Textbook Question
The half-life of 238U is 4.5⨉109 yr. A sample of rock of mass 1.6 g produces 29 dis/s. Assuming all the radioactivity is due to 238U, find the percent by mass of 238U in the rock.
Ch.21 - Radioactivity & Nuclear Chemistry
Tro6th EditionChemistry: A Molecular ApproachISBN: 9780137832217
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Table of contents
- Ch.1 - Matter, Measurement & Problem Solving118
- Ch.2 - Atoms & Elements106
- Ch.3 - Molecules and Compounds131
- Ch.4 - Chemical Reactions and Chemical Quantities56
- Ch.5 - Introduction to Solutions and Aqueous Solutions82
- Ch.6 - Gases114
- Ch.7 - Thermochemistry96
- Ch.8 - The Quantum-Mechanical Model of the Atom60
- Ch.9 - Periodic Properties of the Elements90
- Ch.10 - Chemical Bonding I: The Lewis Model89
- Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory85
- Ch.12 - Liquids, Solids & Intermolecular Forces51
- Ch.13 - Solids & Modern Materials47
- Ch.14 - Solutions90
- Ch.15 - Chemical Kinetics115
- Ch.16 - Chemical Equilibrium68
- Ch.17 - Acids and Bases136
- Ch.18 - Aqueous Ionic Equilibrium153
- Ch.19 - Free Energy & Thermodynamics93
- Ch.20 - Electrochemistry106
- Ch.21 - Radioactivity & Nuclear Chemistry74
- Ch.22 - Organic Chemistry140
Chapter 21, Problem 94
The half-life of 232Th is 1.4⨉1010 yr. Find the number of disintegrations per hour emitted by 1.0 mol of 232Th.
Verified step by step guidance1
Step 1: Understand the concept of half-life and decay constant.The half-life (t1/2) is the time required for half of the radioactive nuclei in a sample to decay. The decay constant (\( \lambda \)) is related to the half-life by the formula: \( \lambda = \frac{\ln(2)}{t_{1/2}} \).
Step 2: Calculate the decay constant (\( \lambda \)).Use the given half-life of 232Th, which is 1.4⨉1010 years, to find \( \lambda \) using the formula: \( \lambda = \frac{\ln(2)}{1.4 \times 10^{10} \text{ yr}} \).
Step 3: Convert the decay constant to per hour.Since the decay constant is initially in terms of years, convert it to hours by using the conversion factor: 1 year = 365.25 days ⨉ 24 hours/day.
Step 4: Calculate the number of disintegrations per hour.Use the formula for activity (A), which is the number of disintegrations per unit time: \( A = \lambda N \), where \( N \) is the number of atoms. For 1.0 mol of 232Th, \( N = 6.022 \times 10^{23} \) atoms (Avogadro's number).
Step 5: Substitute the values into the activity formula.Substitute the decay constant (in per hour) and the number of atoms into the formula \( A = \lambda N \) to find the number of disintegrations per hour.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Half-life
Half-life is the time required for half of the radioactive nuclei in a sample to decay. It is a crucial concept in nuclear chemistry, as it helps determine the stability and longevity of radioactive isotopes. For example, if the half-life of an isotope is known, one can calculate how much of the substance remains after a certain period, which is essential for understanding radioactive decay processes.
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Zero-Order Half-life
Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This can occur in various forms, including alpha, beta, and gamma decay. The rate of decay is characterized by the half-life, and understanding this process is vital for calculating the number of disintegrations over a given time period, such as per hour.
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Avogadro's Number
Avogadro's number, approximately 6.022 × 10²³, is the number of atoms, ions, or molecules in one mole of a substance. This concept is essential for converting between moles and the actual number of particles, allowing chemists to relate macroscopic quantities to atomic-scale phenomena. In the context of the question, it helps determine the total number of thorium atoms in 1.0 mol, which is necessary for calculating the total disintegrations per hour.
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