Chem 162-2010 homework 14th & 15th week

Chem 162-2010 homework 14th & 15th week - Chem...

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Chem 162-2010 14 th th Week Homework Radioactive Decay Processes 21. Each of the following equations represents a radioactive decay process. Supply the missing items. (a) 219 ? Rn → ? + 4 2 He (b) 167 69 ? + 0 -1 e → ? (c) 90 38 Sr → 0 -1 e + ? (d) 79 36 Kr → 0 1 e + ? (a) 219 86 Rn → 215 84 Po + 4 2 He (b) 167 69 Tm + 0 -1 e → 167 68 Er (c) 90 38 Sr → 0 -1 e + 90 39 Y (d) 79 36 Kr → 0 1 e + 79 35 Br 23. What is the final nuclide obtained in each process? (a) Indium-123 decays by electron capture. (b) A succession of three β - emission begins with molybdenum-103. (c) A succession of two α emissions, followed by two β - emissions, begins with astatine-217. (a) Indium-123 decays by electron capture. 123 49 In + 0 -1 e → 123 48 Cd (b) A succession of three β - emission begins with molybdenum-103. 102 42 Mo → 0 -1 e → 0 -1 e → 0 -1 e + 102 45 Rh (c) A succession of two α emissions, followed by two β - emissions, begins with astatine-217. 217 85 At → 4 2 α → 4 2 α → 0 -1 e → 0 -1 e + 209 83 Bi 25. Determine the nuclide that (a) undergoes electron capture to produce 7 Li, (b) undergoes α decay to form 238 U. (a) 7 4 Be + 0 -1 e → 7 3 Li (b) 242 94 Pu → 4 2 α + 238 92 U Rate of Radioactive Decay 29. Chemical reactions can be of different orders—zero, fist, second, and so forth—whereas decay is always a first order process. Why is this so? Decay is a very simple elementary step reaction which involves one molecule decomposing. This is first order. It doesn’t involve rapid-reversible steps, or several molecules colliding. 35. What is the decay rate in atoms/h of (a) a 1-million-atom sample of Pt-200 (t 1/2 = 11.5 h), (b) a 1- million-atom sample of C-11? (a) t-1/2 = 0.693/k 11.5h = 0.693/k k = 0.06026h -1 Rate = kN Rate = 0.06026h -1 x (1 x 10 6 atom/sample) = 6.03 x 10 4 atom/sampleh 1
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(b) t1/2 = 20.39min x (1 hr/60 min) = 0.3398 h 0.3398h = 0.693/k k = 2.039h -1 Rate = kN Rate = 2.039h -1 x (1 x 10 6 atom/sample) = 2.039 x 10 6 atom/sampleh 37. We follow the activity of a sample containing the radioactive isotope sulfur-35. At the start of the experiment, we detect 138 dis per min, and after 20.0 d, the decay rate is 118 dis min -1 . What is the half- life of sulfur-35? 20.0 d x (24h/d) x (60m/h) = 2.88 x 10 4 m ln(N t ) = -kt + ln(N o ) ln(118) = -k x (2.88 x 10 4 ) + ln(138) (((ln(118)) = (((-k) x (2.88 x 10 4 )) + (ln(138)))),X) k = 5.44 x 10 -6 t 1/2 = 0.693/k t 1/2 = 0.693/(5.44 x 10 -6 ) = 1.27 x 10 5 min 1.27 x 10 5 m x (1 h/60 m) x (1 day/24 h) = 88.5 days 39. The half-life of iron-52 is 8.28 hours. Suppose we measure the activity of a sample containing iron- 52 and find it to be 884 dis per min. If we measure its activity again 24.0 hours later, (a) estimate its activity at the later time, and (b) calculate a more exact value of the activity that should be found. 884 dis/min x 60 min/hr = 53040 dis/hr
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Chem 162-2010 homework 14th & 15th week - Chem...

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