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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 m3. With this conversion factor, the annual Rn EEC exposure is = 0.766 MBq h m3, = (2.04 X 104 pCi h L1) 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 nonsmoking male receiving an annual exposure of 1 MBq h m3 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 m3 the probability of death from radon induced lung cancer is 0.016 (MBq h m3)1 x 0.766 (MBq h m3) = 0.012 = 1.2%, or about 1 chance in 83. July 24, 2002...
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This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 CADY

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