Lecture13RadioactDecayKinetics-1_001

Lecture13RadioactDecayKinetics-1_001 - 70 Lecture 13...

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70 Lecture 13: Radioactive Decay Kinetics I. Kinetics of First-Order Processes A. Mechanism : 1. Nucleus has Internal Structure Z A Z A Y QQ X + B ; B  B + RANDOM PROCESS: Identical to Unimolecular Decomposition 2. Chemical Analogy V(r) rx N 2 O 4 2 NO 2 NOTE: Since atoms and molecules are neutral , if T > 0 , they will have (3/2) kT kinetic energy; Collisions may also induce reactions. Collisions are SECOND ORDER ; must distinguish order for chemical reactions F o r nuclei , T 0 and Coulomb barrier prevents collisions ; ONLY FIRST ORDER DECAY 3. Nuclear System a. Nomenclature Change [ ] o N o , the number of nuclei initially (Not Avogadro's number) k , the rate constant = f(Q, I ) First-order Rate Law is Rate =  dN dt N Decay involves internal E a H N 2 O 4 NO 2 Rate = d = k [N 2 O 4 ] k = Instantaneous decay rate rearrangement of system.
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71 b. Example 55 137 Cs -1 0- + 137m Ba + (t 1/2   30 y ) 56 137m Ba + 137 Ba (t TBA) B. Mathematics of First Order Decay Rate constant Probability 1. Solution: dN dT -N dN N - dt N N 0 t 0    n N N 0 = t o r N = N 0 e  t , where is different for every nuclide 2. For a pure sample N nN 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 DEFINITION: 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 , N = N 0 /2 N N 0 e N 0 e - t - t 1/2 1/2  /2 0 1 2 N n 2 = t 1/2 t 1/2 = 1n 2/ = 0.693/ = t 1/2 N 0 e  t = nN 0  = slope
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72 4. Mathematical Shortcuts (but N = N 0 e  t always works). a. If t < < t 1/2 N/N 0 = e  t lim(e x ) x 0 1 x + … N / N 0 = 1  t N = N 0 (1  t) = N 0 N 0 t N 0 N = N 0 t OR N = N 0 t & N t = N 0 ALWAYS TRY TO SEE IF THIS WORKS; RULE : If t < 0.1 t 1/2 , good to 3 sig. figs. b . Problem: How many 238 U nuclei will decay in 1.0 y from a sample that contains 2.38 mg of uranium? remain? abundance = 99.275%, t 1/2 = 4.468 10 9 y Test Rule: 1.0 << 4.468 10 9 y, OK to use N = N 0 t N 0 =  238 10 238 099275 602 10 3 23 . .. g g/mole atoms mole = 5.98 10 18 atoms 238 U = 0.693/4.468 10 9 y = 1.55 10 y 10 1 N = (1.55 10 -10 y 1 ) (5.98 10 18 atoms) (1.0y) = 9.27 10 9 atoms decay N (remaining) = N 0 N = 5.98 10 18 9.27 10 9 = 5.98
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Lecture13RadioactDecayKinetics-1_001 - 70 Lecture 13...

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