Black Body radiation and Uv, x and gamma rays Is the reason why UV, x-rays and gamma rays are non-existent on the blackbody radiation spectrum is because instead of the atoms present in the walls of the blackbody absorbing the energy, the energy is quite simply "too much" and hence they become ionised instead?
If not the case, what really happens?
What would be the answer to this thought experiment?
A x-ray photon suddenly appears and travels towards a blackbbody cavity. What happens to the x-ray photon, what happens to the atoms of the blackbody cavity inline with the x-ray's trajectory???
 A: I dispute your claim. As you can see from the form of the Planck function, governing the intensity of radiation from a perfect blackbody, and from the picture below that plots it, there is radiation at all wavelengths for a blackbody of any temperature. Note the axes have log units and the curves never reach zero.
All that can be said is that there is very little X-ray radiation from a blackbody at low temperatures.
I think you are confused between the spectrum of a real object that might approximate a blackbody and the spectrum of a perfect blackbody. A perfect blackbody absorbs everything incident upon it - including X-rays. Whatever radiative processes cause the absorption of X-rays must also be capable of emitting X-rays. However, a real object may be largely incapable of absorbing or emitting X-rays. Such a thing might have a spectrum that approximates a blackbody at most wavelengths, but it will not be a perfect blackbody.
However, you are describing a cavity. If the cavity absorbs X-rays then that tells you that there are radiative processes also capable of emitting X-rays. e.g. If the photons ionise the atoms via the photoelectric effect, then it is also possible for recombination to occur with the emission of X-ray photons. In a true blackbody, these processes will be in perfect equilibrium.

