The answer is C.
EXPLANATION: Diagnostic x-ray photons interact with tissue in a number of ways, but mostly they are involved in the production of Compton scatter or in the photoelectric effect. Compton scatter is pictured; it occurs when a relatively high-energy (kV) photon uses some of its energy to eject an outer-shell electron. In doing so, the photon is deviated in direction and becomes a scattered photon. Compton scatter causes objectionable scattered radiation fog in large structures such as the abdomen and poses a radiation hazard to personnel during procedures such as fluoroscopy. In the photoelectric effect, a relatively low-energy x-ray photon uses all its energy to eject an inner-shell electron, leaving a "hole" in the K shell. An L-shell electron then drops down to fill the K vacancy, and in so doing emits a characteristic ray whose energy is equal to the difference between the binding energies of the K and L shells. The photoelectric effect occurs with high-atomic-number absorbers such as bone and positive contrast media, and is responsible for the production of radiographic contrast. It is helpful for the production of the radiographic image, but it contributes to the dose received by the patient (because it involves complete absorption of the incident photon).