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Unformatted text preview: 92 Radiation Doses and Hazards Chap. 9 2. A 137Cs ource has an activity of 700jLCi. A gamma photon from 137mBa ith s w energy 0.662 MeV is emitted with a frequencyof 0.845 per decayof 137CS. t A a distance of 2 meters from the source, what is (a) the exposure rate, (b) the kerma rate in air, and (c) the dose equivalent rate? Solution:
First we find the uncollided flux density 2 meters from the source. From Eq. (7.23) Sp (700 x 106 Ci)(3.7 x 1010decays/Ci)(0.845 photons/decay) 4>0 ~ = = 47r(200m)2 c = 43.54cm2s1. (a) Using the interaction coefficient data for air in Ap. C and linearly interpolating betweentabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (,uen/ )air = 0.02931. Then from Eq. (9.9) P we find
JLen x = 1.835X 108E ( Pair ) 4>0 = (1.835 x 108)(0.662)(0.02931)(43.54) = 1.55 X 108 Rls = 55.8 JLR/h. (b) Using the interaction coefficient data for air in Ap. C and linearly interpolating betweentabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (.utrI P)air = 0.02937. Then from Eq. (9.5) we find i< = 1.602 X 1010E
/Ltr Pair ) r/>° = 1 x (1.602 x 101°)(0.662)(0.02937)(43.54) = 1.35 X 1010 Gyjs = 0.488 ILGyjh. (c) Assume charged particle equilibrium so that kerma rate k equalsthe absorbed dose rate D. The dose equivalent equals the quality factor for photons (QF = 1) times the absorbed dose in tissue. We approximate tissue by water. Using the interaction coefficient data for water in Ap. C and linearly interpolating betweentabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (/.lenjP)H30 = 0.0.03260. Then from Eqs. (9.5) and (9.10) we find iT = QF x b = 1.602x 10loE J1.en P )H2O = 1 x (1.602 x 101°)(0.662)(0.03260)(43.54) = 1.51 X 1010 Gy/s = 0.542 JLSv/h.
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