GAMMA- AND X-RAY INTERACTIONS IN 134 all interaction cross sections, even though

Question

GAMMA- AND X-RAY INTERACTIONS IN 134 all interaction cross sections, even though their ence may differ from of the Compton effect. c. Energy-Transfer Cross Section for the Compton Effect also may be The multiplied by a unit thickness of 1 elcm ofmany thought of as the fraction of the incident energy fluence, carried by a beam passin monoenergetic photons, be diverted into Compton interactions in photon through that layer of that will incident (T. It matter. In each interaction the energy of the (hr) is shared between the scattered photon (ha)and the recoiling electron is of interest to know the overall fraction of that is given to the electrons, averaged over all scattering angles, as this energy contributes to the kerma and thence to the dose. That is, we would like to know the value of TIAP, where T is the average kinetic energy of the recoiling electrons. This can be obtained through first modifying the differential K-N cross section in Eq. (7.13) to obtain a quantity referred to as the differential K-N energy-transfer cross section, dn.: a 2.2898 do, do, do. ou 295 san (7.19) Integrating hil overall photon scattering angles from o to 180°.as in Eq (7.15) yields the following statement of nou, the K-N ergy-transfer cross section: 201 a 1 3a aX2a m 2 Tri (1 2a 1 a 2a in (1 2a) 3(1 2a 2a (cm le) (7.20) of This cross section, multiplied by the unit thickness 1dcm2, represents the fraction the energy fluence in a monoenergetic beam that diverted to the recoil electrons by Compton interactions in that layer The Compton (or K-N) energy transfer cross section is also plotted in Fig. 7.6 (ower curve). The vertical diference between the two curves represents the K-N cross section for the energy carried by the d photons, Thus scattere

Explanation / Answer

value of K-N cross section as per equation 7.20 is given by

2*pi*r0^2*((2*(1+alpha)^2/(alpha^2*(1+2*alpha)))-((1+3*alpha)/(1+2*alpha)^2)-((1+alpha)*(2*alpha^2-2*alpha-1)/(alpha^2*(1+2*alpha)^2))-(4*alpha^2/(3*(1+2*alpha)^3))-(((1+alpha)/alpha^3)-(1/(2*alpha))+(1/(2*alpha^3)))*ln(1+2*alpha))

for E=0.662, alpha=1.295

then K-N cross section =9.6511*10^(-26) cm^2/e

for E=1.17 MeV, alpha=2.2896

then K-N cross section=8.9018*10^(-26) cm^2/e

for E=0.835 MeV, alpha=1.634

K-N cross section=9.43*10^(-26) cm^2/e

for E=1.33 MeV, alpha=2.627

K-N cross section=8.6252*10^(-26) cm^2/e

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