Chapter 19: Problem 720
The radioactive decay constant for radium is \(1.36 \times 10^{-11}\). How many disintegrations per second occur in \(100 \mathrm{~g}\) of radium?
Short Answer
Expert verified
The activity of 100 g of radium is approximately \(3.629 × 10^{12}\) disintegrations per second.
Step by step solution
01
Identify the given information
We are given:
- Radioactive decay constant for radium (\(λ\)) = \(1.36 × 10^{-11}\text{s}^{-1}\)
- Mass of radium (\(m\)) = 100 g
- Molar mass of radium (\(M\)) = 226 g/mol (from the periodic table)
- Avogadro's constant (\(N_A\)) = \(6.022 × 10^{23}\) mol⁻¹
02
Calculate the number of moles of radium
Using the mass and molar mass, we can calculate the number of moles of radium:
Number of moles (\(n\)) = \( \frac{m}{M} \)
\(n = \frac{100 \thinspace \text{g}}{226 \thinspace \text{g/mol}}\)
03
Calculate the number of atoms of radium
Using the number of moles and Avogadro's constant, we can calculate the number of atoms of radium:
Number of atoms (\(N\)) = \(n × N_A\)
\(N = \frac{100 \thinspace \text{g}}{226 \thinspace \text{g/mol}} \times 6.022 × 10^{23} \thinspace \text{mol}^{-1}\)
04
Calculate the activity of radium
Using the decay constant and the number of atoms, we can calculate the activity of radium:
Activity (\(A\)) = \(λ × N\)
\(A = 1.36 × 10^{-11} \thinspace \text{s}^{-1} \times \frac{100 \thinspace \text{g}}{226 \thinspace \text{g/mol}} \times 6.022 × 10^{23} \thinspace \text{mol}^{-1}\)
05
Solve for the activity
Now, we can solve for the activity of radium:
\(A ≈ 1.36 × 10^{-11} \thinspace \text{s}^{-1} \times \frac{100 \thinspace \text{g} \times 6.022 × 10^{23} \thinspace \text{mol}^{-1}}{226 \thinspace \text{g/mol}}\)
\(A ≈ 3.629 × 10^{12}\) disintegrations per second
The activity of 100 g of radium is approximately \(3.629 × 10^{12}\) disintegrations per second.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radioactive Decay Constant
Understanding the concept of the radioactive decay constant is pivotal in calculating the stability of a radioactive isotope. The decay constant, denoted as \(\lambda\), represents the probability per unit time that an atom will decay. Essentially, it reflects how quickly a radioactive isotope undergoes decay, a fundamental property inherent to the isotope. The higher the decay constant, the more unstable the isotope, and the faster it decays. In our example with radium, a decay constant of \(1.36 \times 10^{-11} s^{-1}\) indicates that each second, a small fraction of radium atoms will decay, contributing to its overall activity.
To provide context, consider a large group of unstable isotopes: over time, a predictable number of them will decay, and this decay rate is characterized by the decay constant. This constant is crucial for computations in fields like nuclear physics and radiometric dating, where estimating the age of artifacts and geological specimens is required.
To provide context, consider a large group of unstable isotopes: over time, a predictable number of them will decay, and this decay rate is characterized by the decay constant. This constant is crucial for computations in fields like nuclear physics and radiometric dating, where estimating the age of artifacts and geological specimens is required.
Avogadro's Constant
A fundamental aspect of chemistry is Avogadro's constant, denoted as \(N_A\), which is the number of constituent particles, usually atoms or molecules, in one mole of a substance. Avogadro's constant has a value of approximately \(6.022 \times 10^{23} mol^{-1}\), and it plays a pivotal role in converting between atomic scale measurements and macroscopic scale. For example, when we talk about a mole of radium, we are actually referring to \(6.022 \times 10^{23}\) atoms of radium.
Why is this constant so important? In our problem, without Avogadro's constant, we wouldn't be able to translate the mass of radium into the actual number of atoms present, since we experience masses at a macroscopic level, not the atomic level. Thus, understanding Avogadro's constant is essential for anyone working with chemical quantities and reactions, because it provides a bridge between the visible world and the atomic world.
Why is this constant so important? In our problem, without Avogadro's constant, we wouldn't be able to translate the mass of radium into the actual number of atoms present, since we experience masses at a macroscopic level, not the atomic level. Thus, understanding Avogadro's constant is essential for anyone working with chemical quantities and reactions, because it provides a bridge between the visible world and the atomic world.
Activity of Radium
The activity of a radioactive substance is measured in terms of its disintegrations per second (dps), which is a direct indication of how radioactive the substance is. Specifically for radium, or any radioactive material, the activity is calculated by multiplying the number of atoms by the decay constant. This provides a quantitative measurement of the rate at which radium atoms are decaying into other elements or isotopes.
Activity has profound implications in various applications like medical diagnostics and treatment, where specific levels of activity are necessary for imaging and cancer therapy. Additionally, when handling radioactive materials, understanding the activity helps in establishing appropriate safety measures and ensures compliance with regulatory standards. The calculated activity, as shown in our example, tells us how much radium would be undergoing transformation at any given second, providing insight into its stability and potential uses or hazards.
Activity has profound implications in various applications like medical diagnostics and treatment, where specific levels of activity are necessary for imaging and cancer therapy. Additionally, when handling radioactive materials, understanding the activity helps in establishing appropriate safety measures and ensures compliance with regulatory standards. The calculated activity, as shown in our example, tells us how much radium would be undergoing transformation at any given second, providing insight into its stability and potential uses or hazards.
Disintegrations per Second
When it comes to nuclear physics, 'disintegrations per second' is a unit of measurement that quantifies how many atomic nuclei of a radioactive substance decay every second, and it's also known as the Becquerel (Bq) in the International System of Units. This rate of decay is tied directly to the substance's half-life, which is the amount of time it takes for half of the material to decay.
In practical terms, disintegrations per second is how scientists and engineers determine the strength of a radioactive source. It influences not just scientific research, but also practical elements such as the sizing of shields for radiation protection, the design of medical treatments, and even the determination of storage and disposal requirements for radioactive materials. The figure we obtained in our example, \(3.629 \times 10^{12} dps\) for 100g of radium, is a measure of the intensity of its radioactivity, which can be used to assess and manage the risks and benefits of using radium for different purposes.
In practical terms, disintegrations per second is how scientists and engineers determine the strength of a radioactive source. It influences not just scientific research, but also practical elements such as the sizing of shields for radiation protection, the design of medical treatments, and even the determination of storage and disposal requirements for radioactive materials. The figure we obtained in our example, \(3.629 \times 10^{12} dps\) for 100g of radium, is a measure of the intensity of its radioactivity, which can be used to assess and manage the risks and benefits of using radium for different purposes.