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I am looking for better understanding of how light is produced in an incadescent lamp. More specifically: how is the kinetic energy of electrons converted to light?

  • Are we dealing with interband transitions or with intraband relaxation involving photons? Is this Bremsstrahlung (electrons lose their energy as light when colliding with crystal impurities/defects)? Or is this a thetmal radiation resulting from Joule heating?
  • How is the emission affected by presence of impurities and imperfections of the crystal lattice? Do phonons play a role?
  • What properties make a material more suitable for use as a filament: should it be a metal? Should it have a crystalline structure? Will any metal produce light, if a high current is passed through it in vacuum?

Update
The term describing the processes in the incadescent lamp is thermal bremsstrahlung, see the posts on this subject here and here.

Roger V.
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    You don't need metal for a filament. Read on what materials Edison tested. Like cotton. It's whatever causes blackbody radiation which isn't electron transitions I think because it is rather universal and independent of material. It seems to be due to the acceleration and deceleration of charges as they oscillate due to temperature. – DKNguyen May 29 '21 at 19:24
  • @DKNguyen in other words, you suggest that it is the cherenkov radiation/Bremsstrahlung? – Roger V. May 29 '21 at 19:26
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    What? No. I did not intend to. It's not similar to Cherenkov radiation as I know it but reading about Bremsstrahlung, it seems to be the same thing. – DKNguyen May 29 '21 at 19:34
  • You are right, cherenkov means a different thing, I used a wrong term, since I was thinking in Russian. – Roger V. May 29 '21 at 19:42
  • @DKNguyen, re "...like cotton." Cotton doesn't conduct electricity, and it can't exist at white-hot temperatures. Edison experimented with carbon filaments that he fabricated by heating various different organic fibers (apocryphally, including a hair from the beard of a red-headed Scotsman) in an oxygen-free atmosphere until they decomposed to amorphous carbon. – Solomon Slow May 29 '21 at 21:44
  • @SolomonSlow Yes, but still not a metal. – DKNguyen May 29 '21 at 21:57
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    @uhoh thanks, it was a spelling error. – Roger V. May 30 '21 at 06:34
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    @uhoh for more context: in LEDs the light is due to the interbsnd transitions, but it is unlikely in the incadescent lamp (even though one could think of a scenario where strong electric field excites electrons across the band gap, as in zener effect). – Roger V. May 30 '21 at 06:36
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    The canonical question (with 8 answers) is What are the various physical mechanisms for energy transfer to the photon during blackbody emission? Your question is much more specific and not a duplicate, but answers there may provide a good starting point for a specific answer here, with (the 5 answers to) Quantum mechanics of thermal radiation helpful as well. – uhoh May 30 '21 at 07:18
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    @uhoh thank you for the links - they do address many of the questions that I had. The main point that still needs clarification, in my opinion, is whether we are really dealing with thermal radiation here, i.e., whether the role of electric current is only to heat the material (since the current flow itself is not a thermal state). – Roger V. May 30 '21 at 07:59
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    @RogerVadim thank you for the accept. It's of course your choice, but I usually wait (at least) a few days to allow for more answers, comments and insight. Also as a side-effect of delaying the accept, is that it puts it back in the active queue and generally more eyes will see it which may lead to further answers, comments and insight. But I see you've been around a while so I guess it's not news to you :-) – uhoh May 30 '21 at 08:49
  • @R Also, I've updated the answer with some screen shots based on videos linked in Ruslan's comment. – uhoh May 30 '21 at 08:52
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    @uhoh you actually answered the question that bothered me, and supported it by evidence. I am not good in formulating questions - half of the time they end up closed, massively downvoted, or somebody tries to convince me to accept their answer, because what I asked is not what I meant :) – Roger V. May 30 '21 at 09:11
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    @RogerVadim it comes with practice; I've asked over 3,000 SE questions and it gets easier over time. I usually wait until I'm ready to post before writing the title, and I often repeat the question at the end to make sure where I end up still matches where I began. – uhoh May 30 '21 at 09:15

4 Answers4

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Supplementary answer to the OP's clarifying comment:

The main point that still needs clarification, in my opinion, is whether we are really dealing with thermal radiation here, i.e., whether the role of electric current is only to heat the material (since the current flow itself is not a thermal state

Good question, and...

Yes, we are really dealing with thermal radiation here!

The heating produced by the flow of conduction electrons in the bulk of the filament is not related to the thermal radiation coming from the few tens of atoms near the metal surface that are producing the photons that we see.

We know this because we can do some experiments:

  1. When we turn on or off the current, the light produced ramps up or down with a radiative time scale of milliseconds, which is how much time it takes the filament to heat up or cool down from its previous equilibrium "OFF" or "ON" temperature.
  2. We can try to measure the 100/120 Hz flicker of the bulb, and see that it's perhaps a percent. An incandescent bulb does not turn off 100 or 120 times per second. Its light stays relatively constant. We can make light trails with our eyes or cameras with pulsing light sources like some brands of LED lights (e.g. cheap battery operated portables) or florescent lights or some kinds of mercury or sodium vapor street lights, but we can't reproduce those effects with incandescent lights.

Now, this does not mean that electron collisions in metals can't make visible light, but the chances that a conduction electron can get 2 or 3 eV of kinetic energy before hitting another electron and that that also happens within tens of angstroms of the surface so that the light gets out is extremely small.

Basically the tungsten does two totally separate jobs at the same time:

  • acts as a suitable temperature-dependent resistor such that it reaches thermal equilibrium and radiates 100 watts or whatever power it's supposed to
  • acts as a thermal radiator, producing light when heated

update: @Ruslan's comment links to two excellent videos!

intact incandescent filament glowing

Then it breaks, no current flows, and the light continues but starts to dim:

broken incandescent filament still glowing

When it touches another part of the bulb, that part cools more quickly by conduction than by radiation, so it turns dark. But the bit at the right can't cool easily along the filament because it's thermal conductivity is low along the wire, so it's still glowing fairly brightly:

broken incandescent filament still glowing, especially parts that can't cool by conduction

uhoh
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    Indeed, in this video we can see the flicker is not 0 to max, but rather about 90% to max. And in this one we can see that even after the filament breaks down (so that no current flows anymore) it still continues to glow, although losing brightness due to cooling. – Ruslan May 30 '21 at 08:30
  • @Ruslan excellent! I'll add them in, wonderful find :-) – uhoh May 30 '21 at 08:39
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https://en.wikipedia.org/wiki/Thermal_radiation

Depending on the theoretical frame used it may as well be called Bremsstrahlung (free(-ish) electrons in the metal scattering into each other)

As far as I can imagine, impurities and crystal lattice defects will affect electrical properties in the first place.

Do phonons play a role? Not sure, I think electrons dominate the heat exchange in metals. One needs heat exchange in order to bring heat to the surface of the filament. You may as well think about phonons scattering electrons (in other words, the crystal lattice and the electron gas exchanging heat).

Should it be metal? Not really. But it should be at least somewhat conductive for the electricity.

Yes, any metal (and any solid substance in general) will produce light when heated in vacuum (or in any transparent media) as long as it stays solid. One needs some 750K in order to produce some faint visible light or ~3000 to look like a normal incandescent bulb.

The best material for a incandescent lamp filament will be:

  • more or less conductive so it can be heated by electricity
  • be absolutely reflective or transparent for non-visible electromagnetic waves and black for the visible spectrum.
  • stable against decomposition, melting or evaporation at the desired temperature (equal to the desired color temperature of the lamp, for most practical purposes 3000..6000K)

Since we don't have the ideal material, we use tungsten and we try hard to make it evaporate slower, using inert gases and halogens. On the other hand, gases make bulbs less effective because they carry away some of the heat from the filament. That's why we have better or worse gases for filling the bulbs.

Other materials were used or considered in the past, like tantalum or carbonized natural filaments like cotton and wool.

fraxinus
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    When talking about Bremsstrahlung, I meant that the electrons, accelerated by the electric field, lose their speed by hitting impurities and radiate during this collisions. Thermal radiation implies somewhat different scenario: the electron scattering agsinst impurities results in heating and the resulting radiation is no different from that coming from any heated object (i.e., not necessarily heated by electric current.) – Roger V. May 30 '21 at 06:08
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    I'm not sure how you've decided that it's electron-electron scattering (Bremsstrahlung) between free electrons that produces the photons we see as incandescent light for tungsten, rather than band transitions. Is it because it's a metal and so has low emissivity? What about dark (high emissivity) insulators with very few free electrons, would that emission be a different temperature? I think this answer needs some support, except for the first sentence the whole thing is unrelated to photon production. So a temporary, reversible -1 until this is addressed more definitively. Thanks! – uhoh May 30 '21 at 06:18
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    @RogerVadim the filament radiates thermal radiation a great deal independent of the mechanism of heating. Radiation happens at the surface layer and heating happens in the whole volume of the filament - the processes are pretty much independent. What you imagine (scattering of the electrons accelerated by the electric field to directly produce light) happens pretty much in gases (see e.g. glow discharge) and because of this mechanism you get neither thermal spectrum nor ohmic resistance. – fraxinus May 30 '21 at 07:23
  • @uhoh that's why I said "depending on the theoretical frame used". The emission process in a metal can be viewed pretty much as a Bremsstrahlung (free-free scattering) as well as a band transitions. In an insulator, well... you can't use the free-free (plasma-like) approach, because you don't have something resembling plasma there. – fraxinus May 30 '21 at 07:46
  • Is it possible to add a citation or link supporting that "...(t)he emission process in a metal can be viewed pretty much as a Bremsstrahlung (free-free scattering) as well as a band transitions." That's not transparent to me (oblique pun intended) and this is Stack Exchange, so some kind of additional support through explanation or source-citing of why that's so would be very helpful. Thanks! – uhoh May 30 '21 at 07:49
  • I guess radiative recombination is the word normally used for the band transitions that produce light, but I wonder, could be looked at as electron-hole scattering and therefore Bremsstrahlung-like? – uhoh May 30 '21 at 07:54
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    @fraxinus I accepted the answer by at_uhoh, because it addressed the specific point that bothered me the most. However, I acknowledge that your answer is very complete and covered most of the questions in the OP (and you also addressed the difficult point in the comment above.) Thank you. – Roger V. May 30 '21 at 08:43
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    @uhoh holes? What holes in a metal? p.s. mea culpa, what I meant is a transition INSIDE the conduction band. – fraxinus May 30 '21 at 12:45
  • @fraxinus it's hot metal, not absolute zero, so there are holes. And if we're saying inter- or intra-band transitions (from the OP's question) are on the table, then the electron better go into some state where there is not already an electron. If you feel that hot metal can not have any bands with holes that are 2-3 eV lower than some electron, then you can immediately rule out the inter- or intra-band transitions. – uhoh May 30 '21 at 13:49
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The only requirement for radiation to occur, in insulators or conductors, is acceleration of charges or magnetic fields. Bound electrons surrounding a nucleus can be stimulated to radiate by thermal agitation of the nucleus. Rotations, vibrations etc. All atoms have either dipole or multipole magnetic moments, these will also radiate when agitated by heat. This is thermal spectrum radiation, under certain conditions it can have a "black body spectrum" Obviously an electron transitioning from one, non radiating stable atomic state, to another non radiating stable atomic state, will also radiate briefly, this form of radiation is the source of line spectra.

barry
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  • Thanks. In other words, you think that the role if the electric current here is only to heat the material? – Roger V. May 30 '21 at 08:00
  • That is correct. A candle flame radiates because the bluish flame heats the carbon particles to almost white. – barry May 30 '21 at 08:47
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I would like to address something the other answers do not mention, that is, how the electric current heats up the filament on the molecular level and why it can store this energy and why it keeps glowing after you turn it off (no current).

A proportion of the collisions result in excitation of the metallic electrons to higher energy levels, which may produce light emission upon returning to the lower stable energy level. Continuous collisions between electrons produce a resistance to the flow of the mobile electrons, and atoms of the filament are induced to vibrate by the interaction with the moving electrons. The vibrational energy results in the production of a significant amount of heat, and a characteristic of resistive filament lamps is that only about ten percent of their energy input is turned into light, most of the remainder being emitted as heat (infrared electromagnetic radiation).

https://micro.magnet.fsu.edu/primer/java/lightsources/filament/index.html

As you can see, we are dealing with thermal radiation, and the electrons' (in the current, as they scatter off/collide with the molecules) kinetic energy is transferred to the molecules:

  1. vibrational (this is the most important in your case)

  2. rotational

  3. translational energies.

Please note that there are electronic transitions too but I am not mentioning those.

For a molecule with N atoms, the positions of all N nuclei depend on a total of 3N coordinates, so that the molecule has 3N degrees of freedom including translation, rotation and vibration.

https://en.wikipedia.org/wiki/Molecular_vibration

Now when you turn the the lamp off (no current), the molecules are still storing the extra energy that was transferred to them by the current's electrons' (by way of scattering/ collisions), and as the filament tries to reach thermal equilibrium with the environment (cool off), this extra energy causes the molecules in the filament to keep relaxing by emitting photons (the molecules relax to a lower energy level by emitting photons, including the visible range).

Árpád Szendrei
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    Szendrei....I don't believe radiative transitions occur in a tungsten filament at 3000 degrees. If it were so, we would see line spectra in the radiation. The spectrum is a continuous black body spectrum with no discernible narrow band lines. The radiation is induced by mechanical, anharmonic, Newtonian motion. The reason the filament glows after the current is off is simple thermal inertia. Your skin is currently giving off black body radiation by this mechanism as we speak, obviously at a lower temperature. Even liquid helium will radiate a black body spectrum. – barry May 31 '21 at 04:11
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    @barry correct, thank you, the excitation is (as you say induced by) the scattering of electrons. What I am describing is the relaxation of the molecules, that is called radiative transition. But I will edit to clarify. – Árpád Szendrei May 31 '21 at 04:17
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    @ÁrpádSzendrei Thank youf or this addition. Indeed, the mechanism of the energy transfer between the current and the filament is a part of my question. What confuses me is that I think of a current-carrying material as a metal, i.e., a crystalline solid. So the heat transfer is mainly due to the electrons, although lattice vibrations may play a role, and it is not quite clear what are the "molecules" emitting light in this case. – Roger V. May 31 '21 at 07:06