Dear Gaurav Rajak ,

you should get familiar with x-ray production via an X-ray tube. For example via the paper mentioned below.

When doing this, you will learn that x-rays are produced during slow down of accelerated electrons in the anode material of the x-ray tube.

The radiation is called 'bremsstrahlung' *). Please google for that term.

The spectrum of this radiation is very broad ranging from some few keV (some filtration assumed) up to a value, which I will name here Emax.

This Emax is the maximum value of energy loss, which the electrons of the electron beam can experience; and that is the its whole kinetic energy.

Acceleration of the electrons, which are relased from the cathode of the x-ray tube, is done by the high voltage U applied to the tube.

So the energy of the electrons and thus the maximum energy loss Emax is Emax=e*U (e being the electric charge of the electron). For the voltage U in units of V or kV you will have the maximum energy loss und thus the maximum x-ray energy of the x-ray photons in units of eV or keV.

The maximum voltage applied to the electrons is your 'kVp' value (=peak value of kilo-volt).

So you have Emax =e*'kVp'; the unit is keV.

So, for example having a kVp value of 60kV, your maxium x-ray photon energy is 60keV.

But pay attention: this value is only the maximum value. The mean x-ray photon energy value depends on various parameters and is in the range of 1/2 to 2/3 of Emax.

Article X-ray imaging physics for nuclear medicine technologists. Pa...

*) This term is a german one, but is also used in english language.

Final remark: please change your capital 'K' letters in your question into uncapitalized ones 'k'. The international abbreviation for kilo is 'k' .

For example, see Fig. 4 of this paper:

Article Diagnostic x-ray spectra: A comparison of spectra generated ...

Braking radiation is a continuous spectrum with the maximum energy at the electron acceleration voltage, while characteristic X-rays are line spectrum specific to an element. The superposition of both spectra is produced by an X-ray tube.

I am affraid there is no straightforward transformation from kVp to keV, unless you think of the maximum energy of the XRays, as explained already by dr. Martens.

First, since you do not get monoenergetic photons but a rather broad energy spectrum, you need to decide what energy value you seek:

- maximum energy correspond to the kVp

- the energy of the distribution peak is another story, and this depends both on the kVp and the filtration

- the average energy, but this does not make much sense

There is a simple way, however, to describe the beam energy distribution and this is in terms of half value layer (HVL). The HVL (expressed in mm of Al or Cu, usually ) gives a very simple and intuitive indication of both the kVp and the filtration. If you google it, you will find a lot of references on this quantity.

It depends on manufactured conditions , especially the filter. Try to contact with the manufactured company may provide a help.