Department of Physics University of Oslo Interaction Between Ionizing Radiation And Matter, Problems Photons Audun Sanderud

Problem 7.1 • Is the mass Compton attenuation/energy transfer coefficient larger in carbon or lead? Solution: ←Independent of Z Carbon: NAZ/A=0.49954NA Lead: NAZ/A=0.39575NA NA=6.0022 1023

Problem 7.2 • Why is Rayleigh scattering not plotted in Fig. 7.16a,b, although quite significant in Fig. 7.13a,b? Solution:

Problem 7.3 • On the basis of the K-N theory, what is the ratio of the Compton interaction cross section per atom for lead and carbon? Solution:

Problem 7.4 • Calculate the energy of the Compton-scattered photon at q= 0, 45 , 90  and 180 for hn = 50 keV, 500 keV and 5 MeV. Solution: hn´: hn\q 0 45 90 180 50 48.60 45.54 41.82 keV 500 388.6 252.7 169.1 5000 1293 463.6 243.1

Problem 7.5 • What are the corresponding energies and angels of the recoiling electrons for the cases in problem 4? Solution: T: hn\q 0 45 90 180 50 1.393 4.456 8.183 keV 500 111.4 247.3 330.9 5000 3707 4536 4757 j: hn\q 0 45 90 180 50 90 65.55 42.33 keV 500 50.67 26.81 5000 12.62 5.298

Problem 7.6 • Calculate for 1-MeV photons the total K-N cross section from Eq.(7.15), and derive the Compton mass attenuation coefficient for copper in cm2/g and m2/kg. Solution:

Problem 7.7 • What is the maximum energy, what is the average energy, of the Compton recoil electrons generated by 20-keV and 20-MeV g-rays? Solution:

Problem 7.8 • Calculate the energy of a photoelectron ejected from the K-shell in tin by a 40-keV photon. Calculate ttr/r; you may estimate from fig. 7.15. Solution: Tin: K-edge 29.20keV T=hv-Tb=(40-29.20)keV=10.80keV

Problem 7.9 • What is the average energy of the charged particles resulting from pair production in (a) the nuclear field (b) the electron field, for photons of hn = 2 and 20 MeV? Solution: a) b)

Problem 7.10 • A narrow beam containing 1020 photons at 6 MeV impinges perpendicularly on a layer of lead 12 mm thick, having a density 11.3 g/cm3. How many interactions of each type (photoelectric, Compton, pair, Rayleigh) occur in the lead? Solution: Number of interactions ∆Ntot =N0(1-e-∆x µ) ∆N´tot =N0∆x µ = N0∆x (tp.el+sC+kpair +sR) =∆N´p.el+∆N´C+∆N´pair+∆N´R >∆Ntot ∆Np.el =∆N´p.el·∆Ntot/ ∆N´tot Photoelectric Compton Pair Rayleigh 9.999E+17 1.792E+19 2.541E+19 5.375E+16

Problem 7.10 • A narrow beam containing 1020 photons at 6 MeV impinges perpendicularly on a layer of lead 12 mm thick, having a density 11.3 g/cm3. How many interactions of each type (photoelectric, Compton, pair, Rayleigh) occur in the lead? Solution: Number of interactions ∆Ntot =N0(1-e-∆x µ) ∆N´tot =N0∆x µ = N0∆x (tp.el+sC+kpair +sR) =∆N´p.el+∆N´C+∆N´pair+∆N´R >∆Ntot ∆Np.el =∆N´p.el·∆Ntot/ ∆N´tot Photoelectric Compton Pair Rayleigh 9.999E+17 1.792E+19 2.541E+19 5.375E+16