RANDOM PROCESS: Identical to Unimolecular Decomposition. E a

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7 Lecture 13: Radioactive Decay Kinetics I. Kinetics of First-Order Processes. Mechanism: 1. ucleus has Internal Structure Z X Z Decay involves internal + Y + Q ; Q = + rearrangement of system.

Chemical nalogy V(r) 2 O 4 E a O 2 Rate = d = k [ 2 O 4 ] H rx k = 2 O 4 2 O 2 OTE: Since atoms and molecules are neutral, if T >, they will have (3/2) kt kinetic energy; Collisions may also induce

1 7 Lecture 13: Radioactive Decay Kinetics I. Kinetics of First-Order Processes. Mechanism: 1. ucleus has Internal Structure Z X Z Decay involves internal + Y + Q ; Q = + rearrangement of system. RDOM PROCESS: Identical to Unimolecular Decomposition 2. Chemical nalogy V(r) 2 O 4 E a O 2 Rate = d = k [ 2 O 4 ] H rx k = 2 O 4 2 O 2 OTE: Since atoms and molecules are neutral, if T >, they will have (3/2) kt kinetic energy; Collisions may also induce reactions. Collisions are SECOD ORDER; must distinguish order for chemical reactions For nuclei, T and Coulomb barrier prevents collisions; OLY FIRST ORDER DECY 3. uclear System a. omenclature Change [ ] o o, the number of nuclei initially (ot vogadro's number) k, the rate constant = f(q, Iπ) First-order Rate Law is d Rate = = dt Instantaneous decay rate

2 71 b. Example 137 Cs β m a + ν (t1 = 3y) 55-1 / 2. Mathematics of First Order Decay Rate constant Probability 137m a γ a (t = T) 56 1 / 2 1. Solution: d - d t = - dt dt = n or = e t =, where is different for every nuclide t 2. For a pure sample e t = n n = slope t t 3. t 1/2 : The Half-Life Expresses probability in terms of a characteristic time i.e., high probability, short decay time and vice versa DEFIITIO: The half-life (t 1/2 ) of a nucleus is the time required for one-half the nuclei in a sample to decay. i.e., after t = t 1/2, = /2 / 2 = e = 1 = = e 2 - t -t 1/2 1/2 n 2 = t 1/2 t 1/2 = 1n 2/ =.693/ = t 1/2

3 72 4. Mathematical Shortcuts (but = e t always works). a. If t < < t 1/2 / = e t lim(e x ) x 1 x + / = 1 t = (1 t) = t = t OR = t & t = LWYS TRY TO SEE IF THIS WORKS; RULE: If t <.1 t 1/2, good to 3 sig. figs. b. Problem: How many 238 U nuclei will decay in 1. y from a sample that contains 2.38 mg of uranium? remain? abundance = %, t 1/2 = y Test Rule: 1. << y, OK to use = t g = ( ) g / mole 23 atoms mole = atoms 238 U =.693/ y = y 1 = ( y 1 ) ( atoms) (1.y) = atoms decay (remaining) = = = atoms remain i.e., no change at 3 sig fig level c. If t > > t 1/2, trivial result e t = e.693 t/t 1/2 = e = i.e., = and = ; i.e., all (or most of) sample has decayed RULE: IF t > > 1 t 1/2,

4 73 d. Integral half-lives let n = t/t 1/2 = integer, 1, 2, 3 etc. 1 2 = n (same equation, different form). e. verage Lifetime τ τ = = t / = 1 i i td Changing variables as t τ 1 1 td = t (- dt) = - t (- e- = t dt = - te t dt MTH: ax xe ax e dx = ( ax -1) 2 a RESULT: τ = 1/ = 1.44 t 1/2 i.e., average is longer because long times skew distribution. 5. Rate of Energy Loss Importance: Heating of earth's crust ( 4 K, 232 Th, 235,238 U) Miniature power sources ( 238 Pu) Spurious heating in thermochemistry of radioactive elements a. Definitions: Rate = de dt = d dt de d = Q = Q e- t E = t Q Instantaneous Integral

5 74 b. Problem: Calculate the rate of energy loss for 21 Po in kj/mole-min. α-decay ; t 1/2 = d ; Q α = 5.35 MeV t < < t 1/2 E/ t = Q α Q α = 5.35 MeV ( kj/mev) = kj/atom =.693/(138.4 d)(144 min/d) = /min = 1 mole = atoms E/ t = ( kj/atom)( /min)( atoms) E/ t = 187 kj/min-mole fter d E/ t = 935 kj/min-mole. Comparable to chemical reactions ( H), but at end of day, still have same source. II. ctivity Practical spects of Radioactivity Usually measure emitted particles, not (parent or daughter nuclei).. Definition: ctivity is the radiation that is measured from a radioactive source. = ctivity = c d dt = c ~ c t number of particles emitted emission time interval 1. d/dt = SOLUTE DISITEGRTIO RTE (dps, dpm, dph, etc.) c = detection coefficient = Gε. when G is a geometry factor and ε is the detection efficiency 2. Schematic Source radius. r detector Detector area = πr 2 Spherical area = 4πR 2 R G = πr 2 /4πR 2 = r 2 /4R 2 SIC PRICIPLE OF RDITIO SFETY activity decreases at square of distance R

6 75 ε = Fraction of particles that strike the detector and give signal ccurate determination of c is critical to absolute measurements. First-Order Decay Law in Terms of ctivity = e t c = c e t = e t ; uncertainty is ± where is number of counts = = emitted particles C. Practical pplication 1. Plot is t on semilog paper 2. Determine and t 1/2 graphically 3. If t t 1/2, must correct for decay 4. Determine c with standard source (total) = e t1 e t2 (1) > (2) (1) (2) t 1 t 2 < t > = D. Units of Reactivity 1. Curie: Ci 1 Ci = dps 2. equerel: q 1 q = 1 dps E. Minigenerator Experiment m a γ a 2. Measure as a f(t) 3. ackground Counts in detector that originate in surroundings instead of source ; due to 4 K, cosmic rays, etc. sample = source bkg Error = ± number of counts / time To first approximation: = umber of counts/time

7 76 III. Mixture of Independent ctivities PREVIOUS: Parent OW: Parent 1 Parent 2 Parent 3 t 1/2 (1) t 1/2 (2) t 1/2 (3) Stable Daughter ll Daughters Stable [umber of components = umber of slope changes + 1]. Example I: Two Components. General Case total bkg = (1) + (2) + (3) + = (1) e t + (2) e t + (3) e t + RULE: s t, SHORTEST-LIVED COMPOETS DISPPER c i, i, i can all differ C. Rules for Decomposing a 3-Component Decay Curve (or greater) (or why I'm happy the computer can do this now). PLOT O SEMILOGPPER 1. Start: LYZE LOG-TIME PORTIO OF DECY CURVE FIRST. ecause shorter-lived species will have decayed away, only a single pure (linear) component will remain. 2. Draw a straight line (best fit) through this component this line describes decay of this component al all times. slope = ; t = intercept = 3. SUTRCT long lived component (steps 1 & 2) from the total activity. total bkg 3 = Repeat procedure for second longest component, etc. 5. Schematic (3) (2) (2) n (3) t t

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9 78 IV. Parent-Daughter Relationships (The Rate-Determining Step). Case of Radioactive Daughter C e.g. 21 i 21 Po 26 Pb 5.1 d β d α Stable Same problem as a stepwise chemical reaction. Mathematics of the Problem 1. Parent: Decay: d/dt = ; = e t (nothing new) 2. Daughter: Formation: Decay: ET: d /dt = d /dt = d = dt = e - t - Linear first-order differential equation 3. Solution = e - t e - t + e - t If pure initially, = and second term vanishes 4. Daughter: C = + + C

10 79 5. Special Cases: Long time solutions classification time relationship Rate-determining step a. o Equilibrium t > t > t C b. Transient Equilibrium t > t > t c. Secular Equilibrium t >> t > t (c. is case of U-Th decay series) C. o Equilibrium Daughter is rate-determining step t 1/2 () > t 1/2 () Operative Word 1. Fig. 3-4 a. Curve a: Total activity of initially PURE sample of OTE: resembles two-component independent decay curve. OT!!! (total) = () + () b. Curve b: Parent activity, () d. Curve d: Daughter activity, () ; note growth curve 2. Curve c: Long-term behavior: t > > t 1/2 () t/t 1/2 () = and e t OR = - ( t ) e ote: < ; ( )( ) = + 3. Long term curve: ll of has disappeared ; this defines t 1/2 () D. Transient Equilibrium Parent decay is rate-determining step t 1/2 () > t 1/2 () 1. Fig. 3-2 a. Curve a: Total activity of initially pure sample of (total) = () + () decay curve that initially increases with time is a signature of transient or secular equilibrium.

11 8 b. Curve b: Parent ctivity: () c. Curve d: Daughter ctivity: () d. Initial growth of () and then decay according to the half-life of the parent. 2. Maximum Daughter ctivity a. Maximum occurs when d /dt = differentiate equation for and set this equal to zero; this defines t max as t max = n( / ) ; i.e., is maximum when t = t max b. Importance Medical isotopes Milking a cow how long must one wait before extracting the daughter activity again? 137 1β c. Example: 55 Cs 137m γ 5 a 3 y 2. 6m 137 a (stable) t max = n( / ) = n[t = n [ (3y/2.6min) /2 ()/t 1/2 ()] min/y] y (.693 y/2.6m) 2.6 m 3 y 2.6 m t max = 58 min 3. Long-Time Solution: Curve e a. For initially pure, t > > t 1/2 () e t = e or = ( ) e t, OR = i.e., at long time, ratio / is COSTT with time SYSTEM PPERS TO E I EQUILIRIUM

12 81 4. Consequences a. Long term decay is governed by parent b. ctivity: multiply equation in 3a. above by c c c = - = = c. Half-life of can be determined by combining: long term behavior t 1/2 () activity ratio above

RANDOM PROCESS: Identical to Unimolecular Decomposition. ΔE a

RANDOM PROCESS: Identical to Unimolecular Decomposition. ΔE a 7 Lecure 13: Radioacive Decay Kineics I. Kineics of Firs-Order Processes. Mechanism: 1. ucleus has Inernal Srucure Z X Z Decay involves inernal + Y + Q ; Q =+ rearrangemen of sysem. RDOM PROCESS: Idenical