Problem 33
Question
A device used in radiation therapy for cancer contains \(0.50 \mathrm{~g}\) of cobalt \(\frac{60}{27} \mathrm{Co}\) \((59.933819 \mathrm{u}) .\) The half-life of \(\underset{27}{60} \mathrm{C} \circ\) is \(5.27 \mathrm{yr}\). Determine the activity of the radioactive material.
Step-by-Step Solution
Verified Answer
The activity of the radioactive cobalt sample is calculated using decay constant and number of nuclei to find the total radioactive activity.
1Step 1: Understanding the Activity Formula
Activity (A) of a radioactive sample is given by the formula: \[ A = \lambda N \]where \( \lambda \) is the decay constant and \( N \) is the number of radioactive nuclei.
2Step 2: Calculate the Decay Constant
The decay constant \( \lambda \) can be calculated using the formula: \[ \lambda = \frac{\ln(2)}{T_{1/2}} \] where \( T_{1/2} = 5.27 \text{ years} = 5.27 \times 365 \times 24 \times 3600 \text{ seconds} \) for this problem. Compute \( \lambda \).
3Step 3: Convert Mass to Number of Nuclei
The number of nuclei \( N \) can be calculated from the mass of radioactive material using: \[ N = \frac{m}{M} \times N_A \]where \( m = 0.50 \text{ grams} \), \( M = 59.933819 \text{ g/mol} \), and \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \) is Avogadro's number. Compute \( N \).
4Step 4: Calculate Radioactive Activity
Substitute \( \lambda \) and \( N \) into the formula for activity: \[ A = \lambda N \]This will give the activity \( A \), typically in becquerels (Bq). Compute \( A \) to find the activity of the sample.
Key Concepts
Decay ConstantHalf-LifeAvogadro's NumberRadioactive Activity
Decay Constant
In the world of radioactive decay, the decay constant \( \lambda \) plays a crucial role. It is a constant that characterizes the rate at which a radioactive substance undergoes decay. You can think of it as a clock ticking against the radioactive material, dictating the speed of the decay process.
The decay constant is derived from the equation \( \lambda = \frac{\ln(2)}{T_{1/2}} \), where \( T_{1/2} \) represents the half-life of the substance. Here, \( \ln(2) \) is the natural logarithm of 2, a mathematical constant approximately equal to 0.693. By calculating \( \lambda \), we can predict how quickly a radioactive isotope will lose half of its quantity, which is vital in many scientific applications.
The decay constant is derived from the equation \( \lambda = \frac{\ln(2)}{T_{1/2}} \), where \( T_{1/2} \) represents the half-life of the substance. Here, \( \ln(2) \) is the natural logarithm of 2, a mathematical constant approximately equal to 0.693. By calculating \( \lambda \), we can predict how quickly a radioactive isotope will lose half of its quantity, which is vital in many scientific applications.
- A higher decay constant means a faster rate of decay.
- It provides essential data for calculating radioactive activity, which is crucial for understanding a substance's potency.
Half-Life
Half-life, denoted as \( T_{1/2} \), is a core concept in understanding radioactive decay. It is defined as the time taken for half of the radioactive nuclei in a sample to decay. This parameter helps in predicting how long a radioactive material can be effectively used or remain dangerous.
The relationship between half-life and decay constant is given by \( T_{1/2} = \frac{\ln(2)}{\lambda} \). Knowing either \( T_{1/2} \) or \( \lambda \) allows us to determine the other, providing insights into both the speed of decay and the longevity of a radioactive sample.
The relationship between half-life and decay constant is given by \( T_{1/2} = \frac{\ln(2)}{\lambda} \). Knowing either \( T_{1/2} \) or \( \lambda \) allows us to determine the other, providing insights into both the speed of decay and the longevity of a radioactive sample.
- A longer half-life means the substance remains active over a more extended period.
- The half-life is independent of the initial amount of the substance.
Avogadro's Number
Avogadro's number \( N_A \) is one of the fundamental constants in chemistry and physics, defined as \( 6.022 \times 10^{23} \text{ mol}^{-1} \). It is the number of constituent particles, usually atoms or molecules, in one mole of a substance.
In the context of calculating the number of radioactive nuclei \( N \) in a sample when given its mass and molar mass, Avogadro's number is essential. The formula used is \( N = \frac{m}{M} \times N_A \), where \( m \) is the sample's mass and \( M \) is its molar mass.
In the context of calculating the number of radioactive nuclei \( N \) in a sample when given its mass and molar mass, Avogadro's number is essential. The formula used is \( N = \frac{m}{M} \times N_A \), where \( m \) is the sample's mass and \( M \) is its molar mass.
- Avogadro's number bridges the gap between the atomic scale and the macroscopic scale.
- It enables the conversion of a substance's mass to the number of atoms, facilitating understanding of chemical reactions at the molecular level.
Radioactive Activity
Radioactive activity refers to the rate at which a sample of radioactive material decays, measured in decays per second. The unit of activity is the becquerel (Bq), where 1 Bq equals one decay per second.
The activity \( A \) is computed using the formula \( A = \lambda N \), combining the decay constant \( \lambda \) and the number of radioactive nuclei \( N \). Understanding and calculating radioactive activity is vital in radiation therapy, nuclear power, and safety protocols as it gives a measurable indicator of how much radiation a substance emits.
The activity \( A \) is computed using the formula \( A = \lambda N \), combining the decay constant \( \lambda \) and the number of radioactive nuclei \( N \). Understanding and calculating radioactive activity is vital in radiation therapy, nuclear power, and safety protocols as it gives a measurable indicator of how much radiation a substance emits.
- A higher activity indicates a greater number of decays per second.
- Activity helps in determining the "strength" or "potency" of a radioactive source.
Other exercises in this chapter
Problem 30
The half-lives in two different samples, \(A\) and \(B\), of radioactive nuclei are related according to \(T_{1 / 2, \mathrm{~B}}=\frac{1}{2} T_{1 / 2, \mathrm{
View solution Problem 31
How many half-lives are required for the number of radioactive nuclei to decrease to one-millionth of the initial number?
View solution Problem 34
Iodine 131 53 is used in diagnostic and therapeutic techniques in the treatment of thyroid disorders. This isotope has a half-life of \(8.04\) days. What percen
View solution Problem 34
Iodine \(\frac{131}{53} \mathrm{I}\) is used in diagnostic and therapeutic techniques in the treatment of thyroid disorders. This isotope has a half-life of 8.0
View solution