We know, via Wien's law, that a body at 6000K emits an electromagnetic wave at the peak wavelength in the visible spectrum.

How come say the fluorescent tubes which also emit the EM waves that we can see as visible not 6000K?

up vote 26 down vote accepted

Black body, by definition, produces thermal radiation only, which is an EM radiation caused by heat. For such radiation, the temperature of a body defines its radiation spectrum and its peak.

The EM radiation in fluorescent tube is not due to heat, but due to fluorescence, which is a type luminescence, defined as emission of light not caused by heat, but by other processes.

More specifically, in a fluorescent tube, UV photons are emitted by mercury vapor atoms, excited by fast moving charge carriers (sort of electroluminescence), and then visible light photons are emitted by phosphor coating atoms, excited by UV photons (fluorescence). Both steps here are forms of luminescence, not thermal radiation.

Since fluorescent light is not due to thermal radiation, its temperature is not governed by black body radiation curves. Therefore, even though most of the EM radiation emitted by a fluorescent tube is in the visible light spectrum, its temperature is very low.

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    Your definition of blackbody radiation is not correct. The definition is not to do with the motion of particles. However, the gist of your answer is certainly correct. – Rob Jeffries Aug 18 at 16:44
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    @RobJeffries Thanks for your feedback. According to Wikipedia, "black-body radiation is the thermal electromagnetic radiation" and "thermal radiation is electromagnetic radiation generated by the thermal motion of charged particles in matter." So, I am not sure what I am missing here. – V.F. Aug 18 at 16:53
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    The wikipedia you are quoting form is that about thermal radiation - not blackbody radiation. Thermal radiation is a necessary, but insufficient condition to produce blackbody radiation. I also think that the definitions you quote are poor. For example thermal radiation can be produced by systems of atoms in thermal equilibrium. Whilst we could consider this as due to the "thermal motion of charged particles" it is not how quantum mechanics views it. Spontaneous emission must certainly form part of the thermal radiation from atoms. – Rob Jeffries Aug 18 at 17:11
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    @RobJeffries Got it. Thanks again for your comments. – V.F. Aug 18 at 17:25
  • This is correct in spirit but has a bunch of small issues. You seem to think the only thing that counts as thermal emission is that part due to the kinetic energy of the atoms, which is simply false. – knzhou Aug 18 at 17:43

A cylindrical fluorescent tube, say 1 m long and with a cross-sectional diameter of say 3 cm would have an emitting area of 0.094 m$^2$, and if emitting blackbody radiation at 6000 K, would emit $P = \sigma A T^4 = 6.9\ \mathrm{MW}$!

If you looked at such a blackbody, then it would be as bright as the surface of the Sun and you would damage your eyes.

Clearly, the light emitted by fluorescent tubes does not have the spectrum of blackbody radiation.

It would be totally impractical to make a lamp that emitted blackbody radiation at 6000 K for standard use. The materials involved would have to withstand a temperature of 6000 K and you would have to make it out of something that was optically thick to radiation (so probably not a gas on this size scale) and capable of being kept in thermal equilibrium at this temperature.

The light emitted by fluorescent tubes is not a continuum, it consists of a set of bright fluorescent lines at discrete wavelengths. In a standard fluorescent lamp it is an electrical current, rather than the temperature of the gas, that excites atoms and ions in the low-pressure gas into higher energy levels. These atoms and ions then radiate short wavelength radiation that is absorbed by a "phosphor" that coats the wall of the tube and this then re-radiates the energy in the visible part of the spectrum.

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    "It would be totally impractical to make a lamp that emitted blackbody radiation for standard use." - I would suggest adding the caveat "at that temperature" right up front, in that sentence, before getting into the explanation. After all, we do have incandescent light bulbs that emit blackbody radiation, just not at 6000K. – David Z Aug 19 at 4:42

Why V.F.'s answer covers well the fact that fluorescent light is not black body radiation, you should also note that it's highly impractical (some might say nearly impossible) to make black body (incandescent) light sources at 6000K. There just aren't materials that can handle that kind of temperature. It's probably possible with magnetic containment or perhaps under very high pressure, but it's not something you could use in a light bulb.

I assume you are asking why fluorescent tubes are not 6000K yet emit visible light.

The reason is that black-body radiation is not the only possible source of light. There are other mechanisms that result in light emission.

Fluorescent lights use chemicals that can absorb higher frequency light and re-emit at a lower frequency.

  • Blackbody radiation is not a "mechanism". – Rob Jeffries Aug 18 at 16:45
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    @Rob: The word radiation is the correct word for both the process and the (dual wave/particle) resulting from that process. – Ben Voigt Aug 18 at 21:52

protected by ACuriousMind Aug 19 at 19:29

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