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The number of photons emitted per second by a bulb of power 66 W emitting waves of wavelength 600 nm is ______. (Take h = 6.6×10⁻³⁴ J s.)

Asked in GUJCET 2023 · Counting photons

Answer: (4) 2×10²⁰

Step-by-step solution

Given: P = 66 W, λ = 600 nm = 6×10⁻⁷ m, h = 6.6×10⁻³⁴ J s, c = 3×10⁸ m s⁻¹.

Idea: photons per second N = P/E, where each photon has E = (hc)/λ.

E = (6.6×10⁻³⁴×3×10⁸)/(6×10⁻⁷) = 3.3×10⁻¹⁹ J

N = (66)/(3.3×10⁻¹⁹) = 2×10²⁰ per second

So the bulb emits 2×10²⁰ photons per second.

Why the other options are wrong

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