbnail

(P) Power =1 mW=10³ watt
E=P×t=10³×0.1=10² J
λ=1000 nm=10⁶ m
We know,
E=nhv
Here, E= Energy of n photons
n= number of photons
hu = energy of 1 photon
$\begin{array}{l} E = n \times \frac{hc}{λ} \\ n = \frac{Eλ}{hc} = \frac{10^{4} \times 10^{6}}{6.625 \times 10^{- 34} \times 3 \times 10^{8}} \\ = \frac{10^{6}}{6.625 \times 3} = 0.05 \times 10^{6} = 5 \times 10^{14} \end{array}$

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A monochromatic infrared range finder of power 1mW emits photons with wavelength 1000 nm in 0.1 seconds. The number of photons emitted in 0.1 seconds is(Given h =6.626×10-34) Js, c=3×108 ms-1, Avogadro number =6.022×1023)