Why we do not consider electrostatic potential in continuous x ray production? Why we don't consider electrostatic potential in continuous X-ray production?In almost every Book I see they say that by energy conservation, maximum energy X-ray that could be emitted is
$$eV=hυ$$(υ is frequency of x ray produced)
But while putting energy conservation why are authors not considering electrostatic potential energy,Although electrostatic potential is not negligible(as distance between charges is very small).
(Picture from my book.About what it has done)

(Pictures that may help you inderstanding my question)
What i think it can cause a huge diffrence in frequency of wavelength emitted.I think energy conservation formula should be$$eV-(kze²/r_1)=hυ-(kze²/r₂)$$
One more related doubt i have is where exactly this x ray will be released?
Thank, you in advance.
 A: You are describing the scattering process of a single electron and a single proton in a classical point of view. I believe we are not allowed to consider this situation. Instead, we have to consider a single electron scattering with a single atom. The important point is the order of magnitude of the electronic bound-state of the atom:  The energy of the lowest bound-state is approx. ${\bf 10eV}$  -- e.g. take $H$-atom, which has $13.6eV$. 
Now, let's assume an electron has the kinetic energy $E_{kin}$. What is the process, by which the electron emits the photon with the highest energy? Using the conservation of energy, we conclude that if the electron is moving after the "collision", it still carries kinetic energy. Hence, the process which generates the photon with the highest energy is the "collision", where the electron comes to a complete stop. Thus, the electron must be captured and occupy the electron bound-state with the lowest energy.  Therefore, the photon would have the energy $E_{max}\approx E_{kin} + 10eV$. 
Considerations, where we conclude that the electron gains kinetic energy much larger than $10eV$ (due to Coulomb potential), are  wrong: An electron does not "fall" into the nucleus, but it forms an atom. The most energy an electron is allowed to gain due to the Coulomb potential is on the order of $10eV$. However, the additional $10eV$ are never accounted for in the $x$-ray spectrum, because 


*

*the electron is usually not captured, but scattered

*the kinetic energy of the electron is on the order of several $keV$. Hence, the $10eV$ are irrelevant.  

