A137 Cs source has an activity of 700 mu Ci A gamma photon w

A^137 Cs source has an activity of 700 mu Ci. A gamma photon with energy 0.662 MeV is emitted with a frequency of 0.849 per decay. At a distance of 2 meters from the source, what is the exposure rate, the kerina rate in air, and the dose equivalent rate?

Solution

Solution: First we find the uncollided flux density 2 meters from the source.

From Eq. Sp (700 x 10-6 Ci)(3.7 x 1010decays/Ci)(0.845 photons/decay) 4

>0 ~ = = 47r(200m)2 c = 43.54cm-2s-1.

Then from we find JLen x = 1.835X 10-8E ( Pair ) 4&

gt;0 = (1.835 x 10-8)(0.662)(0.02931)(43.54) = 1.55 X 10-8 Rls = 55.8 JLR/h.

(b) Using the interaction coefficient data for air in Ap. C and linearly interpolating between tabulated 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 we find i< = 1.602 X 10-10E /Ltr Pair ) r/> = 1 x (1.602 x 10-1)(0.662)(0.02937)(43.54) = 1.35 X 10-10 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 10-loE J1.en P )H2O = 1 x (1.602 x 10-1)(0.662)(0.03260)(43.54) = 1.51 X 10-10 Gy/s = 0.542 JLSv/h.