Only a fraction of the electrical energy supplied to an incandescent-tungsten li
ID: 534466 • Letter: O
Question
Only a fraction of the electrical energy supplied to an incandescent-tungsten lightbulb is converted to visible light. The rest of the energy shows up as infrared radiation (that is, heat). A 60-W lightbulb converts about 15.0 percent of the energy supplied to it into visible light. Roughly many photons are emitted by the lightbulb per second? (1 W = 1 J/sec) To get you started, you should decide what wavelength you will use for a tungsten lightbulb. Remember, the tungsten lightbulb would be somewhere in the visible range. After deciding what wavelength will be used you should try to calculate the energy of ONE photon which corresponds to that wavelength. After you get to this point, it really just is a matter of plugging in values to the appropriate equations. This question should be fairly easy if you UNDERSTAND the meaning of the equations you're using and, most importantly, the units associated with each variable.Explanation / Answer
Take the average of the wavelengths in visible range i.e 400-700 nm to find number of photons
in the range
= (400 + 700) / 2 = 550 nm ; This will be your wavelength for calculations
energy that is converted to visible light,
0.15*60W = 9 J/sec
For 1 sec 9J of energy is converted to light
E = hc / = [ 6.63*10-34 J.sec * 3*108 m /sec ] / 550*10-9 m
E = 3.6163*10-19 J
1 photon produces 3.6163*10-19 J
? photon produces 9J
Cross multiply to get the answer [9J * 1 photon / 3.6163*10-19 J] = 2.4886*1019 photons
2.4886*1019 photons are emitted by lightbulb per second.
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