CHAPTER 43 9 - 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 =

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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...
View Full Document

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.

Page1 / 2

CHAPTER 43 9 - 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 =

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online