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Ch 23.2 & 23.2: Nuclear Chemistry

Ch 23.2 & 23.2: Nuclear Chemistry. Radioactivity. One of the pieces of evidence for the fact that atoms are made of smaller particles came from the work of Marie Curie (1876-1934).

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Ch 23.2 & 23.2: Nuclear Chemistry

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  1. Ch 23.2 & 23.2: Nuclear Chemistry

  2. Radioactivity • One of the pieces of evidence for the fact that atoms are made of smaller particles came from the work of Marie Curie (1876-1934). • She discovered radioactivity, the spontaneous disintegration of some elements into smaller pieces (spontaneous emission of energy or particles from atomic nuclei).

  3. 3 Types of Radiations mass (amu) Alpha 4.00 amu Beta 0.0005 amu Gamma 0.0000

  4. Penetrating Ability

  5. Types of Radioactive Decay: Balancing Nuclear Equations Reactants = Products Alpha decay - A decreases by 4 and Z decreases by 2. Every element heavier than Pb undergoes a decay. Beta decay - ejection of a b particle from the nucleus from the conversion of a neutron into a proton and the expulsion of 0-1b. The product nuclide will have the same Z but will be one atomic number higher. Positron decay - a positron (01b) is the antiparticle of an electron. A proton in the nucleus is converted into a neutron with the expulsion of the positron. Z remains the same but the atomic number decreases. Electron capture - a nuclear proton is converted into a neutron by the capture of an electron. Z remains the same but the atomic number decreases. Gamma emission - energy release; no change in Z or A.

  6. 1. Alpha emission Note that mass number (A) goes down by 4 and atomic number (Z) goes down by 2. Nucleons (e.g. protons and neutrons) are rearranged but conserved

  7. 2. Beta emission Note that mass number (A) is unchanged and atomic number (Z) goes up by 1. How does this happen?

  8. 23892U ---> 23490Th + 42He + 2 00 6027Co ---> 0-1e + 6028Ni + 00 Virtually all - and -emissions are accompanied with  -ray emission. 3. Gamma ray emission

  9. 4. Positron emission • 116C ---> 115B + 01 How does this happen? 11p ---> 10n + 01 expelled Other types of Nuclear Decay 5. Electron Capture (K-capture) 5526Fe + 0-1 ---> 5525Mn + h (x-ray) 74Be + 0-1 ---> 73Li How does this happen? 11p + 0-1 ---> 10n

  10. Detection of radioactivity by an ionization counter. Figure B24.1

  11. The 238U decay series. Figure 24.3

  12. Practice balancing nuclear reactions Excercises: 22286Rn ---> 21884Po + ? Decay type?___ ? ---> 23190Th + 42He Decay type?___ 2211Na --->? + 0-1 Decay type?___

  13. Practice balancing nuclear reactions Excercises: 4120Ca + 0-1e ---> ? Decay type?___ 137N ---> 136C + ? Decay type?___ 3216S + 10n ---> 11H + ? Decay type?___

  14. A plot of number of neutrons vs. number of protons for the stable nuclides. Figure 24.2

  15. a emission reduces Z 24395Am --> 42a + 23993Np b emission increases Z 6027Co --> 0-1b + 6028Ni Band of Stability and Radioactive Decay Isotopes with low n/p ratio, below band of stability decay, by positron emission or electron capture

  16. PROBLEM: Predict the nature of the nuclear change(s) each of the following radioactive nuclides is likely to undergo: (a)N/Z = 1.4 which is high. The nuclide will probably undergo b decay altering Z to 6 and lowering the ratio. Sample Problem 24.3 Predicting the Mode of Nuclear Decay (a)125B (b)23492U (c)7433As (d)12757La PLAN: Find the N/Z ratio and compare it to the band stability. Then predict which of the modes of decay will give a ratio closer to the band. SOLUTION: (b) The large number of neutrons makes this a good candidate for a decay. (c)N/Z = 1.24 which is in the band of stability. It will probably undergo b decay or positron emission. (d)N/Z = 1.23 which is too low for this area of the band. It can increase Z by positron emission or electron capture.

  17. Binding Energy, Eb Eb is the energy required to separate the nucleus of an atom into protons and neutrons. For deuterium, 21H 21H ---> 11p + 10n Eb = 2.15 x 108 kJ/mol Eb per nucleon = Eb/2 nucleons = 1.08 x 108 kJ/mol nucleons

  18. Calculate Binding Energy For deuterium, 21H: 21H ---> 11p + 10n Mass of 21H = 2.01410 g/mol Mass of proton = 1.007825 g/mol Mass of neutron = 1.008665 g/mol ∆m = (1.007825 + 1.008665) – (2.01410) = 2.01649 - 2.01410 ∆m = 0.00239 g/mol = 2.39 x 10-6 kg/mol From Einstein’s equation: (1 J = 1 kg . m2/s2) Eb = (∆m)c2 = (2.39 x 10-6 kg/mol) x (3 x 108 m/s)2 = = 2.15 x 1011 J/mol = 2.15 x 108 kJ/mol Eb per nucleon = Eb/2 nucleons = 1.08 x 108 kJ/mol nucleons

  19. Calculate Binding Energy For carbon-12: 126C ---> 611p + 610n Mass of 126C = 12.000000 amu or 12.000000 g/mol Mass of proton = 6 x 1.007825 amu = 6.046950 amu Mass of neutron = 6 x 1.008665 amu = 6.051990 amu Total mass = 12.098940 amu ∆m = 12.098940 – 12.000000 = 0.098940 amu/C = 0.098940 g/mol C ∆m = 0.098940 g/mol = 9.8940 x 10-5 kg/mol From Einstein’s equation: (1 J = 1 kg . m2/s2) Eb = (∆m)c2 = (9.8940 x 10-5 kg/mol) x (2.9979 x 108 m/s)2 = = 8.8921 x 1012 J/mol = 8.8921 x 109 kJ/mol Eb per nucleon = Eb/12 mol nucleons = 7.410083 x 108 kJ/mol nucleons

  20. Calculate Binding Energy Eb per mol nucleons = Eb/12 mol nucleons = 7.410083 x 108 kJ/mol nucleons Eb per nucleons = (7.410083 x 108 kJ/mol) x (1 mol/6.022 x 1023 nucleons) = 1.230502 x 10-15 kJ/nucleon = 1.230502 x 10-15 kJ/nucleon = (1.230502 x 10-15 kJ/nucleon) x (1 MeV/1.602 x 10-16 kJ) = 7.681037 MeV/nucleon * (You can use this Eb per nucleon to compare the stability of one nucleus to another type of nucleus.)

  21. The variation in binding energy per nucleon. Figure 24.12

  22. PROBLEM: Iron-56 is an extremely stable nuclide. Compute the binding energy per nucleon for 56Fe and compare it with that for 12C (mass of 56Fe atom = 55.934939 amu; mass of 1H atom = 1.007825 amu; mass of neutron = 1.008665 amu). (0.52846 amu)(931.5 MeV/amu) 56 nucleons Sample Problem 24.6 Calculating the Binding Energy per Nucleon PLAN: Find the mass difference, Dm; multiply that by the MeV equivalent and divide by the number of nucleons. SOLUTION: Mass Difference = [(26 x 1.007825 amu) + (30 x 1.008665 amu)] - 55.934939 Dm = 0.52846 amu Binding energy = = 8.790 MeV/nucleon 12C has a binding energy of 7.680 MeV/nucleon, so 56Fe is more stable.

  23. Half-Life • HALF-LIFE is the time it takes for 1/2 of a sample to disappear. • The rate of a nuclear transformation depends only on the “reactant” concentration. • Concept of HALF-LIFE is especially useful for 1st order reactions. Decay of 20.0 mg of 15O. What remains after 3 half-lives? After 5 half-lives?

  24. Kinetics of Radioactive Decay Activity (A) = Disintegrations/time = kN, where N is the number of atoms Decay is first-order, and so ln (N/No) = -kt The half-life of radioactive decay is t1/2 = 0.693/k SI unit of decay is the becquerel (Bq) = 1 d/s. curie (Ci) = number of nuclei disintegrating each second in 1 g of radium-226 = 3.70x1010 d/s

  25. Find t1/2 Example 1: A sample of radon-222 has an initial a-particle activity of 7.0 x 104 dps. After 6.6 days, its activity drops to 2.1 x 104 dps. Calc the t½ of radon-222.

  26. Find the time it takes to decay Example 2: Gallium-67 has a t½ of 78.2 h. How long will it take for a sample of gallium-67 to decay to 10.0% of its original activity?

  27. Radiocarbon Dating Radioactive C-14 is formed in the upper atmosphere by nuclear reactions initiated by neutrons in cosmic radiation 14N + 1on ---> 14C + 1H The C-14 is oxidized to 14CO2, which circulates through the biosphere. When a plant dies, the C-14 is not replenished. But the C-14 continues to decay with t1/2 = 5730 years. Activity of a sample can be used to date the sample.

  28. Radiocarbon Dating

  29. Find how old an object is Example 3: The activity of carbon-14 of an old wooden object is 7.40 dpm per gram of carbon. Calc the approximate age of the object. Assume the initial activity of cargon-14 (N0) is 12.6 dpm per gram of carbon.

  30. Artificial Nuclear Reactions New elements or new isotopes of known elements are produced by bombarding an atom with a subatomic particle such as a proton or neutron -- or even a much heavier particle such as 4He and 11B. Reactions using neutrons are called n,g reactions because a g ray is usually emitted. Radioisotopes used in medicine are often made by n,g reactions.

  31. Artificial Nuclear Reactions Example of a n,g reaction is production of radioactive 32P for use in studies of P uptake in the body. 3115P + 10n ---> 3215P + g

  32. Transuranium Elements Elements beyond 92 (transuranium) made starting with an n,g reaction 23892U + 10n ---> 23992U + g 23992U ---> 23993Np + 0-1b 23993Np ---> 23994Pu + 0-1b

  33. Transuranium Elements & Glenn Seaborg 106Sg

  34. Nuclear Fission & Lise Meitner 109Mt

  35. Ch 23.6 & 7:Nuclear Fission

  36. Nuclear Fission Fission chain has three general steps: 1. Initiation. Reaction of a single atom starts the chain (e.g., 235U + neutron) 2. Propagation. 236U fission releases neutrons that initiate other fissions 3. Termination.

  37. Ch 23.7: Nuclear Fusion Nuclear Fusion – two or more nuclei react to form a larger nucleus. 31H + 21H ---> 42He + 10n E = -1.7 x 109 kJ/mol (Occurs at extremely high temperature 108K)

  38. Fig. 7.5

  39. Figure 24.16 A light-water nuclear reactor.

  40. Fig. 7.2

  41. Fig. 7.1

  42. Fig. 7.7

  43. Fig. 7.20

  44. Fig. 7.22

  45. Fig. 7.23

  46. Subcritical mass Critical mass Supercritical mass Nuclear reaction in atomic bomb (fission bomb): 23592U + 10n ---> 23592U ---> 14156Ba + 9236Kr + 3 10n

  47. Nuclear Fusion Reactions in the Sun: 11H + 11H  21H + 01e 11H + 21H  32He More examples of Fusion Reactions: 21H + 21H ---> 32He + 10n 4 11H ---> 42He + 2 01e 32He + 32He  42He + 2 11He 32He + 11H  42He + 01e The tokamak design for magnetic containment of a fusion plasma. Figure 24.17

  48. 1. 23592U + 10n ---> 23592U ---> 14156Ba + 9236Kr + 3 10n Nuclear reaction in fusion bomb: 2. 10n + 63Li ---> 31H + 42He + Energy 3. 21H + 31H ---> 42He + 10n + Energy

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