When a hydrogen ($H_2$) lamp is subject to a high voltage hydrogen atoms release energy in the form of light. Somehow $H_2$ is splitted. But...

  • Why high voltage produce the dissociation? Does electrons travel through the gas from one electrode to the other one (as electric discharge suggest)? So is $H^-$ resposible for the emission?


Here I'm citing the first link:

For example, when a high-voltage electrical discharge is passed through a sample of hydrogen gas at low pressure, the resulting individual isolated hydrogen atoms caused by the dissociation of H2 emit a red light.


It seems you are really asking why electric discharge can cause breaking of covalent bonds.

To answer that, it's useful to think about how electric discharge causes ionization. When the tube is 'struck' by application of high voltage, the strong electric field literally pulls electrons away from their nuclei. Once free, the electrons can now accelerate in the field, and they reach relatively high energies, interacting with other atoms and breaking free their electrons too.

Now, about breaking of bonds. Fundamentally, breaking of bonds is similar to how ionization happens - the electrons that form the covalent bond and are shared between atoms jump into higher states and leave the atoms entirely. Without the shared electron(s), the atoms are no longer 'attached' and may separate.

How an H2 lamp operates is by breaking the molecular bonds and exciting the resulting radicals. A deuterium lamp tries to reduce the amount of bond-breaking, so that only de-excitations from molecules produce light. This limits the spectral range to the UV.

  • $\begingroup$ If there were only vacuum, you'd have no way to establish an electric field. To establish an electric field, you need some concentration of particles with electric charge. If you gather a lot of charged particles in one place, they are eventually going to want to fly apart. In a normal charged metal, above a certain local field gradient, electrons will be emitted from the surface. $\endgroup$ – Al Nejati Sep 16 '18 at 1:25
  • $\begingroup$ Didn't thomson realize there was charged particles by 'seeing electrons' going one electrode to the other? $\endgroup$ – santimirandarp Sep 16 '18 at 1:27
  • $\begingroup$ So it is called 'electric discharge' but the is no electric discharge, isnt it? xd $\endgroup$ – santimirandarp Sep 17 '18 at 2:50

Lamps based on gases are not based in molecule splitting. That'd release a lot of heat, but not much light. What's more, there are many noble gases-lamps, which can hardly be splitted.

No, the mechanism is deexcitation of electrons.

The high voltage adds energy to the electrnos, which rise to upper levels (they are higher energy ones). Then, the electrons undergo a deexcitation, which emits light in form of photons.

The frequencuy of light is $$\nu=\frac{\Delta E}{h}$$ where $h$ is the Plack's constant.

This means that each transition emits a different colour (because they correspond to different energy differences, and so different frequencies). The mixture of red, blue, violet, and so on produces that "white-like" light, but it is not pure white.

  • $\begingroup$ Thanks, I know that but that's not the question. Please, look at chem.libretexts.org/Textbook_Maps/General_Chemistry/… first paragraphs. $\endgroup$ – santimirandarp Sep 15 '18 at 23:20
  • $\begingroup$ So what is the question? $\endgroup$ – FGSUZ Sep 15 '18 at 23:28
  • $\begingroup$ If you look at the link, the lamp splits hydrogen $\endgroup$ – santimirandarp Sep 15 '18 at 23:29
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    $\begingroup$ Hmm, now I see. I think that is is probably because dissotiation of $H_2$ means breaking a covalent bond, which results in electrons returning to its original position; but the mechanism is that one: light is only emmited when photons are released in transitions. $\endgroup$ – FGSUZ Sep 15 '18 at 23:30
  • $\begingroup$ Now please re-read the first question... $\endgroup$ – santimirandarp Sep 15 '18 at 23:32

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