A 100 W incandescent light bulb converts approximately 2.5% of the electrical en

Question

A 100 W incandescent light bulb converts approximately 2.5% of the electrical energy supplied to it into visible light. Assume that the average wavelength of the emitted light is ? = 510.0 nm, and that the light is radiated uniformily in all directions.

1. How many photons per second, N, would enter an aperture of area A = 10.0cm2 located a distance D = 3.0 m from the light bulb?

2. Suppose, instead, that the average photon wavelength is 765.0 nm. How many photons per second, N', would enter the aperture?

Explanation / Answer

E = n h f energy of n photons P = E / t = N h f where N = n / t (photons / sec) P = 100 * .025 = 2.5 J / sec since only 2.5 W go into visible light 2.5 = N h f = N h c / y (using y for wavelength) Solve to get N photons / sec radiated by bulb F = a / A = 100 cm^2 / (4 pi * 9 m^2) = .01 / 113 = 8.84 * 10E-5 where F is the fraction of the radiation passing thru the aperature Photons thru aperature = F * N Part II. Just substitute 7.65 * 10E-7 for y

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