Answer to Question #794 Submitted to "Ask the Experts"

When electrons interact with matter they go through various collisional scattering processes. One such interaction is an inelastic scatter by the nucleus of an atom, where the electron can undergo a significant deflection in angle, with a change in energy. Because energy must be conserved, a bremsstrahlung photon results. The photon energy may be as high as that of the incident electron, and the probability of bremsstrahlung production varies with the square of the Z of the target. Also, the efficiency of x-ray production depends on the first power of the Z. In this case the target is tissue, perhaps soft tissue with an effective Z ~ 6, or bone having a Z ~ 20.

For a simple case of a monoenergetic electron field (compared with a beta spectrum) the total photon yield would be higher with bone. In theory, for a given electron energy, the actual initial shapes of the bremsstrahlung energy distributions produced in bone and in soft tisue would be the same, but because of the higher Z with bone the extent of lower energy photon attenuation and photon scattering could be greater; this could produce a shift to the higher energies in the resultant x-ray spectrum. However, if one considers a realistic human-receptor external-exposure scenario, where soft tissue would be over bone, the energy reduction in the incident electron field before it reached the bone would have an effect on the final bremsstrahlung spectrum, i.e., in reduced peak energy and overall yield.

Dave Allard, CHP