Textbook Question
Bone and bony structures contain calcium and phosphorus.
a. Why would the radioisotopes calcium-47 and phosphorus-32 be used in the diagnosis and treatment of bone diseases?
Ch.5 Nuclear Chemistry
Timberlake14th EditionChemistry: An Introduction to General, Organic, and Biological ChemistryISBN: 9781292472249
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Table of contents
- Ch.1 Chemistry in Our Lives19
- Ch.2 Chemistry and Measurements80
- Ch.3 Matter and Energy53
- Ch.4 Atoms and Elements58
- Ch.5 Nuclear Chemistry42
- Ch.6 Ionic and Molecular Compounds57
- Ch.7 Chemical Quantities and Reactions38
- Ch.8 Gases30
- Ch.9 Solutions50
- Ch.10 Acids and Bases and Equilibrium64
- Ch.11 Introduction to Organic Chemistry: Hydrocarbons77
- Ch.12 Alcohols, Thiols, Ethers, Aldehydes, and Ketones82
- Ch.13 Carbohydrates54
- Ch.14 Carboxylic Acids, Esters, Amines, and Amides91
- Ch.15 Lipids61
- Ch.16 Amino Acids, Proteins, and Enzymes84
- Ch.17 Nucleic Acids and Protein Synthesis86
- Ch.18 Metabolic Pathways and ATP Production73
Timberlake14th EditionChemistry: An Introduction to General, Organic, and Biological ChemistryISBN: 9781292472249Not the one you use?Change textbook
Chapter 5, Problem 34b
Fluorine-18, which has a half-life of 110 min, is used in PET scans.
b. If 100. mg of fluorine-18 is shipped at 8:00 a.m., how many milligrams of the radioisotope are still active when the sample arrives at the radiology laboratory at 1:30 p.m.?
Verified step by step guidance1
Determine the total time elapsed between 8:00 a.m. and 1:30 p.m. Subtract the starting time from the ending time. Convert the time into minutes for consistency. (1:30 p.m. is 13:30 in 24-hour format, so the elapsed time is 13:30 - 8:00 = 5 hours and 30 minutes, which equals 330 minutes.)
Calculate the number of half-lives that have passed. Use the formula: \( \text{Number of half-lives} = \frac{\text{Total time elapsed}}{\text{Half-life}} \). Here, the half-life of fluorine-18 is 110 minutes, so substitute the values: \( \text{Number of half-lives} = \frac{330}{110} \).
Use the radioactive decay formula to determine the remaining amount of fluorine-18. The formula is: \( A = A_0 \times (\frac{1}{2})^n \), where \( A \) is the remaining amount, \( A_0 \) is the initial amount, and \( n \) is the number of half-lives. Substitute \( A_0 = 100. \, \text{mg} \) and the calculated value of \( n \).
Simplify the expression \( (\frac{1}{2})^n \) to find the fraction of the original sample that remains after the elapsed time.
Multiply the initial amount \( A_0 \) by the fraction obtained in the previous step to calculate the remaining amount of fluorine-18 in milligrams.
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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 a radioactive substance to decay. In the case of Fluorine-18, which has a half-life of 110 minutes, this means that after 110 minutes, only 50 mg of the original 100 mg would remain. Understanding half-life is crucial for calculating the remaining quantity of a radioactive isotope over time.
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Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This decay occurs at a predictable rate, characterized by the half-life. For Fluorine-18, this decay is important to determine how much of the isotope remains active after a certain period, which is essential for its application in medical imaging.
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Exponential Decay Formula
The exponential decay formula is used to calculate the remaining quantity of a radioactive substance over time. It is expressed as N(t) = N0 * (1/2)^(t/T), where N(t) is the remaining quantity, N0 is the initial quantity, t is the elapsed time, and T is the half-life. This formula allows for precise calculations of how much Fluorine-18 remains active after the time interval from shipment to arrival.
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Related Practice
Textbook Question
For each of the following, indicate if the number of half-lives elapsed is:
1. one half-life
2. two half-lives
3. three half-lives
c. a sample of Au-198 with a half-life of 2.7 days after 5.4 days
Textbook Question
Bone and bony structures contain calcium and phosphorus.
b. During nuclear tests, scientists were concerned that strontium-85, a radioactive product, would be harmful to the growth of bone in children. Explain.
Textbook Question
Technetium-99m emits only gamma radiation. Why would this type of radiation be used in diagnostic imaging rather than an isotope that also emits beta or alpha radiation?
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Textbook Question
The dosage of technicium-99m for a lung scan is 20. µCi/kg of body mass. How many millicuries of technicium-99m should be given to a person weighing 50. kg (1 mCi = 1000 µCi)
Textbook Question
Strontium-85, used for bone scans, has a half-life of 65 days.
b. How long will it take for the radiation level of strontium-85 to drop to one-eighth of its original level?