Why does lead have a higher attenuation coefficient for 5.0 MeV than for 10.0 MeV? I was doing some calculations on radiation, and I noticed that lead has a higher attenuation coefficient for 5.0 MeV than for 10.0 MeV, namely $1.44 \, \mathrm{cm}^{-1}$  for the former and $1.23 \, \mathrm{cm}^{-1}$ for the latter.
For most materials, the attenuation coefficient increases when the energy of the rays increase. The only reason I can think of is that radiation of higher energy interferes more with the electrons and protons in the lead atoms, but I don't see why that would be the case.
 A: There are three main attenuation processes for high energy photons.
The photoelectric effect and Compton scattering are more important at lower energies and their cross-section decreases monotonically with increasing energy. On the other hand, the pair production cross-section takes over and increases towards higher energies.
The sum of these is what is used to calculate the mass attenuation coefficient. (in m$^2$/kg) or absorption coefficient (in $m^{-1}$). The mass attenuation coeffiecient for lead is shown below (multiply by density to get the absorption coefficient), with the contribution from the three processes above shown. As you can see, photoelectric absorption dominates below 1 MeV, leading to declining absorption with increasing energy. The absorption coefficient (just) goes through a minimum at around 5 MeV and is larger and increasing at 10 MeV due to the increasing dominance of pair production. This agrees with what is said in the NIST database.

A: On page 29 of this paper there is a nice graph which answers your question.  There are various possible process by which radiation can be attenuated in lead.  At the high energies that you have quoted pair production is the one and that is more likely for the radiation with the higher energy, so look at page 39. 
