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More Radioactive Decay Calculations

More Radioactive Decay Calculations. Problem #1. The half-life of cobalt-60 is 5.3 years. How much of a 1.000 mg sample of cobalt-60 is left after a 15.9 yr period?. Using A = A o (1/2) n. A = final amount A o = initial amount n = number of half-lives elapsed.

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More Radioactive Decay Calculations

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  1. More Radioactive Decay Calculations

  2. Problem #1 The half-life of cobalt-60 is 5.3 years. How much of a 1.000 mg sample of cobalt-60 is left after a 15.9 yr period?

  3. Using A = Ao(1/2)n A = final amount Ao = initial amount n = number of half-lives elapsed

  4. Using A = Ao(1/2)n A = final amount Ao = initial amount n = number of half-lives elapsed Calculate n: n = T/t1/2 T = total time elapsed t 1/2 = half-life of atom

  5. A = Ao(1/2)n A = ? Ao = 1.000 mg n = ?

  6. A = Ao(1/2)n A = ? Ao = 1.000 mg n = ? T = 15.9 yrs t 1/2 = 5.3 yrs n = T/t1/2 = 15.9/5.3 = 3

  7. A = Ao(1/2)n A = ? Ao = 1.000 mg n = 3 n = T/t1/2 = 15.9/5.3 = 3

  8. A = Ao(1/2)n A = ? Ao = 1.000 mg n = 3 X = 1.000 mg (.5)3 Make certain that you try this with your calculator!!!

  9. A = Ao(1/2)n A = ? Ao = 1.000 mg n = 3 X = 1.000 mg (.5)3 X = .125 mg

  10. Rinky think method, same problem… total time elapsed [---------------15.9 years------------]

  11. Rinky think method… total time elapsed [---------------15.9 years------------] How many half-lives fit in this time frame? [--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]

  12. Rinky think method… total time elapsed [---------------15.9 years------------] How many half-lives fit in this time frame? [--5.3 yrs--][--5.3 yrs--][--5.3 yrs--] 15.9 yrs/5.3 yrs = 3 half-lives

  13. Rinky think method… total time elapsed [---------------15.9 years------------] How many half-lives fit in this time frame? [--5.3 yrs--][--5.3 yrs--][--5.3 yrs--] Decrease mass by half for each half-life.

  14. Rinky think method… total time elapsed [---------------15.9 years------------] How many half-lives fit in this time frame? [--5.3 yrs--][--5.3 yrs--][--5.3 yrs--] Decrease mass by half for each half-life. 1 mg .5 mg .25mg .125 mg!

  15. Rinky think method… total time elapsed [---------------15.9 years------------] [--5.3 yrs--][--5.3 yrs--][--5.3 yrs--] 1 mg .5 mg .25mg .125 mg!

  16. Problem #2 • If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 years. What is the half-life of strontium-90?

  17. Using A = Ao(1/2)n A = .95 g Ao = 1.000 g n = ? Calculate n: n = T/t1/2 T = 2 years t 1/2 = ?

  18. Using A = Ao(1/2)n A = .95 g Ao = 1.000 g n = 2 yrs/x Calculate n: n = T/t1/2 T = 2 years t 1/2 = ?

  19. Using A = Ao(1/2)n A = .95 g Ao = 1.000 g n = 2 yrs/x Plug in the information: .95 g = 1 g (.5)2 yrs/x

  20. Using A = Ao(1/2)n .95 g = 1 g (.5)2 yrs/x Simplify by dividing both sides by 1 g. .95 = .52 yrs/x

  21. Using A = Ao(1/2)n .95 g = 1 g (.5)2 yrs/x .95 = .52 yrs/x To get the exponent in a solvable position, take the logarithm of the problem: log.95 = 2 yrs/x (log.5)

  22. Using A = Ao(1/2)n 95 g = 1 g (.5)2 yrs/x .95 = .52 yrs/x log.95 = 2 yrs/x (log.5) Simplify by dividing both sides by log.5 log.95/log.5 = 2 yrs/x

  23. Using A = Ao(1/2)n .95 g = 1 g (.5)2 yrs/x .95 = .52 yrs/x log.95 = 2 yrs/x (log.5) log.95/log.5 = 2 yrs/x Calculate the logs and divide them: X(.074) = 2 yrs

  24. Using A = Ao(1/2)n .95 g = 1 g (.5)2 yrs/x .95 = .52 yrs/x log.95 = 2 yrs/x log.5 log.95/log.5 = 2 yrs/x X(.074) = 2 yrs Divide by 0.74 to solve for x: X = 2 yrs/.074

  25. Using A = Ao(1/2)n .95 g = 1 g (.5)2 yrs/x .95 = .52 yrs/x log.95 = 2 yrs/x log.5 log.95/log.5 = 2 yrs/x X(.074) = 2 yrs X = 2 yrs/.074 X = 27 years

  26. Rinky think method Quantities too small for this problem! All problems on the test will be okay for rinky thinking…

  27. Problem #3 A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample due to carbon-14 is measured to be .43 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of carbon-14 is 5715 yr. What is the age of the archeological sample?

  28. Using A = Ao(1/2)n A = .43 disintegrations Ao = 15.2 disintegrations n = x/5715 yrs Calculate n: n = T/t1/2 T = ? t 1/2 = 5715 years

  29. Using A = Ao(1/2)n A = .43 disintegrations Ao = 15.2 disintegrations n = x/5715 yrs .43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs .0283 = (.5) x/5715 yrs

  30. Using A = Ao(1/2)n A = .43 disintegrations Ao = 15.2 disintegrations n = x/5715 yrs .43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs .028 = (.5) x/5715 yrs log.028 = x/5715 yrs( log.5)

  31. Using A = Ao(1/2)n A = .43 disintegrations Ao = 15.2 disintegrations n = x/5715 yrs .43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs .028 = (.5) x/5715 yrs log.028/ log.5 = x/5715 yrs

  32. Using A = Ao(1/2)n A = .43 disintegrations Ao = 15.2 disintegrations n = x/5715 yrs .43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs .028 = (.5) x/5715 yrs log.028/ log.5 = x/5715 yrs 5.16 = x/5715 yrs

  33. Using A = Ao(1/2)n A = .43 disintegrations Ao = 15.2 disintegrations n = x/5715 yrs .43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs .028 = (.5) x/5715 yrs log.028/ log.5 = x/5715 yrs 5.16 = x/5715 yrs X = 29395.6 yrs

  34. Rinky think method… 15.2 disintegrations when start so divide by 2 until reach the final amount of .43 (or close to it!). 15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475

  35. Rinky think method… 15.2 disintegrations when start so divide by 2 until reach the final amount of .43 (or close to it!). 15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475 Count how many half-lives have elapsed…

  36. Rinky think method… 15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475 1 2 3 4 5+ Count how many half-lives have elapsed…

  37. Rinky think method… 15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475 1 2 3 4 5+ Multiply half-life of carbon-14 by the number of half-lived elapsed.

  38. Rinky think method… 15.2 --> 7.6 --> 3.8 --> 1.9 --> .95 --> .475 1 2 3 4 5+ a little 5715 years x 5 = 28,575 years + a little This answer is somewhat close to the formula method, thus an acceptable answer on the test!

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