CHAPTER 43 - 10 - (b) Forababy,thetotaldoseinoneyearis dose...

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 Po 84 218   Po 84 218 Pb 82 214 + He 2 4   Po 84 218 At 85 218 + e – 1 0 + ν h 2 h 1 d 2 d 1 ( b ) For a baby, the total dose in one year is dose  = (1.4 × 10 –5  J/yr)(1)/(1.00 J/kg     Gy)(5 kg)  = 2.8 × 10 –6  Sv = 2.8 × 10 –4  rem =       0.3 mrem ˜ 0.06% of allowed dose . 52. ( a ) We find the daughter nucleus by balancing the mass and charge numbers: Z (X) =  Z ( 222 Rn) –  Z ( 4 He) = 86 – 2 = 84; A (X) =  A ( 222 Rn) –  A ( 4 He) = 222 – 4 = 218, so the product nucleus is      . ( b ) From Figure 42–11, we see that   is     radioactive ,     and decays by both  α  and  β    emission: ;      . The half life for both decays is       3.1 min . ( c ) Both daughter nuclei are      chemically reactive .   ( d ) The decay constant is λ  = 0.693/ T 1/2  = 0.693/(3.8 days)(86,400 s/day) = 2.11 × 10 –6  s
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This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at University of California, Berkeley.

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CHAPTER 43 - 10 - (b) Forababy,thetotaldoseinoneyearis dose...

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