Radiation intensity emitted by box full of hydrogene?

Question

We know to calculate the intensity emitted by a black body (total or inside a frequency interval) by Planck's formula. A realization of such is usually a cavity of constant temperature containing almost no matter.

But what do we do if we want to calculate the intensity emitted by matter? F.e. a box of hydrogen at constant temperature and are not fine with a black body approximation?

How could we get at least the intensities of the Hydrogene lines?

Thanks!

Answer 1

I think you are thinking of an idealized question: If you really put hydrogen in a box, a real box will itself have to come to the same temperature T as the hydrogen gas and the radiation field by interacting with them both, and you will then definitely get the BBR result.

So, I think you are interested in a thought-experiment (Gedanken experiment), an approach which has an illustrious track record not only in relativity but also in thermodynamics: We imagine a box that only restricts the motion of the gas and the radiation within a spatial volume, without in any way exchanging energy with them. So it must be 100% reflective, never move or vibrate, etc. Just be impervious.

Even so, I think there are still two issues:
a) the discreteness of the energy levels of the hydrogen atom; and
b) the implications of the Doppler effect in broadening those energy levels.

To explore both issues, let's consider two cases:
a') Imagine infinitely massive hydrogen atoms. In this case, you get discrete energy levels, but the atoms never move.
b') More realistically, allow the hydrogen atoms to have finite mass. Now we have to think about the Doppler effect.

First, case a)/a'):
- If you start with everything at an initial temperature To, it will stay that way. So the BBR will fit the Planck formula, the occupation of the hydrogen energy levels will fit the Boltzmann formula (probability is proportional to exp(-Energy_level/(kTo) ), nothing will change. That probably doesn't interest you.
- So instead, we start with the system at T = 0: all hydrogen atoms at ground state, no photons beyond vacuum noise. Now nothing is happening. So how do we get the temperature above 0? We "magically" populate the hydrogen atoms at energy levels, with a number that matches the Boltzmann formula for To: then the hydrogen gas looks like it has temperature To. The problem, of course, is that there is 0 energy in the radiation field, so hydrogen levels will de-populate as photons are created to increase the entropy by occupying more of the phase space for the electromagnetic field. However, only photons that match transitions between hydrogen energy levels will be created. The result will that, the usual BBR formula for the occupation of photon energy levels will apply, but will only cover photons of allowed frequencies: there will be 0 photons of non-allowed frequencies.
The result is that the Planck formula will apply exactly as before at frequencies that match the energy-differences among the states of hydrogen; but at other frequencies, the spectral density will be 0: No allowed energy density, because no allowed photons. You will get a discrete spectrum of lines, but the intensity of each line will meet Planck's formula.

Second case, b)/b'): Let the hydrogen atoms have finite mass, and take into consideration the Doppler effect:
- Because the hydrogen atoms now move, transitions from one state to another have to specify the state of motion of the atom before the transition. This means that ANY frequency can be reached by Doppler-shifting from a specific energy-level transition. This means that ALL photons need to be included when considering the state of the radiation field. That means that we can't leave out any photon states, and we end up with Planck's original BBR formula, just as before.
- From a "practical" (??) point of view [this is an experiment only God can do], if you were to start out with infinite-mass hydrogen atoms, and then suddenly make their mass merely finite, it would take some time for the discrete-line BBR to transition into the continuum BBR: there are going to be very few high-speed atoms, which is what it will take to bring in the gamma-ray level; and even bridging the gap between the discrete energy levels that already exist will take time, because the variance in the speed giving rise to Doppler shifting is only v ~ sqrt(kTo/m) . But that's not the absolute limit, that's just the point where it becomes harder to find atoms moving. Equilibrium will be reached, however, in the long term.

Conclusion:
- If you consider an ideal experiment, with infinitely massive hydrogen atoms, then you are correct: There will be a change to the BBR: Planck's formula will be true only for frequencies corresponding to hydrogen atom transitions (but you have to include all of them). For other frequencies, the intensity will be 0. So you get a spectrum of discrete lines, which each have intensity matching Planck's formula.
- If you consider hydrogen atoms of finite mass, there will be absolutely no change to the normal BBR. Planck rules!
- To comment briefly on suprastremph's response: zapping hydrogen gas will reveal the spectral lines, but not their intensity. This is because there IS no "normal intensity" of the spectral lines: the energy levels determine only the frequencies of the lines. The intensity issue depends on temperature; which tosses us again back to my discussion above.

Answer 2

By "blackbody" we mean just that, a black body, so there is no reason it cannot be hydrogen in the dark. By blackbody we mean there is no reflected, or external light, and that all incident light is absorbed. This was the paradox of blackbody radiation, even if a body absorbs all light coming into it, it still radiates due to its temperature. This is why cavities were used, to reduce the prevalence of reflected light. The spectra emitted from hydrogen is from a different physical process than blackbody radiation. Blackbody approximations can be used whenever the light produced is due mostly to thermal energy. Some applications would be fire, a red hot stove, and the sun. To see hydrogen spectra, the best thing to do is run a large current through the gas in order to excite the electrons of the gas and not heat the gas quickly (which will result in a continuous blackbody spectrum which would compete with the observation of the spectral lines).

Answer 3

Yes, Mecca emits light, radio waves, and various other lengths of the electromagnetic system. Just as any other town or city on this planet. If you mean "does the black stone of meteoric origin that was worshipped long before mohammed was even a glint in his father's eye emit radiations?" then the answer is that there is probably some residual radiation.