In which type of scatter does an x-ray photon give up all its energy to an inner shell electron?

The correct choice is that in photoelectric scatter, an x-ray photon transfers all of its energy to an inner shell electron. This results in the inner shell electron being ejected from its atom because the energy of the incident photon exceeds the binding energy of that electron. Consequently, the photon ceases to exist after this process occurs, as its energy has fully transferred to the electron.

In the photoelectric effect, the ejected electron is referred to as a photoelectron, and the atom becomes ionized as it loses this inner shell electron. This phenomenon primarily occurs with high-energy photons and materials with high atomic numbers, where the probability of interactions is greater due to the higher binding energies of inner shell electrons.

In contrast, other types of scatter, such as Compton scatter, involve partial energy transfer to a different electron, resulting in the photon being deflected with reduced energy rather than being completely absorbed. Coherent scatter, on the other hand, involves low-energy photons scattering elastically without any energy loss, and radiographic scatter is not typically classified as a fundamental scattering mechanism in radiology. Thus, photoelectric scatter is distinct in its total energy transfer to an inner shell electron.