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Outline: 4/16/07

Outline: 4/16/07. Pick up leftover CAPA outside… Special seminar Wednesday pm Turn in Seminar reports – to me. Today: Chapter 22 (cont’d). Nuclear Chemistry - Modes of decay - Half-life calculations. Quiz #9. Please put away books & papers. Turn your paper over when done.

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Outline: 4/16/07

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  1. Outline: 4/16/07 • Pick up leftover CAPA outside… • Special seminar Wednesday pm • Turn in Seminar reports – to me Today: Chapter 22 (cont’d) Nuclear Chemistry - Modes of decay - Half-life calculations

  2. Quiz #9 • Please put away books & papers. • Turn your paper over when done.

  3. Quiz #9 • Please pass your quiz to the right…

  4. Nuclear Reactions: • Types of radioactive decay: • Alpha (a), Beta (b), and Gamma (g) • fission, fusion, • positron (b+), e- capture (EC)…. • Natural vs. induced reactions: • Worksheet #13 • Balancing equations

  5. Worksheet #13 • Start filling in the blanks: nuclear equation writing Remember: both top and bottom must balance on both sides of the equation (conservation of mass and charge!)

  6. 105 • 39 • Y • 14 • 7 • 60 • 28 • N • Ni Worksheet #13 • (a) 1 + 235 = 131 + ____ • 0 + 92 = 53 + ____ • 105 • 39 • (b) 0 + ____ = 14 • -1 + ____ = 6 • 14 • 7 • (c) 60 = 0 + ____ • 27 = -1 + ____ • 60 • 28

  7. Radioactive Decay: 1st order • First order decay : N = N0 e - k t • Half-life: • ln(N/N0) = - k t • ln(1/2) = - k t1/2 • 0.693 = k t1/2 • t1/2 : a convenient way to refer to k e.g.14C has k = 1.209 × 10-4 yr-1 ; what is its half-life?

  8. N/N0 = e - k t or ln(N/N0) = - k t • ln(34.6/103.7) = - k t • t = 9080 yrs • 0.693 = k t1/2 • k = 1.209 × 10-4 yr-1 • t1/2 = 0.683/k = 5730 yr • e.g. : If 10.0 g of Kennewick Man bone has 34.6 dpm of 14C, and a present-day 10.0 g bone has 103.7 dpm, how old is Kennewick Man?

  9. Decay calculations: • Where have you seen these before? • Chapter 15 (pp. 627-630) = Kinetics • PRACTICE!

  10. 0.143 mol 14C  6.022e23 atoms/mol • = 8.60e21 atoms 14C Worksheet #13 - Question 2 • 0.02% of 1000g = • 0.2 g 14C • 0.2 g  1mol/14g = 0.143 mol 14C • N/N0 = 7.65e20 / 8.60e21 = 0.0889 • ln (0.0889) = - k t • - 2.42 = - 1.209 × 10-4 yr-1 t • t = 20,000 yr

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