I haven't though about this from such a perspective but today while reading wikipedia (yes yes not the best source) I got confused, now the "eV" is said to measure the energy gained by an electron between a potential difference of 1V. I assume particle physicists use this measurement because its quite handy for particle accelerator where if you have say 250kV between cathode and anode and you use an electron beam you can then say the electrons got a final energy of 250 keV, correct? but here's the part that confuses me, in terms of gamma radiation which is among the highest frequency EM radiation and so ionizing, I see a chart where frequency, wavelength and the corresponding energy is given and it is written in eV. it is written that frequencies from the Ghz region up towards the Phz and even EHz, have corresponding energies of only μeV to keV, and only from gamma the energy starts in the MeV region, does that mean that photon Em radiation is not very powerful except starting from X ray to gamma region? And even then I see X rays have on average only about 100KeV, although I assume its enough to damage a cell because the binding energy between an electron and the nucleus is far far less than that between protons and neutrons inside the nucleus correct? What is the typical binding energy between electrons and nucleus in eV for organic materials like flesh? If I for example accelerate a beam of electrons between a cathode and anode to a potential of 100KeV, do they then reach an energy equivalent to the 100KeV X ray range photons for example? I have a feeling this might be the way an X ray tube works by accelerating electrons which are then directed into the anode which serves also as the target creating photon emission of an energy which is proportional to the applied potential difference? I read also that most of the energy of the electron beam gets absorbed in the anode and simply manifests itself as heat, yet the photons that are emitted by bremsstrahlung type of radiation from the target are of the corresponding eV energy of the PD between cathode and anode does this simply mean that of all the electrons from the beam (intensity) most don't make for a photon emission event and get trapped or otherwise end up as heat but those few that get their job done then emit a photon of their corresponding energy? If my assumptions are correct seems like the opposite of the photoelectric effect only with bigger losses in the conversion? Oh and one final question if I may, what determines the wavelength of an EM wave, I have always wondered why the wavelengths of even such high frequencies as Mhz and up to Ghz have such large wavelengths, is the visual picture of a photon flying through space and time making up and down motions much like a dot that would ride a sine wave a correct way of thinking about this? and then between each full up-down-up moment there is some distance that the photon has traveled in the horizontal frame which then become the wavelength? thank you all for taking time to answer.