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When applying DC to a neon lamp, only the negatively-charged electrode glows:


The voltages across the lamps are left: DC (left lead positive), middle: DC (right lead positive), and right: AC.

But... why? The electrodes are the same shape, so the electric field around them should be the same shape, and the gas should break down in the regions at which the electric field strength is above some threshold, which seems like it would be symmetrical. Is there a difference in threshold between positive and negative coronas? If so, do both sides light up at high enough voltage? Or maybe only one type of corona is possible in neon since it's a noble gas? If it contained air would it glow at both electrodes?

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This question has an open bounty worth +50 reputation from endolith ending in 4 days.

Looking for an answer drawing from credible and/or official sources.

Need a complete answer without contradictory information

Quick guess: Light emission occurs once electrons are fast enough that their energy upon a collision is in the range of visible light. With DC, the electric field points in exactly one direction and hence acceleration occurs in one direction. –  Lagerbaer Aug 15 '11 at 3:42
is the third picture made with AC source? –  troyaner Aug 15 '11 at 13:15
@troyaner: yes. –  endolith Aug 15 '11 at 13:41
I'm with the 'charge carrier' explanation, ie its to do with the movement/KE of electrons, this matches the pictures above. –  Nic Aug 15 '11 at 14:44
The ions are, however, way heavier than the electrons and therefore don't get accelerated that much –  Lagerbaer Aug 15 '11 at 15:35

6 Answers 6

The asymmetry comes from the different masses of electrons and neon ions (neon ions are about 36000 times heavier).

This mass asymmetry results in different cross sections for the excitation of neon atoms by electrons and ions. There are some plots of this here (figure 1a for electrons, figure 1b for ions). The interesting processes of excitation occur above around 10eV for electrons and 100eV for ions.

There is a very cool Java simulation of discharges: http://phet.colorado.edu/en/simulation/discharge-lamps . It covers electron excitation, the atomic structure of neon, acceleration and excitation cross-sections very well, and demonstrates why the glow can be localized (it glows at the point that electrons have been accelerated to the necessary energy), and why it is asymmetric (all electrons start from the cathode and can gain energy on the way to the anode).

I hope this answers why only one side can glow. I wish I could explain exactly why the glow is next to the cathode in the picture, but I would just be guessing.

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Yeah, the applet is backwards from reality. The closer you get to the negatively-charged cathode, the less light there is because the electrons haven't been accelerated yet. Reality: "When driven from a DC source, only the negatively charged electrode (cathode) will glow." –  endolith Aug 18 '11 at 2:39

In the Manitoba Grade 9 curriculum handbook, the following explanation is presented: "The neon bulb emits electrons from the negative electrode which crash into the neon atoms, emitting a reddish-orange glow at the negative cathode." I am presently taking my teacher certification and it seems I am expected to let my Grade 9 students figure this out for themselves by applying the Particle Model of electricity. Does anyone find this explanation convincing?

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This is right, but what urges the electrons to leave the cathode? –  Georg Oct 29 '11 at 10:08

It is the excited gas atoms which produce the glow by losing their excitation via the emission of photons.

The reason there are more excited atoms near the cathode is twofold:

  1. The atoms are principally excited via collisions with fast-moving gas ions and to a lesser extend with other (neutral) gas atoms, NOT with electrons (see above comments re. scattering cross-sections). These ions do form from via electron-atom collisions (principally near the anode where the electrons are moving fastest), and these collisions favor the gas losing an electron and becoming positively charged. (This is related to the cascade effect known as "Towsend discharge".) Thus positive ions get accelerated toward the cathode and are moving most quickly near the cathode, thereby transferring more kinetic energy to other gas atoms when they collide with them. These collided-atoms are thus excited and quickly shed photons.

  2. Sputtering. Some of the ions (and/or atoms they collide with) will impact the cathode itself, which will (with some probability) knock off atoms from the cathode itself and release them into the surrounding gas with in some cases significant velocity, and these cathode-ejecta-atoms in turn will collide with the gas and excite the gas atoms, which then emit photons and so forth.

The basic process is described in the early section of this Wikipedia link.

TL/DR: The electrons from the cathode, rather than being responsible for exciting electrons in the surrounding gas atoms, instead free these gas-atom-electrons, leaving positive ions which are accelerated toward the cathode and produce the glow by colliding with and exciting neutral gas atoms near the cathode. That, and sputtering. ;-)

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As I pointed out in another comment, the talk page of Wikipedia:Glow discharge says "Sputtering of the cathode, on the other hand, is not at all important for a glow discharge. It is merely a side effect." Also the neon lamp article says that sputtering wears the electrodes and darkens the lamp over time by coating the surface of the glass with metal. –  endolith yesterday
To be sure, the contribution due to sputtering is a much smaller effect than the ion-atom collisions. That's what I listed it second. But it's not trivial. The wear on the cathode & darkening of the lamp are however irrelevant to the glow and are mere side effects. –  sh37211 19 hours ago

Well, I guess, in the case of AC too, the lamp is glowing in only one electrode, but since its alternating at 50Hz( depends on your geography/country) the persistence of vision is making you see both of them glow at the same time.

In DC case, the lamp is glowing because of Ne ions hitting the cathode(negative electrode). Its simply a gas discharge tube. wiki

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The cathode itself isn't glowing. The gas around it is. –  endolith Aug 17 '11 at 14:20
Also the Wikipedia talk page says that the ions sputtering the cathode has nothing to do with the glow. –  endolith Aug 17 '11 at 14:28

There is an article at Physics of Plasmas,Phys. Plasmas 19, 072113 (2012); http://dx.doi.org/10.1063/1.4737189, which explains the phenomenon discussed here.

The physical mechanism behind neon glow is induced-dipole interaction. Unfortunately, it is too difficult to explain physics without any mathematics; and, one should refer to the article for details.The figure 3 in the article gives pictorial idea behind charged-particle oscillation.

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Answers should contain an explanation, not just a link. meta.stackexchange.com/a/8259/130885 –  endolith Sep 4 '12 at 13:59
It's unfortunate that even the link is behind a paywall. –  adavid Oct 6 '12 at 23:16

It's because the ion and electron densities are highest near the cathode (generating the glow) and flattening out away from the cathode.

enter image description here

Reference from credible and official source

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"A neon display lamp uses the positive column for its light"? –  endolith 5 hours ago

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