Textbook Question
The radioactive isotope 100Tc decays to form the stable iso-tope 100Mo. (a) There are two possible pathways for this decay. Write balanced equations for both.
Ch.20 - Nuclear Chemistry
McMurry8th EditionChemistryISBN: 9781292336145
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Table of contents
- Ch.1 - Chemical Tools: Experimentation & Measurement170
- Ch.2 - Atoms, Molecules & Ions160
- Ch.3 - Mass Relationships in Chemical Reactions169
- Ch.4 - Reactions in Aqueous Solution199
- Ch.5 - Periodicity & Electronic Structure of Atoms102
- Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory92
- Ch.7 - Covalent Bonding and Electron-Dot Structures61
- Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure59
- Ch.9 - Thermochemistry: Chemical Energy122
- Ch.10 - Gases: Their Properties & Behavior155
- Ch.11 - Liquids & Phase Changes54
- Ch.12 - Solids and Solid-State Materials73
- Ch.13 - Solutions & Their Properties120
- Ch.14 - Chemical Kinetics98
- Ch.15 - Chemical Equilibrium125
- Ch.16 - Aqueous Equilibria: Acids & Bases110
- Ch.17 - Applications of Aqueous Equilibria147
- Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium86
- Ch.19 - Electrochemistry154
- Ch.20 - Nuclear Chemistry96
- Ch.21 - Transition Elements and Coordination Chemistry156
- Ch.22 - The Main Group Elements187
- Ch.23 - Organic and Biological Chemistry51
Chapter 20, Problem 87b
The radioactive isotope 100Tc decays to form the stable iso-tope 100Mo. (b) Only one of the pathways is observed. Calculate the energy released by both pathways, and explain why only one is observed. Relevant masses are: 100Tc = 99.907 657, 100Mo = 99.907 48, electron = 0.000 548 6.
Verified step by step guidance1
Identify the type of decay that 100Tc undergoes to form 100Mo. Since 100Tc decays to 100Mo, which has the same mass number but a different atomic number, this suggests a beta decay process.
Write the nuclear equations for the possible decay pathways. For beta-minus decay: \( ^{100}_{43}Tc \rightarrow ^{100}_{42}Mo + e^- + \bar{\nu}_e \). For beta-plus decay: \( ^{100}_{43}Tc \rightarrow ^{100}_{44}Ru + e^+ + \nu_e \).
Calculate the mass defect for each pathway. Mass defect is the difference between the mass of the parent nucleus and the sum of the masses of the products. For beta-minus decay: \( \Delta m = m(^{100}Tc) - (m(^{100}Mo) + m(e^-)) \). For beta-plus decay: \( \Delta m = m(^{100}Tc) - (m(^{100}Ru) + m(e^+)) \).
Convert the mass defect into energy released using Einstein's equation, E=mc^2, where E is the energy, m is the mass defect, and c is the speed of light. This will give the energy released for each decay pathway.
Analyze why only one pathway is observed. Consider factors such as the stability of the resulting nucleus, the energy threshold for positron emission, and conservation laws (like conservation of energy and lepton number) to explain why only one decay pathway is favored or observed.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Radioactive Decay
Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation. This can occur through various pathways, such as alpha decay, beta decay, or gamma decay. In the case of 100Tc, it decays to 100Mo, and understanding the type of decay involved is crucial for calculating the energy released.
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Rate of Radioactive Decay
Mass-Energy Equivalence
Mass-energy equivalence, expressed by Einstein's equation E=mc², indicates that mass can be converted into energy. In radioactive decay, the difference in mass between the parent isotope (100Tc) and the daughter isotope (100Mo) is converted into energy, which can be calculated to determine the energy released during the decay process.
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Pathways of Decay
Different pathways of decay refer to the various modes through which a radioactive isotope can transform into a more stable form. The observation of only one pathway suggests that it is the most energetically favorable or has a higher probability of occurring, often influenced by factors such as the energy barrier and the stability of the resulting isotopes.
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