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.
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[sup]-19[/sup] Joules. So to emit 40 Joules worth of light in a second, the bulb would have to emit about 10[sup]20[/sup] 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.
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 Electro-Optics Handbook, the spectral radiance in photons per second is
n(lambda) = 2c/((lambda)^4)(exp(h(nu)/kT)-1) photons /sec-m^2-steradian-m
(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 Stefan-Boltzman equation
M(Watts per sq. meter) = (sigma)T^4
sigma is 5.6697 X 10^-8 Watt/m^2-K^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[sup]-19[/sup] Joules. So to emit 40 Joules worth of light in a second, the bulb would have to emit about 10[sup]20[/sup] 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.
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[sup]23[/sup]) of photons, so a microeinstein is 6.02 X 10[sup]17[/sup] 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:
A 40 watt cool white fluorescent puts out about 42.4 microensteins per second. That’s 2.55 X 10[sup]19[/sup] photons per second.
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.
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
Interestingly, when you take inefficiency into account, you’ll get more photons, not less. Almost all of the energy which goes into a bulb comes out as light. It’s just that only a small amount of it is visible light. Most of the energy is in infrared light, which has less energy per photon. So you’ll need more photons total to carry the same amount of energy.
Most packaging gives the light output in lumens for the bulb. The OP asked how many photons of light were put out by a light bulb, not how many photons would be contained in the power input.
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.
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.
The page on the lumen (see ‘light’, then ‘light intensity’) also explains the ‘solid angle’ thing with a diagram.
Would it be asking too much for both photons of visible light and all photons?
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 wel–known, 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.
No, some of it will be in the form of heat. Conduction and convection.
Is this true for fluorescent bulbs as well, or just incandescent?
It’ll depend more on the housing, shade, and the like than on the bulb technology itself. Any energy which goes into the bulb will either turn directly into light or into heat (for an incandescent, it’s all into heat). The energy which went into heat will then leave the bulb through one of three mechanisms: Conduction, convection, or radiation. For most light fixtures, radiation would be the dominant form of heat transfer, which would mean that most of the heat energy is leaving via photons (this is in fact the only mechanism by which incandescent bulbs produce photons). What frequencies these photons are produced at will depend on the temperature; for ordinary incandescent bulb temperatures, most of them are infrared.
I’ve looked around some but I haven’t found any data on how much of the heat is convected away. But you are pretty close to right that the majority is radiated in one form or another. Even some of that which is conducted away is emitted as radiation. I’ll withdraw my objection to using power input as a measure of how many photons come from a light bulb.
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.