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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answer is 50.

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Detailed Solution

Energy emitted in 0.1 sec. = 0.1 sec. × 10−3J/s = 10−4 JIf 'n' photons of λ = 1000 nm are emitted, then 10−4 = nhcλ⇒10−4=n×6.63×10−34×3×1081000×10−9⇒ n = 5.02 × 1014 = 50.2 × 1013⇒ 50 (nearest integer)

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