Why do manufacturers use a gas that emits UV light in fluorescent light bulbs?

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Manufacturers use mercury vapor in fluorescent light bulbs because it efficiently produces ultraviolet (UV) light, which is necessary to excite the phosphor coating inside the bulb to emit visible light. Other gases, like neon, do not produce a full spectrum of light and require higher voltages, making them less practical for general lighting. While it is theoretically possible to mix gases to achieve white light, the technical challenges make it impractical compared to using mercury. High-pressure gas options, such as high-pressure sodium lamps, can improve light quality while maintaining efficiency. The discussion also touches on the ionization process in gas discharge, emphasizing that free electrons primarily come from the cathode and collisions within the gas generate UV emissions.
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According to Wikipedia, in fluorescent lamps "[a]n electric current in the gas excites mercury vapor which produces short-wave ultraviolet light that then causes a phosphor coating on the inside of the bulb to glow." Why don't manufacturers use a gas that emits visible light directly when excited by the electric current?
 
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They do. Those are called neon lights. (Or xenon, or sodium vapor...)
 
Other gasses have 2 problems. They don't produce white light and they generally need much higher maintaining voltages. They use mercury because, after warm up, it has a lower impedance so needs less voltage. The high energy UV, or at least blue, photons are needed to excite the phosphors thus making something resembling white light.
 
To elaborate on Drummin's first point, when a gas is excited by an electric current, it will emit discrete wavelengths of light, not a full spectrum of wavelengths. For example, a neon light is typically a very specific color because the gas inside emits primarily at that wavelength and few others. It might be possible to use a combination of gases that emit different wavelengths in order to achieve "white" light, but unfortunately there are many technical details which preclude this from being practical. It is simply much more practical to use a mercury vapor and convert the UV light to visible light using a coating than to mix different gases to make white light.

You can read more here: http://en.wikipedia.org/wiki/Fluorescent_lamp#Principles_of_operation
 
A gas discharge, as people have said, produces a line spectrum (disgusting for illumination) but efficient. This is a major factor when you need to illuminate large outdoor areas and you cannot afford to be fussy about the aesthetics. One way to ameliorate the colour problem is to use a high pressure gas (pressure broadening). High pressure sodium lamps are better because the spectrum is spread out more and they are still more efficient than tubes using phosphors.
I think that phosphor based lighting must just cost more to run.
 
Thanks for all the answers.

Quick follow-up question: in neon lighting, does the neon become a plasma, or are the electrons in the neon atoms just bumped to a higher energy level?
 
I think they are mostly just bumped up to higher energy levels.
 
Drakkith said:
I think they are mostly just bumped up to higher energy levels.

IF the gas is conducting then you have to assume there are plenty of free electrons bumping about, surely (?).
 
sophiecentaur said:
IF the gas is conducting then you have to assume there are plenty of free electrons bumping about, surely (?).

My thinking was that most of the free electrons came from the cathode, not the gas. Besides, at such a low pressure, is the gas really "conducting"?
 
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The filaments are only heated whilst the tube is starting up. Thereafter, I don't think they are hot enough for thermionic emission. In any case, the electrons do not travel far on each cycle of AC mains so I reckon a fair proportion of the gas atoms must be ionised and provide the energetic electrons. The collisions of these electrons against atoms produce the UV emissions. The energy of UV photons will be in the region of 10V or more which could imply a drift distance of perhaps 5% (1/20) of the tube length between collisions.


Besides, at such a low pressure, is the gas really "conducting"?
The 'resistance' of the discharge is so low that a choke / ballast is needed to limit the current. There will be an avalanche effect as electrons cause multiple ionisations on impact. This is how I remember the process happens.
 

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