Problem 2 - Applied Physics

Question

Find the source activity of a 2.64-Ci sample of \({ }_{6}^{14} \mathrm{C}\) (carbon) \(40 \overline{0} 0\) years after certification. Its half-life is 5370 yr.

Step-by-Step Solution

Answer

The initial activity was approximately 4.09 Ci.

Step 1: Understand the problem

We are given a radioactive sample of carbon-14 (\( {}_6^{14} \text{C} \)) with an activity of 2.64 curies (Ci) after 4000 years since certification. The half-life of \( {}_6^{14} \text{C} \) is 5370 years. We need to find the initial activity of the sample when it was certified.

Step 2: Use the decay formula

Radioactive decay can be described using the formula: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]where \( N(t) \) is the remaining activity after time \( t \), \( N_0 \) is the initial activity, and \( t_{1/2} \) is the half-life. We need to solve for \( N_0 \).

Step 3: Apply known values

We know \( N(t) = 2.64 \text{ Ci} \), \( t = 4000 \text{ years} \), and \( t_{1/2} = 5370 \text{ years} \). Substituting these into the decay formula gives:\[ 2.64 = N_0 \left(\frac{1}{2}\right)^{\frac{4000}{5370}} \]

Step 4: Solve for initial activity \( N_0 \)

Rearrange the equation to solve for \( N_0 \):\[ N_0 = \frac{2.64}{\left(\frac{1}{2}\right)^{\frac{4000}{5370}}} \]Calculate:\[ N_0 = \frac{2.64}{\left(\frac{1}{2}\right)^{0.744}} \]

Step 5: Calculate the expression

Evaluate \( \left(\frac{1}{2}\right)^{0.744} \), which is approximately 0.646. Therefore, the initial activity is:\[ N_0 = \frac{2.64}{0.646} \approx 4.09 \text{ Ci} \]

Key Concepts

Carbon-14 Half-Life

Radioactive decay is a natural process where unstable nuclei lose energy by emitting radiation. One of the most well-known isotopes that undergoes this process is Carbon-14. It is instrumental in archaeology, geology, and environmental science. Half-life is a key concept here, defining the time required for half of the radioactive atoms in a sample to decay. For Carbon-14, the half-life is specifically 5,370 years. This means after 5,370 years, half of any given sample will have decayed into a more stable form of carbon.

• This knowledge allows scientists to estimate the age of carbon-containing materials, such as fossils or artifacts, by measuring how much Carbon-14 remains.
• Not only does it provide an age estimation, but it also serves as proof of continuous decay at a known rate.

Understanding the half-life of Carbon-14 is crucial for solving problems related to radioactive decay and age determination.

Initial Activity Calculation

Determining the initial activity of a radioactive sample involves understanding decay over time. Activity, measured in curies (Ci), represents the number of decays per unit time. When given the current activity after a known period, you can calculate the original activity level. This is what the exercise involving a 2.64-Ci sample of Carbon-14 aimed to demonstrate.

To find the initial activity:

• A basic understanding of the decay process is crucial, where the amount of material decays predictably according to the half-life.
• Using the decay formula, as seen in the original step-by-step solution, further helps quantify this process by giving an initial activity value once the current and half-life values are substituted.

This calculation shows how measuring current activity and applying the half-life can return the original quantity of a radioactive material in its past state.

Half-Life Formula

The half-life formula is a powerful mathematical tool in the study of radioactive decay.It provides a way to calculate how much of a radioactive substance remains after a certain period. The formula is written as:\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]Where:

• \( N(t) \) refers to the remaining amount after time \( t \).
• \( N_0 \) is the initial amount of the substance.
• \( t \) is the elapsed time since the initial activity was measured.
• \( t_{1/2} \) denotes the half-life of the substance.

Applying this formula allows you to evaluate the remaining quantity of a substance or to find the original quantity if the current state is known. For example, in the solution to the exercise, we rearranged the formula to find \( N_0 \) (the initial activity) using the present activity \( N(t) \), elapsed time \( t \), and half-life \( t_{1/2} \). This step-by-step application of the formula embodies its importance in practical scenarios involving radiometric dating and decay predictions.