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When solving the hydrogen atom, as a 2 body problem, we have the motion of the center of mass and the motion relative to the center of mass. The well known energy spectrum, $E_n$, that goes like $1/n^2$ is the one resulting from studying the motion relative to the center of mass.

Now the energy of the center of mass of the atom, which is purely kinetic, should be added to $E_n$ to get the full energy. So unless the atom is at rest, there will be an extra (constant) term to $E_n$

My questions are, in practice when people study the hydrogen spectrum experimentally:

1-do people look at the spectrum from 1 atom only? if this is the case, then how can they guarantee that it is not moving (to properly identify the $1/n^2$ behavior)? (very cold atom technology was not known back in the day!)

2-if they are studying a gas of hydrogen, how come the spectrum to be detected is not washed out by the transitions coming from so many atoms in all directions?

3-again for looking at hydrogen gas, how the spectrum can be studied even to see if it goes like $1/n^2$ if the atoms are moving in all directions like crazy?

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People usually do not measure the energy levels directly; they measure the difference by measuring the frequency of light emitted during transition. The extra kinetic energy terms cancels out in the difference. However, the moving atoms will Doppler shift the frequency of the emitted light, part of the reason why every line on the spectrum has a finite width. The effect is small because thermal motion is much much slower than speed of light.

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As Karsus points out, you only ever measure energy differences during transitions between different levels, and it is hard for these to put too much energy into the translational motion. To get a quantitative answer, consider that a photon can only give about $\hbar k$ of momentum to the atom, and this will be weighted down by the total mass ($\approx$ the proton mass), which is far bigger than the reduced mass ($\approx$ the electron mass) which governs the spectrum. The total kinetic energy difference will therefore be of the order of $$\frac{(\hbar k)^2}{2m_p}=\frac{\hbar^2\omega^2}{2m_pc^2}=\frac{(\Delta E)^2}{2m_pc^2}$$ and is therefore smaller than $\Delta E$ by a factor of $\Delta E/2m_pc^2\lesssim7\times10^{-9}$. This effect will therefore cause a broadening of the transition lines by that amount. This is of course independent of how many atoms you're addressing.

However, the translational motion does affect the energies you observe by Doppler shifting them, so that if you're observing a single atom the line will move to higher or lower frequency by a factor of $\sim v/c$. If you have many atoms moving in many different directions at different velocities, then each will radiate at its own natural frequency but you will observe a bunch of different Doppler shifts, and therefore the line will be broadened by a factor of $\sim v_\textrm{th}/c$ where $v_\textrm{th}$ is the thermal velocity given by $m v_\textrm{th}^2\approx k_B T$. This is Dopppler broadening and it is dominant if you do nothing to bring it down. It will certainly be dominant in a high-temperature environment like an arc lamp.

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