Practice portal › Dual Nature of Matter and Radiation › Photon Energy, Number of Photons and Intensity
Asked in GUJCET 2008 · Counting photons
Given: P = 60 W, λ = 660 nm = 6.6×10⁻⁷ m, h = 6.6×10⁻³⁴ J s, c = 3×10⁸ m s⁻¹.
Idea: photons per second n = P/E, where each photon carries E = (hc)/λ (all 60 W taken as emitted at this wavelength).
E = (6.6×10⁻³⁴×3×10⁸)/(6.6×10⁻⁷) = 3×10⁻¹⁹ J
n = (60)/(3×10⁻¹⁹) = 2×10²⁰ s⁻¹
So the bulb emits 2×10²⁰ photons per second.
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