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#1




how many photons emitted from a light bulb
Lets say its a standard office type fluorescent bulb about 4 ft long. I don't know how many watts....how about whatever is standard 40? 60? How many photons would be emitted by such a white light source per second?
Is it easy to figure out...If so how would one do this? Estimates are fine, I just want a general ball park number that has some credible thought behind it. 
#2




For a rough ballpark estimate: if your bulb uses 40 watts, that's 40 Joules in a second. A photon of visible light has a wavelength of around 500 nanometres; since the energy of a photon is given by E = hc/l, where h is Planck's constant, c is the speed of light, and l is the wavelength, each photon has... ::: calculates ::: about 4 x 10^{19} Joules. So to emit 40 Joules worth of light in a second, the bulb would have to emit about 10^{20} photons in a second.
This is a very rough calculation, of course. To refine it, you'd have to take the efficiency of the bulb into account (not all 40 watts go into light energy), and look at the spectrum of the bulb to figure out exactly what the average photon energy is. I suspect that these factors might cause the final answer to differ by a factor of 10 or so, but not much more than that. 
#3




Actually, incandescent bulbs are easier. You can ignore the variations in emissivity of the tungsten filament and trweat it like a pure blackbody radiator. According to the RCA ElectroOptics Handbook, the spectral radiance in photons per second is
n(lambda) = 2c/((lambda)^4)(exp(h(nu)/kT)1) photons /secm^2steradianm (p. 36) That gives you the number of photons per wavelength increment. You gotta integrate over wavelengths to get the total number of photons. It's easier to get the radiant emmittance integrated over all wavelengths and angles in terms of power rather than photons  it follows the simple StefanBoltzman equation M(Watts per sq. meter) = (sigma)T^4 sigma is 5.6697 X 10^8 Watt/m^2K^4 
#4




For a rough ballpark estimate: if your bulb uses 40 watts, that's 40 Joules in a second. A photon of visible light has a wavelength of around 500 nanometres; since the energy of a photon is given by E = hc/l, where h is Planck's constant, c is the speed of light, and l is the wavelength, each photon has... ::: calculates ::: about 4 x 10^{19} Joules. So to emit 40 Joules worth of light in a second, the bulb would have to emit about 10^{20} photons in a second.
This is a very rough calculation, of course. To refine it, you'd have to take the efficiency of the bulb into account (not all 40 watts go into light energy), and look at the spectrum of the bulb to figure out exactly what the average photon energy is. I suspect that these factors might cause the final answer to differ by a factor of 10 or so, but not much more than that. 


#5




counting photons
This site has a table of the photon emission rates of various lamps expressed in microeinsteins per second. An Einstein represents 1 mole (6.02 X 10^{23}) of photons, so a microeinstein is 6.02 X 10^{17} photons. Being a plant site, the values in the table only count photosynthetically active radiation, so there'll be some photons missed at the red and blue ends of the spectrum:
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#6




So far everyone has based their answer on the electrical power input to the bulb. As soon as I can get around to it I'll try to figure out how many photons are in the light output in lumens.

#7




You're referring to efficacy, which is usually an empirically derived number for real world purposes, but can be theoretically approximated in many cases. Here's a page with some good random lighting links and info.
The page states a 100W incandescent has an efficay of 17.1 lumens/watt Light bulb links galore 
#8




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#9




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So far I've managed to dig up that 1 candela is 1/643 W/unit solid angle, and 1 candela is 4Pi lumens. Now as soon as I can figure out just what unit solid angle they are talking about (1 steradian?) the rest is a downhill pull. 


#10




But all of the power output of an incandescent is in photons. The OP wasn't clear about 'visible light' or not.
It's still a worthy calculation if you're doing it The candela takes into account the sensitivity of the eye. From here. Quote:

#11




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Would it be asking too much for both photons of visible light and all photons? 
#12




Your question is complicated by the fact that photons of different color (wavelength) have different energies. So you can't simply convert output in Watts to numbers of photons without knowing how many photons of which color are present. In the case of blackbody radiation the relativer components of each color are welknown, but in other cases, like a fluorescent lamp with visible phosphors, the problem become more complex. You gotta know how much light of each color is present.

#13




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#17




Continuing what David Simmons & Chonos were just discussing, the filament gives off almost 100% of it's energy input as photons. Very little heat gets conducted back into the base of the bulb down the filament supports. Now when those photons hit the frosted glass globe, a bunch of them get absorbed and converted to heat that convects or oonducts.
So the answer depends a bunch on whether we're talking about filament output or bulb output.
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