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 = A o (1/2) n A = final amount A o = initial amount n = number of half-lives elapsed A = final amount A o = initial amount n = number of half-lives elapsed

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

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

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

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

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

9. A = A o (1/2) n A = ? A o = 1.000 mg n = 3 X = 1.000 mg (.5) 3 X =.125 mg A = ? A o = 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------------] 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--] 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 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. 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! 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! 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 = A o (1/2) n A =.95 g A o = 1.000 g n = ? Calculate n: n = T/t 1/2 T = 2 years t 1/2 = ? A =.95 g A o = 1.000 g n = ? Calculate n: n = T/t 1/2 T = 2 years t 1/2 = ?

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

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

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

21. Using A = A o (1/2) n.95 g = 1 g (.5) 2 yrs/x.95 =.5 2 yrs/x To get the exponent in a solvable position, take the logarithm of the problem: log.95 = 2 yrs/x (log.5).95 g = 1 g (.5) 2 yrs/x.95 =.5 2 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 = A o (1/2) n 95 g = 1 g (.5) 2 yrs/x.95 =.5 2 yrs/x log.95 = 2 yrs/x (log.5) Simplify by dividing both sides by log.5 log.95/log.5 = 2 yrs/x 95 g = 1 g (.5) 2 yrs/x.95 =.5 2 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 = A o (1/2) n.95 g = 1 g (.5) 2 yrs/x.95 =.5 2 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.95 g = 1 g (.5) 2 yrs/x.95 =.5 2 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 = A o (1/2) n.95 g = 1 g (.5) 2 yrs/x.95 =.5 2 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/.074.95 g = 1 g (.5) 2 yrs/x.95 =.5 2 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 = A o (1/2) n.95 g = 1 g (.5) 2 yrs/x.95 =.5 2 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.95 g = 1 g (.5) 2 yrs/x.95 =.5 2 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… 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? 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 = A o (1/2) n A =.43 disintegrations A o = 15.2 disintegrations n = x/5715 yrs Calculate n: n = T/t 1/2 T = ? t 1/2 = 5715 years A =.43 disintegrations A o = 15.2 disintegrations n = x/5715 yrs Calculate n: n = T/t 1/2 T = ? t 1/2 = 5715 years

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

30. Using A = A o (1/2) n A =.43 disintegrations A o = 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) A =.43 disintegrations A o = 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 = A o (1/2) n A =.43 disintegrations A o = 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 A =.43 disintegrations A o = 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 = A o (1/2) n A =.43 disintegrations A o = 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 A =.43 disintegrations A o = 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 = A o (1/2) n A =.43 disintegrations A o = 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 A =.43 disintegrations A o = 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 --.475 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 --.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 --.475 Count how many half-lives have elapsed… 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 --.475 Count how many half-lives have elapsed…

36. Rinky think method… 15.2 --> 7.6 --> 3.8 --> 1.9 --.475 1 2 3 4 5+ Count how many half-lives have elapsed… 15.2 --> 7.6 --> 3.8 --> 1.9 --.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 --.475 1 2 3 4 5+ Multiply half-life of carbon-14 by the number of half-lived elapsed. 15.2 --> 7.6 --> 3.8 --> 1.9 --.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! 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!