1. 24-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 24 Nuclear Reactions and Their Applications

2. 24-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Nuclear Reactions and Their Applications 24.1 Radioactive Decay and Nuclear Stability 24.2 The Kinetics of Radioactive Decay 24.3 Nuclear Transmutation: Induced Changes in Nuclei 24.4 The Effects of Nuclear Radiation on Matter 24.5 Applications of Radioisotopes 24.6 The Interconversion of Mass and Energy 24.7 Applications of Fission and Fusion

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4. 24-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Nuclear Chemistry Radioactivity: An unstable nucleus spontaneously disintegrates or decays by emitting radiation. Nucleus contains protons and neutrons. They are composed of quarks. Nucleus contains nucleons. Atomic number: Z: Number of protons in a nucleus. Atomic mass number: A: Number of protons + Number of neutrons. Isotopes: Atoms having same atomic number and different atomic mass numbers. Nuclide: A unique atom represented by:

5. 24-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Types of Radioactive Decay 1. Alpha particle: Helium nucleus. Dense, positively charged particles. A decreases by 4, Z decreases by 2. Every element heavier than Pb and some lighter ones also.

6. 24-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Types of Radioactive Decay 2. Beta particle production: Mass number remains constant. An electron is released. Ex: Unstable nuclide creates an electron as it releases energy. A remains unchanged, Z increases by 1. Next higher atomic number is formed.

7. 24-7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Types of Radioactive Decay 3.Gamma ray: High energy photon. Nucleus from excited state goes to ground state. They have no mass or charge, emission does not change A or Z.

8. 24-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Types of Radioactive Decay 4. Positron Production: Has same mass as e, but opposite charge. Antiparticle of an electron. Proton converted to a neutron. A remains unchanged, Z decreases by 1. Lower atomic number is formed.

9. 24-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Types of Radioactive Decay 5. Electron Capture: One of the inner orbital electrons is captured by nucleus. Nuclear proton is transferred into a neutron. A is unchanged, Z decreases by 1. Gamma rays are always produced in e capture.

10. 24-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.1 The behavior of three types of radioactive emissions in an electric field.

11. 24-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Types of Radioactive Decay: Balancing Nuclear Equations Total A Total Z Reactants = Total A Total Z Products Alpha decay - A decreases by 4 and Z decreases by 2. Every element heavier than Pb undergoes  decay. Beta decay - ejection of a  particle from the nucleus from the conversion of a neutron into a proton and the expulsion of 0 -1 . The product nuclide will have the same Z but will be one atomic number higher. Positron decay - a positron ( 0 1  ) 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.

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13. 24-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problems: Write balanced equations for the following: C produces a positron. Bi produces a beta particle. Np produces an alpha particle. 4) Supply the missing particles:

14. 24-14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 24.1Writing Equations for Nuclear Reactions PLAN: SOLUTION: PROBLEM:Write balanced equations for the following nuclear reactions: (a) Naturally occurring thorium-232 undergoes  decay. (b) Chlorine-36 undergoes electron capture. Write a skeleton equation; balance the number of neutrons and charges; solve for the unknown nuclide. A = 228 and Z = 88 232 90 Th 228 88 Ra + 4 2 He (a) 232 90 Th 228 88 Ra + 4 2 He (b) 36 17 Cl + 0 -1 e A Z X A = 36 and Z = 16 36 17 Cl + 0 -1 e 36 16 S

15. 24-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Nuclear stability and the Mode of Decay : N/Z= 1, for lighter nuclei to be stable. In heavier nuclei N/Z ratio > 1. Very few stable nuclides exist with N/Z < 1 (H, He) Lighter nuclides are stable. N/Z ratio increases as Z increases. Z > 83 all nuclides are stable. Elements with even Z have larger number of stable nuclides.

16. 24-16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.2 A plot of neutrons vs. protons for the stable nuclides.

17. 24-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Nuclear Stability and Mode of Decay Very few stable nuclides exist with N/Z < 1. The N/Z ratio of stable nuclides gradually increases a Z increases. All nuclides with Z > 83 are unstable. Elements with an even Z usually have a larger number of stable nuclides than elements with an odd Z. Well over half the stable nuclides have both even N and even Z. Predicting the Mode of Decay Neutron-rich nuclides undergo  decay. Neutron-poor nuclides undergo positron decay or electron capture. Heavy nuclides undergo  decay.

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19. 24-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 24.2Predicting Nuclear Stability PLAN: SOLUTION: PROBLEM:Which of the following nuclides would you predcit to be stable and which radioactive? Explain. (a) 18 10 Ne(b) 32 16 S(c) 236 90 Th(d) 123 56 Ba Stability will depend upon the N/Z ratio, the value of Z, the value of stable N/Z nuclei, and whether N and Z are even or odd. (a) Radioactive. N/Z = 0.8; there are too few neutrons to be stable. (b) Stable. N/Z = 1.0; Z < 20 and N and Z are even. (c) Radioactive. Every nuclide with Z > 83 is radioactive. (d) Radioactive. N/Z = 1.20; the diagram on shows stability when N/Z ≥ 1.3.

20. 24-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 24.3Predicting the Mode of Nuclear Decay PLAN: SOLUTION: PROBLEM:Predict the nature of the nuclear change(s) each of the following radioactive nuclides is likely to undergo: (a) 12 5 B(b) 234 92 U(c) 74 33 As(d) 127 57 La 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. (a) N/Z = 1.4 which is high. The nuclide will probably undergo  decay altering Z to 6 and lowering the ratio. (b) The large number of neutrons makes this a good candidate for  decay. (c) N/Z = 1.24 which is in the band of stability. It will probably undergo  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.

21. 24-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Kinetics of Radioactive Decay: Rate of decay: Negative of the change in the number of nuclides per unit time. rate = kN, K is the decay constant. The rate of decay is proportional to the number of nuclides. This represents a first-order process. Decay rate (A) = -∆N/∆t SI unit of decay is the becquerel (Bq) = 1d/s. curie (Ci) = number of nuclei disintegrating each second in 1g of radium-226 = 3.70x1010d/s Large k means a short half-life and vice versa.

22. 24-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.3 The 238 U decay series.

23. 24-23 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Decay rate (A) =  N/  t SI unit of decay is the becquerel (Bq) = 1d/s. curie (Ci) = number of nuclei disintegrating each second in 1g of radium-226 = 3.70x10 10 d/s Nuclear decay is a first-order rate process. Large k means a short half-life and vice versa.

24. 24-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.4 Decrease in the number of 14 C nuclei over time.

25. 24-25 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problems Half-Life:...the time required for the number of nuclides to reach half the original value (N0/2). Technetium-99m(m=excited nuclear state to GS emits gamma particle) has a rate constant 1.16 x 10-1 /h. What is the half life of the nuclide Tc? Halflife of Mo 99 is 67.0 h. How much of a 1.000mg sample of 99 Mo is left after 335 h?

26. 24-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 24.4Finding the Number of Radioactive Nuclei PLAN: SOLUTION: PROBLEM:Strontium-90 is a radioactive by-product of nuclear reactors that behaves biologically like calcium, the element above it in Group 2A(2). When 90 Sr is ingested by mammals, it is found in their milk and eventually in the bones of those drinking the milk. If a sample of 90 Sr has an activity of 1.2x10 12 d/s, what are the activity and the fraction of nuclei that have decayed after 59 yr (t 1/2 of 90 Sr = 29 yr) The fraction of nuclei that have decayed is the change in the number of nuclei, expressed as a fraction of the starting number. The activity of the sample (A) is proportional to the number of nuclei (N). We are given the A 0 and can find A from the integrated form of the first-order rate equation. t 1/2 = ln2/k so k = 0.693/29 yr= 0.024 yr -1 ln N 0 /N t = ln A 0 /A t = ktln A t = -kt + ln A 0 ln A t = -(0.024yr -1 )(59yr) + ln(1.2x10 12 d/s) ln A t = 26.4A t = 2.9x10 11 d/s Fraction decayed = (1.2x10 12 -2.9x10 11 ) (1.2x10 12 ) Fraction decayed = 0.76

27. 24-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Detection and Uses of Radioactivity: GM counter: Probe filled with Ar gas which is ionized by moving particles. Scintillation counter: ZnS give off light when they are struck by high-energy radiation.

28. 24-28 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Dating by Radioactivity: Radiocarbon dating/C-14 dating: Based on radioactivity of C which decays by beta particle production. Reactions: C-14 can be used to date wood and cloth aircrafts. Half life of C is 5730 years. As a plant is cut C/ C ratio decreases. C has to be burnt to give CO2. A small amount can be used in mass spectrometer, C atoms are ionized and accelerated through a magnetic field that deflects their path. Earth’s crust formed 4.3 bya.

29. 24-29 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problem 6) The remains showed a C decay rate of 3.1 counts per min per gram of C. Assuming that decay rate of C in freshly cut wood is 13.6 counts per min per gram of C, calculate the age of remnants. The half life of C is 5730 years.

30. 24-30 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.5 Radiocarbon dating for determining the age of artifacts.

31. 24-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 24.5Applying Radiocarbon Dating SOLUTION: PROBLEM:The charred bones of a sloth in a cave in Chile represent the earliest evidence of human presence in the southern tip of South America. A sample of the bone has a specific activity of 5.22 disintegrations per minute per gram of carbon (d/min*g). If the ratio of 12 C: 14 C in living organisms results in a specific activity of 15.3 d/min*g, how old are the bones? (t 1/2 of 14 C = 5730 yr) PLAN:Calculate the rate constant using the given half-life. Then use the first-order rate equation to find the age of the bones. k = ln 2/t 1/2 = 0.693/5730yr= 1.21x10 -4 yr -1 t = 1/k ln A 0 /A t =1/(1.21x10 -4 yr -1 ) ln (15.3/5.22)= 8.89x10 3 yr The bones are about 8900 years old.

32. 24-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Particle accelerators and the Transuranium elements: Neutrons are used as projectiles, no charge, not repelled. Particle accelerators were used to give particles a higher K.E. Linear accelerator: Series of tubes of increasing lengths that through a source of voltage change their polarities. Cyclotron: Uses electromagnet to give the particle a spiral path.

33. 24-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.6 A linear accelerator. The linear accelerator operated by Standford University, California

34. 24-34 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.7 The cyclotron accelerator.

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36. 24-36 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Effects of Nuclear Radiation of Matter: 1. Excitation: Radiation of relatively low energy interacts with an atom of a substance, which absorbs some energy and reemits it. Radiation that causes excitation is called nonionizing radiation. 2. Ionization: Radiation collides with an atom to remove an electron. Called ionizing radiation. Cation-electron pairs result.

37. 24-37 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Effects of Ionizing Radiation: 1. Gray: (Gy) = 1 Joule of energy absorbed per kg of body tissue. Rad(radiation-absorbed dose) =0.01 Gy Rem (roentgen equvivalent for man) no of rems= no of radiations x RBE. SI unit for dosage equvivalent is Sv- sievert. 1 rem= 0.01 sv.

38. 24-38 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Effects of Ionizing Radiation: 2. Penetrating power of Emissions: penetrating power is inversely related to mass and charge of the emission. Gamma rays are highly penetrating, β particles can penetrate approx. 1cm. α particles are stopped by the skin.

39. 24-39 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.8 Penetrating power of radioactive emissions Penetrating power is inversely related to the mass and charge of the emission. Nuclear changes cause chemical changes in surrounding matter by excitation and ionization.

40. 24-40 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Effects of Ionizing Radiation: 3. Molecular interactions: Ionizing radiation causes free electrons, which produces free radicals in water molecules. Lipids are attacked by radicals. This can lead to cell damage as well as cancer. Genetic mutations can occur when bonds in the DNA of sperm and egg cells are altered by free radicals. 4. People living near test centers, nuclear energy facilities, and waste disposal areas are exposed to more radiation.

41. 24-41 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Applications of Radioisotopes: Tracer- emits nonionizing radiation that signals the presence of a substance. NAA(Neutron activation analysis) –to detect tracers of ammunition on a suspects hand traces of hair in a victim of poisoning. Meaure friction in the engine.

42. 24-42 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Applications of Radioisotopes: Used in detection of thyroid- patient drinks radiotracers of Na_I. Technetium-99 is also used for imaging heart, lungs and liver. Fe-59 used to detect hemoglobin in blood cells. PET is used for imaging brain and its functions.

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46. 24-46 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.11 PET and brain activity. normal Alzheimer’s Figure 24.10 The use of radioisotopes to image the thyroid gland. asymmetric scan indicates disease normal

47. 24-47 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.12 The increased shelf life of irradiated food.

48. 24-48 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Thermodynamics Stability of the Nucleus: When a system gains or loses energy it also gains or loses a quantity of mass.  E =  mc2  E/c2=  m  m = mass defect  E = change in energy

49. 24-49 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Thermodynamics Stability of the Nucleus: If  E =  (exothermic), mass is lost from the system. Binding Energy: is the energy required to decompose the nucleus into its components. Iron-56 is the most stable nucleus. It has a binding energy per nucleon of 8.79 MeV

50. 24-50 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Interconversion of Mass and Energy E = mc 2  E =  mc 2  m =  E / c 2 The mass of the nucleus is less than the combined masses of its nucleons. The mass decrease that occurs when nucleons are united into a nucleus is called the mass defect. The mass defect (  m) can be used to calculate the nuclear binding energy in MeV. 1 amu = 931.5x10 6 eV = 931.5 MeV

51. 24-51 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Nuclear Fission and Nuclear Fusion: Fusion: Combining two light nuclei to form a heavier, more stable nucleus. Fission: Splitting a heavy nucleus into two nuclei with smaller mass numbers. A self-sustaining fission process is called a chain reaction.

52. 24-52 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.9 RCOH O HOR'+ RCO O R' + HOH H + Which reactant contributes which group to the ester?

53. 24-53 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 24.6Calculating the Binding Energy per Nucleon PLAN: SOLUTION: PROBLEM:Iron-56 is an extremely stable nuclide. Compute the binding energy per nucleon for 56 Fe and compare it with that for 12 C (mass of 56 Fe atom = 55.934939 amu; mass of 1 H atom = 1.007825 amu; mass of neutron = 1.008665 amu). Find the mass defect,  m; multiply that by the MeV equivalent and divide by the number of nucleons. Mass Defect = [(26 x 1.007825 amu) + (30 x 1.008665 amu)] - 55.934939  m = 0.52846 amu Binding energy == 8.790 Mev/nucleon (0.52846 amu)(931.5 MeV/amu) 56 nucleons 12 C has a binding energy of 7.680 MeV/nucleon, so 56 Fe is more stable.

54. 24-54 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.13 The variation in binding energy per nucleon.

55. 24-55 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.14 Induced fission of 235 U.

56. 24-56 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.15 A chain reaction of 235 U.

57. 24-57 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.16 Diagram of an atomic bomb.

58. 24-58 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.17 A light-water reactor.

59. 24-59 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 24.18 The tokamak design for magnetic containment of a fusion plasma.

60. 24-60 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B24.1 Detection of radioactivity by an ionization counter.

61. 24-61 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B24.2 Vials of a scintillation “cocktail” emitting light

62. 24-62 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B24.3 Element synthesis in the life cycle of a star

63. 24-63 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B24.4 A view of Supernova 1987A.