Hi Here is the number for the measured frequency for hydrogen 1s to 2s transition: 2 466 061 413 187 035 Hz By way of interest, what is our most accurate theoretical calculation of this number? I've tried the ordinary Bohr formula and it is only accurate to about 4 places. I'm also wondering if there is any Doppler frequency shift factor involved in the theoretical formula. Because the light emanates from (I presume) a moving source (electron in orbital).
To make a comparison with calculations, you need to put an uncertainty on that. For instance, the value used in calculating the Rydberg constant for the CODATA 2010 is from Fischer et al., 2 466 061 413 187.080(34) kHz which is not as precise as implied by the number you wrote, which appears to be precise to one hertz. The best I could find after a quick search dates back a bit, and the uncertainty then was at the kHz level. I don't know what the current status is. You have to introduce relativistic corrections, such has hyperfine interaction, as well as QED corrections, such as the Lamb shift. First, these are electronic bound states, so the electron does not "move." Second, it is not the electron that emits radiation, but rather the combination of the electron and the proton (nucleus). Calculations are made in the reference frame of the center of mass of the atom, and as such correspond to the frequency of light emitted by an atom at rest with respect to the lab frame.
Many thanks for your very comprehensive answers to my query. The reference for the number I quoted is: http://edoc.ub.uni-muenchen.de/13943/2/Parthey_Christian.pdf Do I take it that because of the 'bound state' one does not consider the recoil energy of the electron either in the calculation of transition energy and hence of emission frequency. In my simple way I imagined that within the context of the system, the electron would have independent freedom of movement in any direction which did not change its distance from the nucleus fixed by kq^2/r^2 = mv^2/r. And that the photon would emit in a random direction so some proportion of the emissions would impart energy to the electron rather than to the system as a whole. This (very small) energy would need to be deducted from transition energy to give hf for the emitted photon.
The uncertainty there is 10 Hz, of the same order of magnitude as the reference I gave. In quantum mechanics, an electron in a bound state does not have a defined trajectory, nor is it at a fixed distance from the nucleus. It is in an orbital, and its position around the nucleus is only known probabilistically; you can picture it as a "fuzzy" cloud around the nucleus. The entire atom will recoil from the emission of a photon. The photon frequency calculated is the central frequency of the peak that would be measured for an atom initially (i.e., before emission) at rest (with respect to the detector), considering that, because of the Heisenberg uncertainty principle, the peak in the emission spectrum is not infinitely narrow, but has a certain width (called natural linewidth). For an actual spectrum, the peak would be even broader due to the Doppler effect.
Can you perhaps indulge a quantum mechanical ignoramus a little further ? The principle of photon/electron interactions resulting in electron recoil (as opposed to atom recoil) is demonstrated in the Compton effect. How would the physical situation differ in that instance - is it because the electrons are considered "free or loosely bound" (ie high energy)? In which case are there 'degrees' of binding to the point where there is some transition point between atoms that can demonstrate the Compton effect and those that can't? Or just a gradual decrease in the effect until it effectively disappears altogether. Or if I can put it a slightly different way - due to wave particle duality - is it not possible for a wave type quantity (electron in bound orbital) to demonstrate particle type properties (such as recoil) to a degree however minute that might be ? Hope that's not a completely dumb question !