32
$\begingroup$

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

http://commons.wikimedia.org/wiki/File:Neonlamp3.JPG

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?

Do neon signs work in a different manner, since they have a long region of glowing gas, rather than just glowing near the cathode?

Improve this question
$\endgroup$
11
  • 5
    $\begingroup$ 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. $\endgroup$ – Lagerbaer Aug 15 '11 at 3:42
  • $\begingroup$ I'm with the 'charge carrier' explanation, ie its to do with the movement/KE of electrons, this matches the pictures above. $\endgroup$ – Nic Aug 15 '11 at 14:44
  • $\begingroup$ @Nic: Aren't there also neon ions carrying charge in the opposite direction of electrons? Is it a difference in mass thing? $\endgroup$ – endolith Aug 15 '11 at 14:47
  • $\begingroup$ Actually, I guess this is called "glow discharge", not "corona discharge". Not sure if they are very different, or if one is a subset of the other. $\endgroup$ – endolith Aug 15 '11 at 14:49
  • 1
    $\begingroup$ The ions are, however, way heavier than the electrons and therefore don't get accelerated that much $\endgroup$ – Lagerbaer Aug 15 '11 at 15:35
12
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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." $\endgroup$ – endolith Aug 18 '11 at 2:39
  • $\begingroup$ Combining you answer with sh37211's, it seems the electrons are accelerated toward the anode, where they have sufficient energy to ionize gas atoms, and then the gas ions are accelerated toward the cathode, until they gain enough energy to excite neutral gas atoms in collisions, and those neutral gas atoms relaxing to non-excited state is what emits most of the light, so the ions being faster at the cathode end is why there is more light there? Maybe the spectrum varies from end to end because the neutral atoms are excited to different states? $\endgroup$ – endolith Dec 2 '14 at 16:45
2
+50
$\begingroup$

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 extent with other (neutral) gas atoms, NOT with electrons (see above comments regarding scattering cross-sections). These ions do form 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 the Glow Discharge Wikipedia article.

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. ;-)

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ – endolith Nov 27 '14 at 22:43
  • $\begingroup$ 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. $\endgroup$ – sh37211 Nov 28 '14 at 5:55
1
$\begingroup$

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?

Improve this answer
$\endgroup$
1
  • $\begingroup$ This is right, but what urges the electrons to leave the cathode? $\endgroup$ – Georg Oct 29 '11 at 10:08
0
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ The cathode itself isn't glowing. The gas around it is. $\endgroup$ – endolith Aug 17 '11 at 14:20
  • $\begingroup$ Also the Wikipedia talk page says that the ions sputtering the cathode has nothing to do with the glow. $\endgroup$ – endolith Aug 17 '11 at 14:28
0
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Answers should contain an explanation, not just a link. meta.stackexchange.com/a/8259/130885 $\endgroup$ – endolith Sep 4 '12 at 13:59
  • 1
    $\begingroup$ It's unfortunate that even the link is behind a paywall. $\endgroup$ – adavid Oct 6 '12 at 23:16
0
$\begingroup$

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

Improve this answer
$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.