What is the relationship between $\rm keV$ and $\rm kVp$? On X-ray topics we get used to talk about "energy", but what is keV? and what is the relationship between keV and kVp?
 A: In terms of radiography / radiology, kVp is the tube voltage / tube potential between the cathode and the anode, set by the operator. The unit eV (or keV as used in the range of general radiography, MeV as used by Radiation Therapy) describes the energy of the particles - in this case, the electrons in the x-ray tube, and the x-ray photons coming off the anode and towards the patient. 
So as the electrons in the tube are accelerated across the tube voltage, they will gain the same kinetic energy as the tube potential - so, if for example the kVp was 100kV, the electron energy will be 100keV as they reach the anode.
However, the conversion from electron energy to photon (x-ray) energy in the anode is of varying efficiency. So the x-ray beam emitted will have an energy spectrum (measured in keV in radiography for convenience, although it would normally be measured in Hz elsewhere in physics). The bottom of the range is zero keV, the top of the range cannot be higher than the energy supplied by the tube potential, in this example 100kVp - so the most energetic x-rays will be (almost) 100keV, but there won't be very many of them. 
For a regular tungsten anode, where the majority of the radiation is generated by bremsstralung interactions, and given typical exit window filtration, the average keV will be roughly 1/3rd of the kVp. This will change depending on the examination though. For mammography where most of the radiation is generated through characteristic radiation, and specific filters are used, the average x-ray keV could be quite different from 1/3rd of the kVp.
Hence the term "kVp" stands for "kilovolts peak", because it defines the top keV value the x-rays can have.
A: kVp is the maximum voltage in an X-Ray tube. When electrons travel through the voltage $V$, they will gain the kinetic energy of $E = e V$. Since $e$ is very small in SI units, and does not really tell you much, the energy of an electron traveling though $1 \, \mathrm V$ is used as a energy unit, namely $\mathrm{eV}$ with $1 \mathrm{eV} = e \cdot 1 \, \mathrm V$.
The energy and frequency of your X-Ray photon will be related with $E = h f = \hbar \omega$. Where $f$ is the frequency and $\omega$ is the angular frequency. $2 \pi f = \omega$.
A: Is 100 kVp = 100 keV?
No.
Voltage is defined as Energy/$q$, where
$q$ is the charge of an electron.
An eV is charge$\times$voltage to get energy back:
$\mathrm{eV} = q \times (E/q) =$ Energy in units of electron volts.
But isn't Energy in units of Joules?
Yes.
You convert eV which is Energy,
to Joules which is Energy,
with the conversion factor
which is the charge of an electron:
$1.60217662 x 10^{-19} \text{ J/eV} = q$.
