4 Two air-filled cavity ionization chambers are identical except that one has an aluminium wall and the other graphite. The walls are thicker than the maximum secondary-electron range for 1 MeV photons, which are negligi- bly attenuated. Calculate the approximate ratio of the charge generated in the two chambers, assumed to be BG cavities. Answer: 4A1/ qc = 1.093 Solution: Assuming that only Compton interactions occur, the mean initial electron energy from the Compton effect for photons with energy * = 1 MeV is obtained from kxoKN/6KN. From the photon Klein-Nishina cross sections in the Data Tables, OKN E=kx " 0.9294 x 10-25 OKN = 1 MeV X 0.2112 x 10-24 = 0.440 MeV The mean energy of the equilibrium spectrum may be crudely estimated as Eo/2; thus E = E /2 = 0.22 MeV For a given photon energy fluence , the absorbed dose to a chamber wall can be written, under CPE conditions, as Dwall = 4 (Hen/ P)wall and can also be obtained as wall = Pair Swallair = (W/e)air wall,air VPair where v is the chamber volume and vair = Mair From these two expressions, we obtain (Men/ P)wall VP air (W /e)air $wall,air and for the ratio of the charges in our two chambers, identical except for their wall material, we obtain HAI = [(Men/ P)AlCh MeV [SC.Allo.22 MEV = 0.0269 2.3657 0.0279 X 2.0854 = 1.093 4c