1. 3-1 Quiz 1 1.(5 Points) Provide a route for the production of 238 Pu? 5 point bonus: What is the use of this isotope? Give an example of where it has been used. 2.(5 Points) Below is a mass parabola from the table of the isotopes for A=40. Use this figure to show which isotopes of A=40 are stable?

2. 3-2 Quiz 1 3.(15 Points) Identify the daughter and the decay mode for the following isotopes Parent isotopeDecay modeDaughter Isotope 210 Po 196 Pb 204 Bi 222 At 99m Tc 99 Tc 14 C 239 Pu 261 Rf

3. 3-3 Quiz 1 4.(5 Points) Provide the number of naturally occurring isotopes for the elements below. This includes long lived radioactive isotopes ElementNumber of Stable Isotopes Re V K La Sn Sb In H Pm Lr

4. 3-4 Quiz 1 5.(5 Points) Provide the spin and parity of the following isotopes IsotopeSpinParity 242m Am 239 Pu 240 Pu 119m Sn 65 Ni 99 Tc 242 Am 179m Hf

5. 3-5 Quiz 1 6.(10 Points) Provide the cumulative fission yields for the A isobars from 233 U, 235 U, and 239 Pu. A 233 U 235 U 239 Pu 116 95 72 160 99

6. 3-6 Quiz 1 7.(10 Points) Provide the decay constants (in s -1 ) for the following isotopes 8.(5 Points) List the 4 stable odd-odd isotopes Isotope  (s -1 ) 50 Ca 81 Zn 95 Zr 137 Cs 151 Sm 241 Am 242 Pu

7. 3-7 Quiz 1 9. (15 Points) Determine the Q value for the following reactions 10.(10 Points) Calculate the mass of 1 mL of nuclear matter. Base the calculation on an 56 Fe nucleus, with r o = 1.3 fm to determine the radius of 56 Fe. Reaction or decayQ value (MeV) Alpha decay of 239 Pu 45 Sc + 4 He  5 H+ 44 Ti b - decay of 137 Cs n+ 235 U  232 Th+α n+ 235 U  236 U EC of 192 Hg Formation of stable Au from bombarding 208 Pb with a proton

8. 3-8 Quiz 1 11.(5 Points) Please provide the gamma decay intensities for the data below 12. (10 Points) What is nuclear skin thickness? How can one measure nuclear radii? Isotope Gamma Energy (keV) %% 46 Sc1120.5 103 Ru497.1 198 Au411.8 241 Am59.1 152 Eu121.8 152 Eu1408.0 152 Eu344.3

9. 3-9 CHEM 312: Lecture 3 Radioactive Decay Kinetics Outline Readings: Modern Nuclear Chemistry Chapter 3; Nuclear and Radiochemistry Chapters 4 and 5 Radioactive decay kinetics §Basic decay equations §Utilization of equations àMixtures àEquilibrium àBranching àCross section §Natural radiation §Dating

10. 3-10 Expected Standard Deviation Solve with: Apply to radioactive decay §M is the number of atoms decaying àNumber of counts for a detector Relative error =  -1 What is a reasonable number of counts  More counts, lower error Countserror% error 103.1631.62 10010.00 100031.623.16 10000100.001.00

11. 3-11 Important Equations! N t =N o e - t NN =number of nuclei, = decay constant, t=time àAàAlso works for A (activity) or C (counts) *A*A t =A o e - t, C t =C o e - t A= N 1/ =1/(ln2/t 1/2 )=1.443t 1/2 =  Error §M§M is number of counts

12. 3-12 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

13. 3-13 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 *0.150 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

14. 3-14 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 counts/second

15. 3-15 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 t =A 1 +A 2 = 1 N 1 + 2 N 2 §any complex decay curve may be analyzed into its components àGraphic analysis of data is possible =0.554 hr -1 t 1/2 =1.25 hr =0.067 hr -1 t 1/2 =10.4 hr

16. 3-16 Parent – daughter decay Isotope can decay into radioactive isotope §Uranium and thorium decay series àAlpha and beta *A change from alpha decay Different designation §4n ( 232 Th) §4n+2 ( 238 U) §4n+3 ( 235 U) For a decay parent -> daughter §Rate of daughter formation dependent upon parent decay rate- daughter decay rate

17. 3-17 Parent-daughter Integrate over t Multiply by e - 2 t and solve for N 2 Growth of daughter from parent Initial daughter

18. 3-18 Parent daughter relationship Find N, 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

19. 3-19 Many Decays Can use the Bateman solution to calculate entire chain Bateman assumes only parent present at time 0 Program for Bateman http://www.ergoffice.com/downloads.aspx

20. 3-20 Branching decay Branching Decay §partial decay constants must be considered àIsotope has only one half life §if decay chain branches and two branches are later rejoined, 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 is the % of the decay branch

21. 3-21 Branching Decay For a branching decay of alpha and beta  t =  +   Branching ratio =  i  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 min -1 ; 0.36=   t ; 0.36=   0.0114 min -1  =0.0041 min -1 àt 1/2 alpha = 169 min  t =  +   0.0114 min -1 =0.0041 min -1 +   0.0073 min -1 =  àt 1/2 beta = 95.0 min

22. 3-22 Cross sections Accelerator: beam of particles striking a thin target with minimum beam attenuation When a sample is embedded in a uniform flux of particles incident on it from all direction, such as in a nuclear reactor, the cross section is defined: §R i = # of processes of type under consideration occurring in the target per unit time §I= # of incident particles per unit time §n= # of nuclei/cm 3 §x=target thickness (cm) §  =flux of particles/cm 2 /sec §N=number of nuclei contained in sample 10 -24 cm 2 =1 barn

23. 3-23 Production of radionuclides   =cross section   =neutron flux §t=time of irradiation   1-e -( t) ) *maximum level (saturation factor) Activity of radioactive product at end bombardment is divided by saturation factor, formation rate is obtained  R=A/  1-e -( t) ) half life% 150 275 387.5 493.75 596.875

24. 3-24 Nuclei production: Short irradiation compared to half-life Find amount of 59 Fe (t 1/2 =44.5 d, = 1.803E-7 s -1 ) from irradiation of 1 g of Fe in a neutron flux of 1E13 n/cm 2 /s for 1 hour  58 Fe(n,  ) 59 Fe: 58 Fe+ n   + 59 Fe  1.3E-24 cm 2 §N o = 1g/55.845 g/mol *6.02E23 atom/mol*0.00282 § N o =3.04E19 atom R= 1E13 n/cm 2 /s *1.3E-24 cm 2 * 3.04E21 atom R=3.952E8 atoms/sec 1.423E12 atoms 59 Fe in 1 hour

25. 3-25 Nuclei production: Long irradiation compared to half-life Find amount of 56 Mn (t 1/2 =2.578 hr, = 7.469E-5 s -1 ) from irradiation of 1 g of Mn in a neutron flux of 1E13 n/cm 2 /s for 1 hour  55 Mn(n,  ) 56 Mn: 55 Mn+ n   + 56 Mn  13.3E-24 cm 2 §N o = 1g/54.93804 g/mol *6.02E23 atom/mol § N o =1.096E22 atom R= 1E13 n/cm 2 /s *13.3E-24 cm 2 * 1.096E22 atom R=1.457E12 atoms/sec 5.247E15 atoms 56 Mn in 1 hour (does not account for decay)

26. 3-26 Formation rate from activity R=A/  1-e -( t) ) 4.603E15 atoms 56 Mn (t 1/2 =2.578 hr, = 7.469E- 5 s -1 ) from 1 hour irradiation A= N= 4.603E15* 7.469E-5 =3.436E11 Bq R=A/  1-e -( t) ) R= 3.436E11/(1-exp(- 7.469E-5 *3600)) R=1.457E12 atom/sec

27. 3-27 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

28. 3-28 Dating P t =P o e - t Measuring ratio of daughter to parent atoms §No daughter atoms present at t=0 §All daughter due to parent decay § No daughter lost during time t A mineral has a 206 Pb/ 238 U =0.4. What is the age of the mineral? à2.2E9 years

29. 3-29 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?

30. 3-30 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

31. 3-31 Topic review Utilize and understand the basic decay equations Relate half life to lifetime Understand relationship between count time and error Utilization of equations for mixtures, equilibrium and branching Use cross sections for calculation nuclear reactions and isotope production Utilize the dating equation for isotope pair

32. 3-32 Study Questions Compare and contrast nuclear decay kinetics and chemical kinetics. If M is the total number of counts, what is the standard deviation and relative error from the counts? Define Curie and Becquerel How can half-life be evaluated? What is the relationship between the decay constant, the half-life, and the average lifetime? 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? A 0.150 g sample of 248 Cm has a alpha activity of 0.636 mCi. What is the half-life of 248 Cm? What is the half life for each decay mode for the isotope 212 Bi? How are cross sections used to determine isotope production rate? Determine the amount of 60 Co produced from the exposure of 1 g of Co metal to a neutron flux of 10 14 n/cm 2 /sec for 300 seconds. What are the basic assumptions in using radionuclides for dating?

33. 3-33 Pop Quiz You have a source that is 0.3 Bq and the source is detected with 50 % efficiency. It is counted for 10 minutes. Which total counts shown below are not expected from these conditions? 95, 81, 73, 104, 90, 97, 87 Submit by e-mail or bring to class on 24 September Comment on Blog

34. 3-34 Useful projects Make excel sheets to calculate §Mass or mole to activity àCalculate specific activity §Concentration and volume to activity àDetermine activity for counting §Isotope production from irradiation §Parent to progeny àDaughter and granddaughter *i.e., 239 U to 239 Np to 239 Pu

35. 3-35 Alpha Decay Readings §Nuclear and Radiochemistry: Chapter 3 §Modern Nuclear Chemistry: Chapter 7 Energetics of Alpha Decay Theory of Alpha Decay Hindrance Factors Heavy Particle Radioactivity Proton Radioactivity Identified at positively charged particle by Rutherford §Helium nucleus ( 4 He 2+ ) based on observed emission bands §Energetics àAlpha decay energies 4-9 MeV àOriginally thought to be monoenergetic, fine structure discovered A Z  (A-4) (Z-2) + 4 He + Q 

36. 3-36 Alpha Decay Energetics Q value positive for alpha decay §Q value exceeds alpha decay energy  m  T  = m d T d §m d and T d represent daughter From semiempirical mass equation §emission of an α-particle lowers Coulomb energy of nucleus §increases stability of heavy nuclei while not affecting overall binding energy per nucleon àtightly bound α-particle has approximately same binding energy/nucleon as original nucleus *Emitted particle must have reasonable energy/nucleon *Energetic reason for alpha rather than proton Energies of alpha particles generally increase with atomic number of parent

37. 3-37 Energetics Calculation of Q value from mass excess  238 U  234 Th +  + Q àIsotope Δ (MeV) 238 U47.3070 234 Th40.612 4 He2.4249  Q  =47.3070 – (40.612 + 2.4249) = 4.270 MeV §Q energy divided between α particle and heavy recoiling daughter àkinetic energy of alpha particle will be slightly less than Q value Conservation of momentum in decay, daughter and alpha are equal  d =   §recoil momentum and  -particle momentum are equal in magnitude and opposite in direction §p 2 =2mT where m= mass and T=kinetic energy 238 U alpha decay energy

38. 3-38 Energetics Kinetic energy of emitted particle is less than Coulomb barrier α-particle and daughter nucleus §Equation specific of alpha §Particles touching §For 238 U decay Alpha decay energies are small compared to required energy for reverse reaction Alpha particle carries as much energy as possible from Q value, For even-even nuclei, alpha decay leads to ground state of daughter nucleus §as little angular momentum as possible §ground state spins of even-even parents, daughters and alpha particle are l=0

39. 3-39 Alpha decay theory Distance of closest approach for scattering of a 4.2 MeV alpha particle is ~62 fm §Distance at which alpha particle stops moving towards daughter §Repulsion from Coulomb barrier Alpha particle should not get near nucleus §should be trapped behind a potential energy barrier Wave functions are only completely confined by infinitely highpotential energy barriers §With finite size barrier wave function has different behavior §main component inside barrier §finite piece outside barrier Tunneling §trapped particle has component of wave function outside potential barrier §Some probability to go through barrier àRelated to decay probability §Higher energy has higher tunneling probability Alpha decay energy VcVc

40. 3-40 Alpha Decay Theory Closer particle energy to barrier maximum more likely particle will penetrate barrier More energetic alpha will encounter barrier more often §Increase probability of barrier penetration due to Geiger Nuttall law of alpha decay §constants A and B have Z dependence. simple relationship describes data on α-decay §over 20 orders of magnitude in decay constant or half-life  1 MeV change in  -decay energy results in a change of 10 5 in half-life

41. 3-41 Alpha Decay Calculations Alpha particle barrier penetration from Gamow §T=e -2G Determination of decay constant from potential information Using square-well potential, integrating and substituting §Z daughter, z alpha

42. 3-42 Gamow calculations From Gamow Calculated emission rate typically one order of magnitude larger than observed rate §observed half-lives are longer than predicted §Observation suggest a route to evaluate alpha particle pre-formation factor

43. 3-43 Alpha Decay Theory Even-even nuclei undergoing l=0 decay §average preformation factor is ~ 10 -2 §neglects effects of angular momentum àAssumes α-particle carries off no orbital angular momentum (ℓ = 0) §If α decay takes place to or from excited state some angular momentum may be carried off by α-particle §Results in change in decay constant when compared to calculated

44. 3-44 Hindered  -Decay Previous derivation only holds for even-even nuclei §odd-odd, even-odd, and odd-even nuclei have longer half-lives than predicted due to hindrance factors Assumes existence of pre-formed  -particles §Ground-state transition from nucleus containing odd nucleon in highest filled state can take place only if that nucleon becomes part of  -particle àtherefore another nucleon pair is broken àless favorable situation than formation of an  -particle from already existing pairs in an even-even nucleus *may give rise to observed hindrance §  -particle is assembled from existing pairs in such a nucleus, product nucleus will be in an excited state àthis may explain higher probability transitions to excited states Hindrance from difference between calculation and measured half-life §Hindrance factors between 1 and 3E4 §Hindrance factors determine by àratio of measured alpha decay half life over calculated alpha decay half life àratio of calculated alpha decay constant over measured alpha decay constant

45. 3-45 Hindrance Factors Transition of 241 Am (5/2-) to 237 Np §states of 237 Np (5/2+) ground state and (7/2+) 1 st excited state have hindrance factors of about 500 (red circle) §Main transition to 60 keV above ground state is 5/2-, almost unhindered

46. 3-46 Hindrance Factors 5 classes of hindrance factors based on hindrance values §Between 1 and 4, transition is called a “favored” àemitted alpha particle is assembled from two low lying pairs of nucleons in parent nucleus, leaving odd nucleon in its initial orbital §Hindrance factor of 4-10 indicates a mixing or favorable overlap between initial and final nuclear states involved in transition §Factors of 10-100 indicate that spin projections of initial and final states are parallel, but wave function overlap is not favorable §Factors of 100-1000 indicate transitions with a change in parity but with projections of initial and final states being parallel §Hindrance factors of >1000 indicate that transition involves a parity change and a spin flip

47. 3-47 Topic Review Understand and utilize systematics and energetics involved in alpha decay Calculate Q values for alpha decay §Relate to alpha energy and fine structure Correlate Q value and half-life Models for alpha decay constant §Tunneling and potentials Hindered of alpha decay Understand proton and other charged particle emission

48. 3-48 Homework Questions Calculate alpha decay Q value and Coulomb barrier potential for following, compare values § 212 Bi, 210 Po, 238 Pu, 239 Pu, 240 Am, 241 Am What is basis for daughter recoil during alpha decay? What is relationship between Q a and alpha decay energy (T a ) What are some general trends observed in alpha decay? Compare calculated and experimental alpha decay half life for following isotopes § 238 Pu, 239 Pu, 241 Pu, 245 Pu §Determine hindrance values for odd A Pu isotopes above What are hindrance factor trends? How would one predict half-life of an alpha decay from experimental data?

49. 3-49 Pop Quiz Calculate alpha decay energy for 252 Cf and 254 Cf from mass excess data below. Which is expected to have shorter alpha decay half-life and why? Calculate alpha decay half-life for 252 Cf and 254 Cf from data below. (use % alpha decay)