Chapter 24 Nuclear Reactions and Their Applications 24-1
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 24-2
24-3 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. 24-4
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 α decay. Beta decay - ejection of a β particle from the nucleus from the conversion of a neutron into a proton and the expulsion of 0-1β. The product nuclide will have the same A but will be one atomic number higher. Positron decay - a positron ( 0 1 β) is the antiparticle of an electron. A proton in the nucleus is converted into a neutron with the expulsion of the positron. A remains the same but the atomic number decreases. Electron capture - a nuclear proton is converted into a neutron by the capture of an electron. A remains the same but the atomic number decreases. Gamma emission - energy release; no change in Z or A. 24-5
24-6 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 24.1 Writing Equations for Nuclear Reactions PROBLEM: PLAN: Write balanced equations for the following nuclear reactions: (a) Naturally occurring thorium-232 undergoes α decay. (b) Chlorine-36 undergoes electron capture. Write a skeleton equation; balance the number of neutrons and charges; solve for the unknown nuclide. SOLUTION: (a) 232 90 Th 228 88 Ra + 4 2 He A = 228 and Z = 88 232 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 24-7
Figure 24.2 A plot of neutrons vs. protons for the stable nuclides. 24-8
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 β decay. Neutron-poor nuclides undergo positron decay or electron capture. Heavy nuclides undergo α decay. 24-9
24-10 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 24.2 Predicting Nuclear Stability 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 PLAN: 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. SOLUTION: (a) Radioactive. N/Z = 0.8; there are too few neutrons to be stable. (c) Radioactive. Every nuclide with Z > 83 is radioactive. (b) Stable. N/Z = 1.0; Z < 20 and N and Z are even. (d) Radioactive. N/Z = 1.20; the diagram on shows stability when N/Z 1.3. 24-11
Sample Problem 24.3 Predicting the Mode of Nuclear Decay 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 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: (a) N/Z = 1.4 which is high. The nuclide will probably undergo β decay altering Z to 6 and lowering the ratio. (c) N/Z = 1.24 which is in the band of stability. It will probably undergo β decay. (b) The large number of neutrons makes this a good candidate for α decay. (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. 24-12
Figure 24.3 The 238 U decay series. 24-13
Decay rate (A) = N/ t SI unit of decay is the becquerel (Bq) = 1d/s. curie (Ci( 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. 24-14
Figure 24.4 Decrease in the number of 14 C nuclei over time. 24-15
Sample Problem 24.4 Finding the Number of Radioactive Nuclei 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) PLAN: 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. SOLUTION: 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 = kt ln A t = -kt + ln A 0 ln A t = -(0.024yr -1 )(59yr) + ln(1.2x10 12 d/s) ln A t = 26.4 A t = 2.9x10 11 d/s 24-16 Fraction (1.2x1012-2.9x1011 ) decayed = (1.2x10 12 ) Fraction decayed = 0.76
Figure 24.5 Radiocarbon dating for determining the age of artifacts. 24-17
Sample Problem 24.5 Applying Radiocarbon Dating 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: SOLUTION: 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. 24-18
Figure 24.6 A linear accelerator. The linear accelerator operated by Standford University, California 24-19
Figure 24.7 The cyclotron accelerator. 24-20
24-21 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. 24-22
24-23 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
24-24 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
24-25 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 24.9 Which reactant contributes which group to the ester? O R C OH H + O + H OR' R C O R' + H O H O R C OH 18 OH O 18 + R' R C O R' + H O H O 18 R C OH O + R' OH R C O 18 R' + H O H 24-26
Figure 24.10 The use of radioisotopes to image the thyroid gland. asymmetric scan indicates disease normal Figure 24.11 PET and brain activity. 24-27 normal Alzheimer s
Figure 24.12 The increased shelf life of irradiated food. 24-28
The Interconversion of Mass and Energy E = mc 2 E E = mc 2 m m = E 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 24-29
Sample Problem 24.6 Calculating the Binding Energy per Nucleon PROBLEM: PLAN: SOLUTION: 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 = (0.52846 amu)(931.5 MeV/amu) 56 nucleons = 8.790 Mev/nucleon 12 C has a binding energy of 7.680 MeV/nucleon, so 56 Fe is more stable. 24-30
Figure 24.13 The variation in binding energy per nucleon. 24-31
Figure 24.14 Induced fission of 235 U. 24-32
Figure 24.15 A chain reaction of 235 U. 24-33
Figure 24.16 Diagram of an atomic bomb. 24-34
Figure 24.17 A light-water reactor. 24-35
Figure 24.18 The tokamak design for magnetic containment of a fusion plasma. 24-36
Figure B24.1 Detection of radioactivity by an ionization counter. 24-37
Figure B24.2 Vials of a scintillation cocktail emitting light 24-38
Element synthesis in the life cycle of a star Figure B24.3 24-39
Figure B24.4 A view of Supernova 1987A. 24-40