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

Question

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?

Explanation / Answer

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:

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.

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. ;-)

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