 Solution

For photons, the total mass attenuation coefficient sums contributions from photoelectric absorption, Compton scattering, coherent Rayleigh scattering, and pair production in nuclear or electronic fields when above their thresholds. At sufficiently high photon energies, photonuclear reactions can also contribute. Attenuation counts removal from the incident narrow beam, including scattering without complete local deposition of the original energy.

In water at both specified energies, Compton scattering dominates. At $140\mathrm{keV}$, photoelectric absorption and coherent Rayleigh scattering make smaller contributions; at $511\mathrm{keV}$ these are still less important. Nuclear-field pair production is impossible below $2m_e{c}^2=1.022\mathrm{MeV}$, so it occurs in neither example. The positrons used to produce the fluorine-18 annihilation radiation do not change that threshold for interactions of its $511\mathrm{keV}$ photons in water.

Strictly, $\mu$ is a linear attenuation coefficient, whereas $\mu /\rho$ is a mass attenuation coefficient; the source's wording blurs these units. The mass energy-absorption coefficient ${\mu }_{\mathrm{en}}/\rho$ weights the fraction of incident photon energy transferred to charged particles and retained after radiative losses, rather than simply counting all removals. Coherent scattering contributes attenuation with negligible energy transfer. A photon deflected by Compton scattering can carry much of the original energy away. One often writes ${\mu }_{\mathrm{en}}=(1-g){\mu }_{\mathrm{tr}}$, with $g$ the fraction of secondary charged-particle kinetic energy lost radiatively. Under charged-particle equilibrium, multiplying photon energy fluence by ${\mu }_{\mathrm{en}}/\rho$ gives the absorbed dose; that equilibrium condition matters near interfaces and in buildup regions.^[ref]