In an X−ray tube, a heated filament releases electrons, a potential difference accelerates them and a dense target stops them. Vacuum reduces collisions before they reach the target, and cooling removes substantial heating. Braking electrons emit a continuous bremsstrahlung spectrum. If one electron transfers all its kinetic energy to one photon, hfmax = eV, giving λmin=(hceV).
Characteristic peaks arise from transitions between target-atom energy levels after inner-shell vacancies form. Their energies depend on target material rather than accelerating voltage, provided the voltage is sufficient to create the vacancies. Increasing filament current mainly increases photon production rate; increasing tube voltage raises the cutoff photon energy. Higher-energy X-rays generally penetrate more strongly. Attenuation by absorption and scattering depends on photon energy and material, allowing contrast between tissue, bone and metal.
Example 1
A tube voltage is 20.0 kV. What is the maximum photon energy?
An electron gains eV=20.0 keV. Thus the maximum photon energy is 20.0 keV, if all that energy becomes one photon.
Example 2
Use h=6.63×10−34Js,c=3.00×108ms−1,e=1.60×10−19C and V=40.0 kV. Find λmin.
λmin=(hceV)=3.11×10−11m.
Watch out: eV is the maximum possible photon energy; braking events produce a continuous range, with characteristic peaks superimposed.