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4. Radioactive Decay

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  1. 4. Radioactive Decay • radioactive transmutation and decay are synonymous expressions • 4 main series 4n 232Thorium 4n + 2 238Uranium-Radium 4n + 3 235Actinium 4n + 1 237Neptunium 4.1 Decay Series

  2. 4.2a Law and Energy of Radioactive Decay • radioactive decay law follows Poisson statistics behaves as where: N is the number of atoms of a certain radionuclide; -dN/dt is the disintegration rate; and  is the disintegration constant in sec-1

  3. 4.2a Law and Energy of Radioactive Decay • law of radioactive decay describes the kinetics of a reaction Where A is the mother radionuclide; B is the daughter nuclide; X is the emitted particle; and E is the energy set free by the decay process (also known as Q-value)

  4. 4.2a Law and Energy of Radioactive Decay • radioactive decay only possible when E > 0 which can be calculated as • however decay may only arise if nuclide A surmounts an energy barrier with a threshold ES or through quantum mechanical tunneling

  5. 4.2b Kinetics of Radioactivity Half-Life • the time for any given radioisotope to decrease to 1/2 of its original quantity • range from a few microseconds to billions of years

  6. 4.2b Kinetics of Radioactivity t1/2 = 5 years

  7. 4.2b Kinetics of Radioactivity • each isotope has its own distinct half-life (t1/2) and in almost all cases no operation, physical or chemical, can alter the transformation rate 1st half-life  50% decay 2nd half-life  75% decay 3rd half-life  87.5% decay 4th half-life  93.75% decay 5th half-life  96.87% decay 6th half-life  98.44% decay 7th half-life  99.22% decay

  8. 4.2c Probability of Disintegration • number of nuclei dN in a time interval dt will be proportional to that time interval and to the number of nuclei N that are present; or at any time t there are N nuclei  dN = - Ndt • where  is the proportionality constant and the -ve sign is introduced because N decreases

  9. 4.2c Probability of Disintegration • at t = 0: N = N0 therefore lnN0 = C  • the fraction of any radioisotope remaining after n half-lives is given by

  10. 4.2c Probability of Disintegration • where No is the original quantity and N is the quantity after n half lives

  11. 4.2c Probability of Disintegration • if the time t is small compared with the half-life of the radionuclide ( t<<t1/2)then we can approximate

  12. 4.2c Probability of Disintegration Average Life of an Isotope • it is equally important to know the average life of an isotope 

  13. 4.2c Probability of Disintegration Decay Constant Problems • what is the  constant 52V which has a t1/2 = 3.74 min.?

  14. 4.2c Probability of Disintegration • what is the  constant for 51Cr which has a t1/2 = 27.7 days? • what is the  constant for 226Ra which has t1/2 = 1622 yrs

  15. 4.2c Probability of Disintegration Decay Problem • what % of a given amount of 226Ra will decay during a period of 1000 years? 1/2 life of 226Ra = 1622 yr

  16. 4.2c Probability of Disintegration • therefore the percentage transformed during the 1000 year period is: 100% - 64.5% = 35.5%

  17. 4.2d Activity • Curie (Ci), originally defined as the activity of 1 gm of Ra in which 3.7  1010 atoms are transformed per sec • in S.I. units activity is measured in Becquerel (Bq), where 1 Bq = 1 tps -> the quantity of radioactive material in which one atom is transformed per sec

  18. 4.2d Activity activity of a radionuclide is given by its disintegration rate

  19. 4.2d Activity • equal weights of radioisotopes do not give equivalent amounts of radioactivity • 238U and its daughter 234Th have about the same no. of atoms per gm. However their half- lives are greatly different • 238U = 4.5  109 yr; 234Th = 24.1 days • therefore, 234Th is transforming 6.8  1010 faster than 238U

  20. 60Co , 0.314 MeV , 1.1173 MeV 60Ni , 1.332 MeV 4.2d Activity 1 Bq with 3 emissions

  21. 42K , 2.04 MeV 18% , 1.53 MeV 42Ca 4.2d Activity 1 Bq with 1.18 emissions

  22. 4.2d Activity 1 kilobecquerel (kBq) = 103 Bq 1 megabecquerel (MBq) = 106 Bq 1 gigabecquerel (GBq) = 109 Bq 1 terabecquerel (TBq) = 1012 Bq 1 millicurie (mCi) = 10-3 Ci 1 microcurie (μCi = 10-6 Ci 1 nanocurie (nCi) = 10-9 Ci 1 picocurie (pCi) = 10-12 Ci 1 femtocurie (fCi) = 10-15 Ci 1 Ci = 3.7  1010 Bq

  23. 4.2d Activity • since activity A is proportional to N, the number of atoms, we get A = A0e-t • the mass m of radioactive atoms can be calculated from their number N; activity A; M mass of nuclide; and Nav Avogadro’s number • ( 6.02 X 1023)

  24. 4.2d Activity Problem ● how much time is required for 5 mg of 22Na (t1/2 = 2.60 y) to reduce to 1 mg? ● since the mass of a sample willbe proportional to the no. of atoms in the sample get

  25. 4.2d Activity Specific Activity • the relationship between mass of the material and activity or  AS (SA) = no. of Bq's/unit mass or volume

  26. 4.2d Activity • SA can also be represented in combined mathematical known terms

  27. 4.2d Activity • SA may also be derived by using the fact that there are 3.7  1010 tps in 1 gm of 226Ra

  28. 4.2d Activity Problem • calculate the specific activity of 14C (t1/2 = 5730 yrs)

  29. 4.2d Activity Problem • potassium (atomic weight = 39.102 AMU) contains: • 93.10 atom % 39K, having atomic mass 38.96371 AMU • 0.0118 atom % 40K, which has a mass of 40.0 AMU and is radioactive with: t1/2 = 1.3  109 yr • 6.88 atom % 41K having a mass of 40.96184 AMU

  30. 4.2d Activity • estimate the specific activity of naturally occurring potassium • specific activity refers to the activity of 1 g material • 1 g of naturally occurring potassium contains: 1.18  10-4 g 40K plus non-radioactive isotopes

  31. 4.2d Activity

  32. 4.2d Activity Problem • prior to use of nuclear weapons, the SA of 14C in soluble ocean carbonates was found to be 16 dis/min ·g carbon • amount of carbon in these carbonates has been estimated as 4.5  1016 kg • how many MCi of 14C did the ocean carbonates contain?

  33. 4.2d Activity Problem • a mixture of 239Pu and 240Pu has a specific activity of 6.0  109 dps • the half-lives of the isotopes are 2.44  104 and 6.58  103y, respectively • calculate the isotopic composition

  34. 4.2d Activity • for 239Pu • for 240Pu

  35. 4.2d Activity • number of seconds in a year is • for 239Pu: A = 2.27 109/s  g • for 240Pu: A = 8.37 109/s  g • let the fraction of 239Pu = x; then the fraction 240Pu = 1 - x

  36. 4.2d Activity (2.27  109)x+(1 – x)(8.37  109) = 6.0  109 (8.37  109) – (6.10 109) x = 6.0  109 2.37  109 = (6.1  109) x x = 0.39 = 39% 239Pu

  37. 4.2d Activity Problem • if 3  10-9 kg of radioactive 200Au has an activity of 58.9 Ci, what is its half-life? • no. of atoms in 3  10-9 kg of 200Au is

  38. 4.2d Activity • decay constant is found from A = N • finally

  39. 4.3 Radioactive Equilibria • net production of nuclide 2 is given by decay rate of nuclide 1 less the decay rate of nuclide 2

  40. 4.3 Radioactive Equilibria • given that: • solution of first order differential equation

  41. 4.3 Radioactive Equilibria • if nuclide 1 and 2 are separated at t = 0; then nuclide 2 is not produced and

  42. 4.3 Radioactive Equilibria • after substitution for λ: • the exponent term can be written to show the influence the ratio of

  43. 4.3 Radioactive Equilibria

  44. 4.4 Secular Equilibrium • in secular equilibrium t1/2 (1)>> t1/2 (2) so • reduces