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CHAPTER 43 - 9 - 45 %oftheintercepted particles,wehave 6 10...

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45. If the counter counts 90% of the intercepted  β  particles, we have n  = (0.90)(0.20)(0.025 × 10 –6  Ci)(3.7 × 10 10  decays/s     Ci) =      1.7 × 10 2  counts/s . 46. If the decay rate were constant, the time required would be  t 1  = (36 Gy)/(1.0 × 10 –2  Gy/min)(60 min/h)(24 h/day) = 2.50 days. This time is (2.50 days)/(14.3 days) = 0.175 half-lives. The activity of the source at this time would be (1.0 × 10 –2  Gy/min)( ! ) 0.175  = 0.886 × 10 –2  Gy/min. If we approximate the exponential decay as linear, and use the average activity, we get t 2  = (36 Gy)/ ! (1.0 × 10 –2  Gy/min + 0.886 × 10 –2  Gy/min)(1440 min/day) = 2.65 days        ˜ 2.7 days . 47. If we start with the current definition of the roentgen, we get 1 R = (0.878 × 10 –2  J/kg)/(1.60 × 10 –19  J/eV)(1000 g/kg)(35 eV/pair)  = 1.57 × 10 12  pairs/g ˜ 1.6 × 10 12  pairs/g. 48. Because each decay gives one gamma ray, the rate at which energy is emitted is P  = (1.85 × 10 –6  Ci)(3.7
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