4. A bulb of 40 W is producing a light of wavelength
620 nm with 80% of efficiency then the number
of photons emitted by the bulb in 20 seconds are
(ley = 1.6 x 10-19 J, họ = 12400 eV )
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Answer:
2 × 10`21.
Explanation:
Power of a bulb (P) = 40 W (Given)
Wavelength of light (λ) = 620 nm (Given)
Efficiency = 80% (Given)
Energy emitted by a bulb = Power x Time(s) = 40 x 20 = 800 J
Energy of photons being emitted by a bulb = (80/100) x 800 = 640 J
Energy of a photon = hcλ
=12400 ×1.6×10`−19×10`−10
= 620 × 10−9
= 3.2 ×10−19 J
Number of photons emitted = Total energy emitted × Energy of the photon
= 640 3.2×10−19
= 2 × 10`21
Thus, the number of photons emitted by the bulb in 20 seconds are 2 × 10`21.
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