Problem 1 - Applied Physics

Question

Find the source activity of a 1.24-Ci sample of \({ }^{13} \mathrm{~N}\) (nitrogen) \(20.0 \mathrm{~min}\) after certification. Its half-life is \(10.0 \mathrm{~min}\).

Step-by-Step Solution

Answer

The remaining activity of the 1.24-Ci \(^{13}\text{N} \) sample after 20 minutes is 0.310 Ci.

Step 1: Identify the given values

Given: - Initial activity \( A_0 = 1.24 \) Ci (curies)- Time elapsed \( t = 20.0 \) min- Half-life \( T_{1/2} = 10.0 \) min. We will use these values to find the remaining activity.

Step 2: Calculate the decay constant

The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) by the formula:\[ \lambda = \frac{\ln(2)}{T_{1/2}}.\]Substituting the given half-life:\[ \lambda = \frac{\ln(2)}{10.0} \approx 0.0693 \text{ min}^{-1}.\]

Step 3: Use the decay formula

The activity \( A \) remaining after time \( t \) can be calculated using:\[ A = A_0 \times e^{-\lambda t}.\]Substitute the values: \( A_0 = 1.24 \) Ci, \( \lambda = 0.0693 \text{ min}^{-1} \), and \( t = 20.0 \) min:\[ A = 1.24 \times e^{-0.0693 \times 20}.\]

Step 4: Calculate the remaining activity

Compute the value:\[ A = 1.24 \times e^{-1.386} \approx 1.24 \times 0.250 \approx 0.310 \text{ Ci}.\]

Step 5: Conclusion: Analyze the result

After 20 minutes, the remaining activity of the \(^{13}\text{N} \) sample is approximately 0.310 Ci.

Key Concepts

Half-life

The concept of half-life in radioactive decay is fundamental and quite intuitive once broken down.
The half-life of a radioactive substance is the time required for half of the radioactive nuclei in a sample to decay. It’s a fixed time period, unique to each radioactive isotope. In our example, the half-life of nitrogen-13 ( { }^{13} ext{N} ) is 10 minutes.
This means that every 10 minutes, half of the remaining radioactive nitrogen nuclei transform into a stable form or into another element through a decay process, halving the sample's radioactivity.
If you start with a sample activity of 1.24 curies (Ci), after one half-life (10 minutes), only 0.62 Ci of nitrogen-13 will remain active. After another 10 minutes, this will halve again, as shown in our original problem solution, leaving approximately 0.31 Ci active.
Being able to predict exactly how a radioactive sample decays over time makes it easier to understand nuclear reactions, date ancient artifacts, and even treat certain medical conditions.

Decay Constant

The decay constant (\( \lambda \) ) is another key concept in understanding radioactive decay.
It represents the probability per unit time that a single nucleus will decay.
Mathematically, the decay constant is linked to the half-life through the formula:

• \( \lambda = \frac{\ln(2)}{T_{1/2}} \)

Where \( \ln(2) \) is the natural logarithm of 2. In our exercise, with a half-life of 10 minutes, the decay constant for nitrogen-13 was calculated to be approximately 0.0693 min\(^{-1}\).
This tells us how quickly or slowly the substance is decaying.
A larger decay constant indicates a faster decay process, while a smaller value would indicate slower decay.
The decay constant is crucial for accurately modeling how a radioactive sample diminishes over time.

Radioactive Sample Activity

Activity is a term used to describe the rate at which a sample of radioactive material decays.
In simpler terms, it's the number of decays per second in a given sample.To calculate the remaining activity of a sample after a certain period, we use the decay equation:

• \( A = A_0 \times e^{-\lambda t} \)

Where:

• \( A_0 \) is the initial activity.
• \( \lambda \) is the decay constant.
• \( t \) is the time elapsed.

In our exercise, the initial activity was 1.24 Ci, and using this formula over a 20-minute span, we find the remaining activity drops to approximately 0.310 Ci.
Understanding this formula allows us to predict how active a radioactive sample will be at any given time. It’s critical in nuclear medicine where precise dosages of radioactive materials must be used effectively, ensuring safety and efficacy.