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3-1 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

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3-2 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

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3-3 Parent - daughter How does daughter isotope change with parent decay §isotope 1 (parent) decays into isotope 2 (daughter) 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

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3-4 Parent-daughter Integrate over t Multiply by e - 2 t and solve for N 2 Growth of daughter from parent Initial daughter

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3-5 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

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3-6 Half life relationships Can simplify relative activities based on half life relationships No daughter decay §No activity from daughter §Number of daughter atoms due to parent decay Daughter Radioactive No Equilibrium §If parent is shorter-lived than daughter ( 1 2 ) àno 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

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3-7 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

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3-8 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

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3-9 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

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3-10 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

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3-11 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

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3-12 Equations can be used to evaluate radionuclide activity and dose over time

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3-13 Equations for production reactions: Cross Sections Probability of a nuclear process is generally expressed in terms of a cross section §dimensions of an area Originates from probability for reaction between nucleus and impinging particle is proportional to the cross-sectional target area presented by the nucleus §Doesnt hold for charged particles that have to overcome Coulomb barriers or for slow neutrons Total cross section for collision with fast particle is never greater than twice the geometrical cross- sectional area of the nucleus cross section is close to 1 barn for this case 10 -24 cm 2 =1 barn

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3-14 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

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3-15 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

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3-16 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

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3-17 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)

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3-18 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

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3-19 Environmental radionuclides and dating Primordial nuclides that have survived since time elements were formed §t 1/2 >1E9 a §Decay products of these long lived nuclides à 40 K, 87 Rb, 238 U, 235 U, 232 Th shorter lived nuclides formed continuously by 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 nuclides introduced into the environment by activities of man §Actinides and fission products § 14 C and 3 H

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3-20 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

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3-21 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

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3-22 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?

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3-23 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

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3-24 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

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3-25 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?

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3-26 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

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3-27 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

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Mr. ShieldsRegents Chemistry U02 L03 Nuclear Decay Series Uranium has an atomic number greater than 83. Therefore it is naturally radioactive. Most abundant.

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