How does a discharge UV lamp works

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A discharge UV lamp operates by exciting a rare gas, such as helium, through accelerated free electrons within an electric field. These electrons gain sufficient kinetic energy to cause inelastic collisions with gas atoms, leading to their excitation. The excited atoms then return to a lower energy state, emitting UV photons in the process. These UV photons interact with the lamp's coating, resulting in the emission of visible light. The method of electron generation depends on whether the lamp is a hot or cold cathode, influencing the ionization process and conductivity of the gas.
poul
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I think i have looked at like 20-30 web pages, and i can't find a simple step-by-step explanation of the operational principle.

I know, that when a rare gas like Helium, relaxes from an excited state two characteristic emission lines occur. But how do you excite the gas in the first place?

I know you can excite it, by shooting some electrons through the gas, but we have no free electron in the system, since the electric field should not be strong enough to ionize the helium atom by itself.
 
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There are free electrons in the system. The electrons get accelerated by the electric field. Once the electrons are a high enough kinetic energy, the atoms absorb a discrete quantity from them in an inelastic collision, becoming excited. The unstable, excited atoms quickly fall back to a lower state, emitting a UV photon in the process. This UV photon interacts with the coating of the tube in a similar manner, transferring energy to excited/ground states, emitting visible light.

I read your post again, and I am not sure what you mean. You have no free electrons in a system you are doing? I am not sure how that would work in this case. All gas discharge lamps I know of have free electrons flowing through an ionized gas.
 
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The (electrons in the) atoms can also be excited through inter-atom collsions.
 
TheFerruccio said:
There are free electrons in the system. The electrons get accelerated by the electric field. Once the electrons are a high enough kinetic energy, the atoms absorb a discrete quantity from them in an inelastic collision, becoming excited. The unstable, excited atoms quickly fall back to a lower state, emitting a UV photon in the process. This UV photon interacts with the coating of the tube in a similar manner, transferring energy to excited/ground states, emitting visible light.

I read your post again, and I am not sure what you mean. You have no free electrons in a system you are doing? I am not sure how that would work in this case. All gas discharge lamps I know of have free electrons flowing through an ionized gas.



okay so for a normal gas discharge lamp, we have free electrons you say. Are these made, by heating a filament or similar and then accelerated?
 
poul said:
okay so for a normal gas discharge lamp, we have free electrons you say. Are these made, by heating a filament or similar and then accelerated?

It depends on if it is a hot cathode or a cold cathode. Hot cathode lamps use thermionic emission to release the electrons from the electrodes to maintain conduction in the gas. Cold cathodes need an initial (much higher than hot cathode) voltage to ionize the gas and make it conductive.

Atmospheric gas is typically non-conductive until this "breakdown voltage" is applied. The voltage necessary is reduced in a vacuum and with certain gases, which is why these lamps have the gases they do.
 
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