![]() In most discharge tubes you are supplying energy by accelerating electrons and ions with electric fields. ![]() $^1$ This is only really true for thermal excitation. If in thermal equilibrium, then the radiation it produces will be continuous and approximate to a blackbody spectrum. It is thus opaque to radiation at a wide range of frequencies corresponding to the continuum of transition energies between these states. The solid filament already has a continuum of conduction band energy states available to populate and depopulate. I doubt there is enough hydrogen gas in a discharge tube to do this, since by design they are meant to display clean line spectra.Ī filament lamp is different. ![]() If you then ramp up the density of the gas, the recombination continuum would get stronger, gradually filling in the gaps between the emission lines until you approached a blackbody spectrum. That could be done (and is done in the Sun for example) by raising the temperature and having significant amounts of hydrogen photoionisation, which then allows a photorecombination continuum (in practice this actually involves H $-$ ions in the Sun). To get it to emit as a blackbody (why would you want to do that - you use discharge lamps to see the line spectrum?) you need a source of absorption over a continuum of wavelengths and you need to make sure there are enough absorbers (by making the gas dense, or having a large amount of gas) that it is opaque to radiation. However, it cannot do the latter - it could absorb radiation at the discrete frequencies governed by the hydrogen energy levels, but is otherwise quite transparent (by design). The hydrogen discharge lamp can do the former - the hydrogen energy levels are populated according to the Boltzmann factors $^1$ and there may even be a little ionisation. To be a blackbody a source must be in thermal equilibrium and be capable of absorbing all radiation that is incident upon it.
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