Problem 32 - Chemistry Structure and Dynamics

Question

Calculate the rate constants in \(\mathrm{s}^{-1}\) for the decay of the following nuclides from their half-lives. (a) \({ }^{18} \mathrm{~F}, 110\) minutes (b) \({ }^{54} \mathrm{Mn}, 312\) days (c) \({ }^{3} \mathrm{H}, 12.26\) years (d) \({ }^{14} \mathrm{C}, 5730\) years (e) \({ }^{129} \mathrm{I}, 1.6 \times 10^{7}\) years

Step-by-Step Solution

Answer

Rate constants are: (a) \(1.05 \times 10^{-4}\,\mathrm{s}^{-1}\), (b) \(2.57 \times 10^{-8}\,\mathrm{s}^{-1}\), (c) \(1.79 \times 10^{-9}\,\mathrm{s}^{-1}\), (d) \(3.83 \times 10^{-12}\,\mathrm{s}^{-1}\), (e) \(1.37 \times 10^{-18}\,\mathrm{s}^{-1}\).

Step 1: Understand the Relationship Between Half-life and Rate Constant

The rate constant \( k \) for radioactive decay can be found using the formula \( k = \frac{\ln(2)}{t_{1/2}} \), where \( t_{1/2} \) is the half-life of the nuclide. The natural logarithm of 2, \( \ln(2) \), is approximately 0.693.

Step 2: Convert Half-lives to Seconds

Since the rate constant needs to be in \( \mathrm{s}^{-1} \), convert the given half-lives into seconds.(a) \( 110 \) minutes = \( 110 \times 60 \) seconds(b) \( 312 \) days = \( 312 \times 24 \times 3600 \) seconds(c) \( 12.26 \) years = \( 12.26 \times 365.25 \times 24 \times 3600 \) seconds(d) \( 5730 \) years = \( 5730 \times 365.25 \times 24 \times 3600 \) seconds(e) \( 1.6 \times 10^{7} \) years = \( 1.6 \times 10^{7} \times 365.25 \times 24 \times 3600 \) seconds.

Step 3: Calculate Rate Constant for Each Nuclide

Using the formula \( k = \frac{\ln(2)}{t_{1/2}} \), calculate the rate constant for each nuclide after converting the half-lives to seconds.(a) \( k \approx \frac{0.693}{110 \times 60} \approx 1.05 \times 10^{-4} \; \mathrm{s}^{-1} \)(b) \( k \approx \frac{0.693}{312 \times 24 \times 3600} \approx 2.57 \times 10^{-8} \; \mathrm{s}^{-1} \)(c) \( k \approx \frac{0.693}{12.26 \times 365.25 \times 24 \times 3600} \approx 1.79 \times 10^{-9} \; \mathrm{s}^{-1} \)(d) \( k \approx \frac{0.693}{5730 \times 365.25 \times 24 \times 3600} \approx 3.83 \times 10^{-12} \; \mathrm{s}^{-1} \)(e) \( k \approx \frac{0.693}{1.6 \times 10^{7} \times 365.25 \times 24 \times 3600} \approx 1.37 \times 10^{-18} \; \mathrm{s}^{-1} \).

Key Concepts

Half-life Calculation

Calculating the half-life of a substance helps in understanding how long it will take for half of a radioactive material to decay. It is crucial in fields like nuclear chemistry, medicine, and archaeology. The half-life value is always constant for a given isotope, meaning that no matter how much of the substance is present, half will always decay in the same time period. The formula used to define half-life is:

• Half-life, denoted as \( t_{1/2} \), is the time required for the quantity to reduce to half its initial amount.
• The constant \( k \) can be calculated using the formula: \( k = \frac{\ln(2)}{t_{1/2}} \).
• \( \ln(2) \) is a constant equal to approximately 0.693.

Understanding how to compute the half-life prepares us to predict the behavior of radioactive substances over time. This predictability is crucial when using radioactive materials in any practical application.

Rate Constant

The rate constant, often represented as \( k \), gives important information about the speed of radioactive decay. This constant is essential in determining how fast a particular nuclide will decay into another element or isotope. It represents the probability per unit time that an atom will decay, which plays a significant role in nuclear chemistry.

• To find \( k \), use the formula given by \( k = \frac{\ln(2)}{t_{1/2}} \). This shows the inverse relationship between half-life and rate constant: a longer half-life means a smaller rate constant, indicating slower decay.
• The units of \( k \) are typically \( \,s^{-1} \), indicating the decay rate per second.

Knowing the rate constant allows scientists and engineers to design systems and processes that rely on the predictable decay of radioactive materials, ensuring safety and efficiency.

Nuclear Chemistry

Nuclear chemistry is a field of chemistry dealing with radioactivity, nuclear processes, and properties. One aspect of nuclear chemistry is to understand how nuclear reactions occur and the rates at which radioactive materials decay. This knowledge is applied in various fields, including medical treatments, nuclear power generation, and radiation safety.

• Radionuclides, or radioactive isotopes, decay over time, which is a key focus in nuclear chemistry.
• Radioactive decay involves several modes, such as alpha, beta, and gamma decay, each releasing different particles or radiation.
• Nuclear chemistry also explores the mechanisms behind these decay processes and their outcomes on the involved elements.

By comprehending these processes, scientists can manipulate nuclear reactions for energy production or medical applications, taking advantage of the principles predicting decay behaviors.

Exponential Decay Rate

Exponential decay describes the process where radioactive substances decrease at a rate proportional to their current value. This type of decay follows an exponential model, making it predictable and mathematically manageable.

• The mathematical model for exponential decay is expressed as \( N(t) = N_0 e^{-kt} \), where:
  • \( N(t) \) is the quantity remaining after time \( t \).
  • \( N_0 \) is the initial quantity of the substance.
  • \( k \) is the rate constant of decay.
• This formula highlights the predictable decline over time, with the rate constant \( k \) shaping the decay curve.
• An understanding of exponential decay is crucial in accurately predicting how radioactive isotopes change over time, aiding in waste management, safety measures, and practical applications.

The exponential nature of decay ensures that while the process continues indefinitely, the amount remaining always trends toward zero, a feature important in ensuring long-term safety in applications involving radioactivity.