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If we imagine a object made up of Hydrogen gas that is optically thick to all radiation, and is in thermal equilibrium, then, microscopically, photons will be emitted and absorbed as emission/absorption lines.

However, the overall object should emit radiation according to Planck’s Law, which describes intensity as a continuous function of wavelength (and temperature).

How does this occur and where do the photons we detect at wavelengths between spectral lines of hydrogen come from originally?

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An object made up from a specific single molecule gas? Photons absorbing photons? Of what variable is your non-continous curve a function of? Such uncharged boson numbers (e.g., in the electromagnetic theory, photons) are not conserved btw., so that's no problem. –  NikolajK Dec 6 '12 at 11:09

3 Answers 3

Hydrogen gas is not a Hydrogen atom. During some atomic collisions one has more complicated systems than just a Hydrogen atom, i.e. they have more complicated emission/absorption spectrums. As well, there is a Doppler broadening of lines that smears sharp lines. But the principal reason is, of course, (temporarily) creating complex many-atomic (and even plasma) configurations with richer spectrum. In a thick object there is sufficient number of such complex systems.

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Another point would probably be that an object made up entirely of hydrogen atoms is not necessarily a black body and hence does not necessarily obey Planck’s law. –  Claudius Dec 6 '12 at 11:25
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In that case it cannot be optically thick (i.e., absorb everything). Such is rare gas around some cosmic objects. It demonstrates absorption lines in the spectrum and is transparent "between lines". –  Vladimir Kalitvianski Dec 6 '12 at 11:29
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Ah, correct, thank you very much! :-) –  Claudius Dec 6 '12 at 11:30

I think that there is a simple contradiction in the question: either you define your object as a black body that absorbs and emits every wavelength or you talk about spectral lines.

Hydrogen is not a black body. If you imagine it so then don't be surprised if you reach contradictory conclusions.

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You probably mean a gas of hypothetical atoms which can only interact with photons at one given frequency, i.e. atoms with the only one excitation energy level.

If such excited atom has finite lifetime, which looks required, this automatically means it's energy level have non-zero width. And this means in turn, that absolutely cold gas will emit one non-zero width line.

While going into equilibrium with radiation, such atoms become to move as result of photon propulsion. The more propulsion acts, the more chaotic movements. Finally such gas will be in equilibrium with radiation only when it's Doppler widening of emission lines reach the situation when all widen lines overlapping will coincide with Planck spectrum.

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