Why aren't we seeing $\Delta n = 0$ transitions in the spectrum of an X-ray-tube?

Question

Well basically title. The characteristic / discrete radiation from an x-ray tube comes from electrons falling down into a vacancy which was created by an incoming electron from the acceleration voltage. This electrons always comes from an upper shell and emits some radiation in the length of a few dozen pm.

My question is: Why does this vacancy need to be filled by an electron from an upper shell? Can't it happen that e.g. if an electron of an $n^2s_\frac{1}{2}$ state gets knocked away from the atom and the vacancy gets filled by an electron of the corresponding $n^2p_\frac{1}{2}$ state, which would in turn result in a wavelength (obviously depending on the material and n) that we maybe could see? Is it just a simple "well it's more favourable energy-wise for the electron from an higher shell to fall down" argument?

Obviously the "simple" explanation is that for x-rays we're using Bohrs which just isn't accurate, but then where's the catch that hinders us from seeing brightly lit x-ray tubes?

Answer 1

As discussed in "Is the 2p→2s transition possible?" and "Can an electron make a transition between sub energy states of the same energy level?", $\Delta n =0$ transitions do happen. The reason characteristic x-rays "always" come from a higher shell is because photons from same shell transitions are much less obvious because both their energies and transition rates are much lower. Since they don't stand out, they are not "characteristic" of the element.

For example, perhaps the most common x-ray tube anode metal is tungsten. According to the NIST Tables for W LXXIV, tungsten's K lines due to transitions to the 1S state have wavelengths and transition rates:

compared to the $2p-2s$ transition:

So the $2p-2s$ $\Delta n =0$ transition has a rate a thousandth that of the K lines, and an energy ($0.754\,\mathrm{nm} \rightarrow 1.64\,\mathrm{KeV}$) 40 times less. The low energy means that even the relatively few $2p-2s$ photons that are produced are hard to observe. They are much more attenuated than the K lines when escaping the tungsten, and any $2p-2s$ line is also buried under the bremmstrahlung background that is greatest at lower energies. This background is usually undesirable, so in most cases these low energy bremmstrahlung x-rays are actively filtered out, which also removes any $2p-2s$ photons.

Answer 2

I believe the reason that you don't see long wavelength radiation from high voltage electron irradiation of metals like tungsten is simply opacity. Any light produced in the metal cannot escape. If light can escape, we call the material a "phosphor", and we use it when the production of light, rather than x-rays, is desired.