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HW17-15 - Thus for this problem we have =(4.6...

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(d) From Table 9.10, the radiation induced risk to the first generation, per rad gonad dose to the parents is seen to be: risk(BEIR)= (6 to 43)/106 and risk(UNSCEAR)= (18)/106. Thus for a parental gonad dose of 0.95 rem (or 0.95 rad since the exposure was from gamma rays) the radiogenic hereditary risk for the male operator is .risk(BEIR) = 0.95(6 to 43)/106 ~ (6 to 41) X 10-4% . .risk(UNSCEAR) = 0.95(18)/106 ~ 17 X 10-4%. 14. An individual is exposed 75% of the time to radon with a physical concentration of 4.6 pCifL and an equilibrium factor of F = 0.6. The remaining 25% of the time, the individual is exposed to radon at a concentration of 1.3 pCifL and with an equilibrium factor of F = 0.8. What is the annual radon exposure (on an EEC basis) in MBq h m-3? Solution: The annual radon exposure is calculated as R = Ei FiCihi, where Ci is the radon concentration at location i, Fi is the equilibrium factor at that location, and hi is the number of hours per year spent in location i.
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Unformatted text preview: Thus, for this problem we have = (4.6 pCi/L)(o.6)(o.75 x 365.25 d/y x 24 hid) +(1.3 pCi/L)(o.8)(0.25 x 365.25 d/y x 24 hid) = 2.04 x 104 pCi hit. From page 253, we find that an EEC of 4 pCifL is equivalent to 150 Bq m-3. With this conversion factor, the annual Rn EEC exposure is = 0.766 MBq h m-3, = (2.04 X 104 pCi h L-1) 150 Bq/m3 4 pCi/L If the individual in the previous problem is a nonsmoking male and receives the same annual radon exposure for his entire life, what is the probability he will die from lung cancer as a result his radon exposure? Solution: From Table 9.14, we find that for a non-smoking male receiving an annual exposure of 1 MBq h m-3 for his entire life the probability he dies from radon induced lung cancer is 0.016. Thus, for a lifetime annual exposure of 0.766 MBq h m-3 the probability of death from radon induced lung cancer is 0.016 (MBq h m-3)-1 x 0.766 (MBq h m-3) = 0.012 = 1.2%, or about 1 chance in 83. July 24, 2002...
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