A 100 W incandescent light bulb converts approximately 2.5% of the electrical en
ID: 1304141 • Letter: A
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 ? = 520 nm, and that the light is radiated uniformily in all directions.
E =
How many photons per second, N, would enter an aperture of area A = 5cm2 located a distance D = 4 m from the light bulb?
N =
Suppose, instead, that the average photon wavelength is 780 nm. How many photons per second, N', would enter the aperture?
N' =
100 WExplanation / Answer
E = hc/? = 6.626x10E-34x3E8/520E-9
= 3.82x10-19 J
no. of photons = Energy generated per second/photon energy
= 6.5x1018 photons
now if wavelength = 780 nm then no. of photons = 6.5x1018x780/520
= 9.75x1018 photons
no. of photons will increase as eergy per photon is less as compared to before, because of the increase in the wavelength of photon(energy is inversly proportional to the wavelenght)
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