I am a high school student, so I know only the basics of X-Rays. I simply know about continuous X-rays, cutoff wavelength and threshold wavelength.
Now if I increase the potential applied to the X-ray tube, I am certain that the minimum wavelength of emitted radiation decreases. I am unsure about its intensity. How does that change?
[Physics] Effect of Change of Potential Difference applied to an X-Ray tube
electromagnetic-radiationintensitypotentialwavelengthx-rays
Related Solutions
An X-ray tube creates a broad spectrum of X-rays due to Bremsstrahlung radiation and specific frequencies due to X-ray fluorescence. There's an example spectrum on the Wikipedia page about X-ray tubes. I assume you're talking about the fluorescence peaks since you refer to characteristic X-rays.
The fluorescence peaks are generated because the incident electrons collide with and eject electrons from the inner orbitals of the atoms in the target leaving the atoms ionised. When the ion recombines with an electron the energy of recombination is emitted as a photon. The energy of the emitted photon depends only on the energy of the atomic orbital that the electron is relaxing into, so it's a property of the atom. As long as the incident electrons have enough energy to ionise the atom the emitted photon will have the same energy.
So changing the energy of the incident photons may well change the intensity of the fluorescence peak, but it will not change the wavelength. Typically a commercial X-ray tube, e.g. for use in a diffractometer, will have the electron energy tuned to maximise the intensity of the required fluorescence peak (usually the $K^\alpha$.
Regarding the core question,
is it certain that the inner electron is ejected completely out of the atom, and not to another vacant, higher energy level?
this is the most likely scenario but it is not completely necessary. It is perfectly possible (at least in the sense that it's not forbidden) for the electronic collision to leave the atom in an excited state with a core hole and that electron occupying some valence or Rydberg orbital, and for that state to decay radiatively to the ground state of the neutral atom, or indeed to some slightly excited state that then decays by emitting a visible or IR photon.
If that is the case, then the frequency of the emitted photon will be slightly below the shell edge. This is good and bad: on one side, the core-valence energy difference tends to be big enough that the effect of where the original electron ends up will be minimal, but on the other side, any emission below the edge will stick out like a sore thumb.
However, I suspect that the probability of this process is essentially negligible, because it's competing with the Auger effect, in which the valence electron drops back into the core hole and uses the energy to ionize another electron from the valence band, with a very fast timescale.
It is hard to make general statements because there is a wide variety of possible situations covered by the question, but my impression is that pre-edge peaks tend to be somewhere between negligible and weak, though they can be observed if your apparatus is sensitive enough:
Best Answer
An X-ray tube makes X-rays by Bremsstrahlung (quickly decelerating the electrons), but also by exciting the atoms of the anode. Just as an electric discharge makes a red glow in neon gas, so spectrum features of the anode metal are prominent in the X-rays produced.
So, to make a higher energy of X-ray, an effective procedure is to change the anode material to something with higher atomic number (the more charge in the nucleus, the higher the binding energy of the inner electrons, thus the higher the possible energy of fluorescence).
In order to stand up to high currents, an anode is usually a good heat conductor (copper) or resistant to high temperature (molybdenum, tungsten). Copper's highest fluorescence is 8.98 keV, molybdenum 20 keV, and tungsten 69 keV.
So if you want a high intensity fluorescence, you need to provide voltages higher than the fluorescence to excite that intense X-ray emission, or provide very high voltages and currents and hope the Bremstrahllung (continuous spectrum) is intense enough (and the anode doesn't melt).
Because the Bremstrahllung includes a lot of low energy radiation, it is usual to use an X-ray tube window that acts to filter out the unwanted low energy radiation. Both Bremsstrahlung and fluorescence energies can be no greater than the energy per electron.
To convert from energy $E$ (keV) to frequency $F$ (Hz) use Planck's constant $h = 4.1\times 10^{-18}\ \mathrm{keV\,sec}$, $$F = E /h$$