1. 3-1 Radioactive Decay Kinetics RDCH 702 Outline Radioactive decay kinetics §Basic decay equations §Utilization of equations àMixtures àEquilibrium àBranching §Natural radiation §Dating

2. 3-2 Introduction Number of radioactive nuclei that decay in a sample decreases with time §Exponential decrease §Independent of P, T, mass action àConditions associated with chemical kinetics *Electron capture and internal conversion can be affected by conditions §Specific for isotope §Irreversible Decay of given radionuclide is random §Statistical àEvaluate behavior of group

3. 3-3 Basic Decay Equations Decay is 1 st order Rate proportional to amount of parent isotope §Equal to the rate of isotope disintegration §Proportional to number of radioactive nuclei àrate of decay=decay constant*# radioactive nuclei *A=  Decay constant is average decay probability per nucleus for a unit time Represented by

4. 3-4 Basic decay equations probability of disintegration independent past history and present circumstances §depends only on length of time interval Probability of decay: p=  t Probability of not decaying: 1-p=1-  t  (1-  t) n =probability that atom will survive n intervals of  t  n  t=t, therefore (1-  t) n =(1- t/n) n lim n  ∞ (1+x/n) n =e x, (1- t/n) n =e - t is limiting value Considering N o initial atoms §fraction remaining unchanged after time t is àN/N o = e - t *N is number of atoms remaining at time t N=N o e - t

5. 3-5 Radioactivity as Statistical Phenomenon: Error from Counting Binomial Distribution for Radioactive Disintegrations §Probability W(m) of obtaining m disintegrations in time t from N o original radioactive atoms For radioactive disintegration  p=1-e- t  Average number M of atoms disintegrating in time t is M=N o (1-e - t )  For small t, M=N o t and disintegration R=M/t=N o  Small t means count time is short compared to half life  Corresponds to -dN/dt= N Expected Standard Deviation M is number of counts Relative error =  -1 What is a reasonable number of counts for an experiment?  More counts, lower error

6. 3-6 Measured Activity In practicality, activity (A) is used instead of the number of atoms (N). A= c t, m §where c is the detection coefficient §A=A O e - t Units §Curie à3.7E10 decay/s *1 g Ra §Becquerel à1 decay/s

7. 3-7 Half Life and decay constant Half-life is time needed to decrease nuclides by 50% Relationship between t 1/2 and N/N o =1/2=e - t ln(1/2)=- t 1/2 ln 2= t 1/2 t 1/2 =(ln 2)/

8. 3-8 Half lives Large variation in half-lives for different isotopes §Short half-lives can be measured àEvaluate activity over time *Observation on order of half-life §Long half-lives àBased on decay rate and sample *Need to know total amount of nuclide in sample *A= n ØA is activity, n is number of nuclei

9. 3-9 Exponential Decay Average Life (  ) for a radionuclide §found from sum of times of existence of all atoms divided by initial number of nuclei  1/ =1/(ln2/t 1/2 )=1.443t 1/2 =  àAverage life greater than half life by factor of 1/0.693 àduring time 1/ activity reduced to 1/e it’s initial value

10. 3-10 Lifetime Total number of nuclei that decay over time §Dose §Atom at a time Couple with Heisenberg uncertainty principle   E  t ≥h/2    t is  àwith energy in eV   E ≥(4.133E-15 eV s/2     is decay width *Resonance energy  t 1/2 =1 sec,  1.44 s,  =4.56E-16 eV

11. 3-11 Width and energy Need very short half-lives for large widths Useful in Moessbauer spectroscopy  Absorption distribution is centered around E   E  emission centered E    E. overlapping part of the peaks can be changed by changing the temperature of the source and/or the absorber.

12. 3-12 Equations N t =N o e - t  N=number of nuclei, = decay constant, t=time àAlso works for A (activity) or C (counts) *A t =A o e - t, C t =C o e - t A= N 1/ =1/(ln2/t 1/2 )=1.443t 1/2 = 

13. 3-13 Half-life calculation Using N t =N o e - t For an isotope the initial count rate was 890 Bq. After 180 minutes the count rate was found to be 750 Bq §What is the half-life of the isotope  750=890exp(- *180 min)  750/890=exp(- *180 min)  ln(750/890)= - *180 min  -0.171/180 min= -   min -1  =  ln2/t 1/2 àt 1/2 =ln2/9.5E-4=729.6 min

14. 3-14 Half-life calculation A= N A 0.150 g sample of 248 Cm has a alpha activity of 0.636 mCi. §What is the half-life of 248 Cm? àFind A *0.636 E-3 Ci (3.7E10 Bq/Ci)=2.35E7 Bq àFind N *.150 g x 1 mole/248 g x 6.02E23/mole= 3.64E20 atoms   A/N= 2.35E7 Bq/3.64E20 atoms=6.46E-14 s -1 *t 1/2 =ln2/  6.46E-14 s -1 =1.07E13 s *1.07E13 s=1.79E11 min=2.99E9 h=1.24E8 d =3.4E5 a

15. 3-15 Counting A= N Your gamma detector efficiency at 59 keV is 15.5 %. What is the expected gamma counts from 75 micromole of 241 Am? §Gamma branch is 35.9 % for 241 Am  C=(0.155)(0.359) N §t 1/2 =432.7 a* (3.16E7 s/a)=1.37E10 s  =ln2/1.37E10 s=5.08E-11 s -1 §N=75E-6 moles *6.02E23/mole=4.52E19 atoms C=(0.155)(0.359)5.08E-11 s -1 *4.52E19 =1.28E8 Bq

16. 3-16 Decay Scheme

17. 3-17 Specific activity Activity of a given amount of radionuclide  Use A= N àUse of carrier should be included SA of 226 Ra §1 g 226 Ra, t 1/2 = 1599 a §1 g * 1 mole/226 g * 6.02E23 atoms/mole = 2.66E21 atom = N  t 1/2 =1599 a *3.16E7 s/a = 5.05E10 s  =ln2/ 5.05E10 s =1.37E-11 s -1  A= 1.37E-11 s -1 * 2.66E21=3.7E10 Bq

18. 3-18 Specific Activity 1 g 244 Cm, t 1/2 =18.1 a §1 g * 1 mole/244 g * 6.02E23 atoms/mole = 2.47E21 atom = N  t 1/2 =18.1 a *3.16E7 s/a = 5.72E8 s  =ln2/ 5.72E8 s =1.21E-9 s -1  A= 1.21E-9 s -1 * 2.47E21=2.99E12 Bq Generalized equation for 1 g §6.02E23/Isotope mass *2.19E-8/ t 1/2 (a) §1.32E16/(Isotope mass* t 1/2 (a))

19. 3-19 Specific Activity Isotopet 1/2 aSA (Bq/g) 14C57151.65E+11 228Th1.91E+003.03E+13 232Th1.40E+104.06E+03 233U1.59E+053.56E+08 235U7.04E+087.98E+04 238U4.47E+091.24E+04 237Np2.14E+062.60E+07 238Pu8.77E+016.32E+11 239Pu2.40E+042.30E+09 242Pu3.75E+051.45E+08 244Pu8.00E+076.76E+05 241Am4.33E+021.27E+11 243Am7.37E+037.37E+09 244Cm1.81E+012.99E+12 248Cm3.48E+051.53E+08

20. 3-20

21. 3-21 Specific Activity Activity/mole §N=6.02E23 SA (Bq/mole) of 129 I, t 1/2 =1.57E7 a  t 1/2 =1.57E7 a *3.16E7 s/a = 4.96E14 s  =ln2/ 4.96E14 s =1.397E-15 s -1  A= 1.397E-15 s -1 *6.02E23=8.41E8 Bq Generalized equation §SA (Bq/mole)=1.32E16/t 1/2 (a)

22. 3-22 Specific Activity Isotopet 1/2 a SA (Bq/mole) 3H12.31.07E+15 14C57152.31E+12 22Na2.65.08E+15 55Fe2.734.84E+15 228Th1.91E+006.91E+15 232Th1.40E+109.43E+05 233U1.59E+058.30E+10 235U7.04E+081.88E+07 238U4.47E+092.95E+06 237Np2.14E+066.17E+09 238Pu8.77E+011.51E+14 239Pu2.40E+045.50E+11 242Pu3.75E+053.52E+10 244Pu8.00E+071.65E+08 241Am4.33E+023.05E+13 243Am7.37E+031.79E+12 244Cm1.81E+017.29E+14 248Cm3.48E+053.79E+10

23. 3-23 SA with carrier 1E6 Bq of 152 Eu is added to 1 mmole Eu. §Specific activity of Eu (Bq/g) §Need to find g Eu à1E-3 mole *151.96 g/mole = 1.52E-1 g à=1E6 Bq/1.52E-1 g =6.58E6 Bq/g *=1E9 Bq/mole What is SA after 5 years §t 1/2 =13.54 a à= 6.58E6*exp((-ln2/13.54)*5)= *5.09E6 Bq/g

24. 3-24 Lifetime Atom at a time chemistry 261 Rf lifetime §Find the lifetime for an atom of 261 Rf àt 1/2 = 65 s   =1.443t 1/2   =93 s Determines time for experiment Method for determining half-life

25. 3-25 Mixtures of radionuclides Composite decay §Sum of all decay particles àNot distinguished by energy Mixtures of Independently Decaying Activities §if two radioactive species mixed together, observed total activity is sum of two separate activities: A=A 1 +A 2 =c 1 1 N 1 +c 2 2 N 2 §any complex decay curve may be analyzed into its components àGraphic analysis of data is possible

26. 3-26 Can determine initial concentration and half-life of each radionuclide =0.554 t 1/2 =1.25 hr =0.067 t 1/2 =10.4 hr

27. 3-27 Parent – daughter decay Isotope can decay into radioactive isotope §Uranium decay series §Lower energy §Different properties àA àZ àSpin àParity For a decay parent -> daughter §Rate of daughter formation dependent upon parent decay rate- daughter decay rate

28. 3-28 Parent - daughter For the system 1 decays into 2 Rearranging gives Solve and substitute for N 1 using N 1t =N 1o e - t §Linear 1 st order differential equation àSolve by integrating factors Multiply by e 2t

29. 3-29 Parent-daughter Integrate from t 0->t Multiply by e - 2t and solve for N 2 Growth of daughter from parent Decay of initial daughter

30. 3-30 Parent daughter relationship Can solve equation for activity from A=  Find maximum daughter activity based on dN/dt=0 Solve for t For 99m Tc (t 1/2 =6.01 h) from 99 Mo (2.75 d), find time for maximum daughter activity  Tc =2.8 d -1, Mo =0.25 d -1

31. 3-31 Half life relationships Can simplify relative activities based on half life relationships No daughter decay §No activity of daughter §Number of daughter atoms due to parent decay No Equilibrium §if parent is shorter-lived than daughter ( 1  2 ), no equilibrium attained at any time §daughter reaches maximum activity when 1 N 1 = 2 N 2 àAll parents decay, then decay is based on daughter

32. 3-32 Half life relationships Transient equilibrium §Parent half life greater than 10 x daughter half life  ( 1 < 2 ) Parent daughter ratio becomes constant over time §As t goes toward infinity

33. 3-33

34. 3-34 Half life relationship Secular equilibrium §Parent much longer half-life than daughter à1E4 times greater ( 1 << 2 ) §Parent activity does not measurably decrease in many daughter half-lives

35. 3-35

36. 3-36 Many Decays Bateman solution Only parent present at time 0

37. 3-37 Branching decay Branching Decay §partial decay constants must be considered àA has only one half life §if decay chain branches and two branches are later rejoined, the two branches are treated as separate chains àproduction of common member beyond branch point is sum of numbers of atoms formed by the two paths Branching ratio is based on relative constants  i  t

38. 3-38 Branching Decay For a branching decay of alpha and beta  A t =A  +A   A= N, so * t N =  N+  N  t =  +   1=A   A t  +A   A t   1=   t  +   t Consider 212 Bi, what is the half life for each decay mode? §Alpha branch 36 %, beta branch 64 % §t 1/2 =60.55 min  t =0.0114; 0.36=   t ; 0.36=   0.0114;  =0.0041 àt 1/2 alpha = 169 min  t =  +   0.0114=0.0041+   0.0073=  àt 1/2 beta = 95.0 min

39. 3-39

40. 3-40 Natural Radionuclides 70 naturally occurring radioactive isotopes §Mainly decay from actinides §Tritium § 14 C § 40 K 70 kg “reference man,” §4400 Bq of 40 K §3600 Bq of 14 C US diet §1 pCi/day of 238 U, 226 Ra, and 210 Po air § ~ 0.15 pCi/L of 222 Rn earth’s crust §~10 ppm and ~4 ppm of the radioelements Th and U. interior heat budget of the planet Earth is dominated by the contributions from the radioactive decay of U, Th, and K

41. 3-41 Production of radionuclides R=N 0    =cross section   =neutron flux To full consider produced nuclei need saturation factor §Account for decay during production   1-e -( t) ) àt=time of irradiation, decay constant of product If the disintegration rate of radioactive product at end bombardment is divided by saturation factor, the formation rate obtained  R=A/  1-e -( t) ) half life% 150 275 387.5 493.75 596.875

42. 3-42 Environmental radionuclides primordial nuclides that have survived since the time the elements were formed §t 1/2 >1E9 a §Decay products of these long lived nuclides à 40 K., 87 Rn, 238 U, 235 U, 232 Th cosmogenic are shorter lived nuclides formed continuously by the interaction of comic rays with matter § 3 H., 14 C, 7 Be à 14 N(n, 1 H ) 14 C (slow n) à 14 N(n, 3 H ) 12 C (fast n) anthropogenic are nuclides introduced into the environment by the activities of man §Actinides and fission products § 14 C and 3 H

43. 3-43 Dating Radioactive decay as clock  Based on N t =N o e - t àSolve for t N 0 and N t are the number of radionuclides present at times t=0 and t=t §N t from A = λN t the age of the object §Need to determine N o àFor decay of parent P to daughter D total number of nuclei is constant

44. 3-44 Dating P t =P o e - t Measuring ratio of daughter to parent atoms §no daughter atoms present at t=0 §that they are all due to the parent decay §none have been lost during time t A mineral has a 206 Pb/ 238 U =0.4. What is the age of the mineral? §t=(1/(ln2/4.5E9))ln(1+0.4) à2.2E9 years

45. 3-45 Dating 14 C dating §Based on constant formation of 14 C àNo longer uptakes C upon organism death 227 Bq 14 C /kgC at equilibrium What is the age of a wooden sample with 0.15 Bq/g C? §t=(1/(ln2/5730 a))*ln(0.227/0.15)=3420 a

46. 3-46 Dating Determine when Oklo reactor operated §Today 0.7 % 235 U §Reactor 3.5 % 235 U  Compare 235 U/ 238 U (U r ) ratios and use N t =N o e - t

47. 3-47 Questions Make excel sheets to calculate §Activity to mass àCalculate specific activity Should also consider §Concentration and volume to activity àDetermine activity for counting §Parent to progeny àDaughter and granddaughter *i.e., 239 U to 239 Np to 239 Pu