Chapter 23: Problem 2
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.
Short Answer
Expert verified
The initial activity was approximately 4.09 Ci.
Step by step solution
01
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.
02
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 \( N_0 \).
03
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:\[ 2.64 = N_0 \left(\frac{1}{2}\right)^{\frac{4000}{5370}} \]
04
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:\[ N_0 = \frac{2.64}{\left(\frac{1}{2}\right)^{0.744}} \]
05
Calculate the expression
Evaluate \( \left(\frac{1}{2}\right)^{0.744} \), which is approximately 0.646. Therefore, the initial activity is:\[ N_0 = \frac{2.64}{0.646} \approx 4.09 \text{ Ci} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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:
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.
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.