2) The conversion efficiency of a lightbulb represents the fraction of the elect

Question

2) The conversion efficiency of a lightbulb represents the fraction of the electrical power that the bulb converts into visible light. It is thus a measure of the effectiveness of the bulb as a light source. Due to their greater efficiency in converting electrical power into visible light, compact fluorescent lightbulbs (CFLs) have replaced incandescent lightbulbs (Ws) as light sources. Consider a 100. watt ( = 100. J/s) lightbulb. Calculate the number of photons per second emitted by a) A CFL (conversion efficiency = 8%) b) An IL (conversion efficiency = 2%) For ease in calculation, assume that all of the emitted photons have a wavelength lambda = 500. nm, roughly the center of the visible region of the spectrum.

Explanation / Answer

Energy of each photon = h*f = 6.63*10^-34 * 500 * 10^-9 = 1.98*10^-40 Joules

1) Total Energy emitted by CFL in one sec = 100 J.

Total Energy emitted in visible spectrum = 0.08*100 = 8 Joules

Number of photons emitted in visible spectrum = 8 Joules/1,98*10^-40 ~ 4 * 10^40

2) Energy emitted by IL in visible spectrum in one sec = 0.02*100 = 2 Joules

Number of photons emitted in visible spectrum = 2 / 1.98*10^-40 ~ 10^40

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