A radioactive sample decays with a half-life of 71.19 minutes. If the original activity of the sample was 16.352 mCi, what is the activity of the sample after 2.548 hours have passed? Express your answer in terms of mCi using at least three significant figures.
A radioactive sample decays with a half-life of 71.19 minutes. If the original activity of the sample was 16.352 mCi, what is the activity of the sample after 2.548 hours have passed? Express your answer in terms of mCi using at least three significant figures.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Problem Statement:**
A radioactive sample decays with a half-life of 71.19 minutes. If the original activity of the sample was 16.352 mCi, what is the activity of the sample after 2.548 hours have passed?
Express your answer in terms of mCi using at least three significant figures.
**Solution Template:**
- Calculate total elapsed time in minutes: \[ \text{Elapsed time} = 2.548 \, \text{hours} \times 60 \, \text{minutes/hour} \]
- Total elapsed time: \[ 2.548 \times 60 = 152.88 \, \text{minutes} \]
- Use the decay formula: \[ A(t) = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
- Where:
- \( A(t) \) = Activity at time \( t \)
- \( A_0 \) = Original activity (16.352 mCi)
- \( t \) = Elapsed time (152.88 minutes)
- \( T_{1/2} \) = Half-life of the sample (71.19 minutes)
- Substitute the values:
\[ A(152.88) = 16.352 \times \left( \frac{1}{2} \right)^{\frac{152.88}{71.19}} \]
- Calculate:
\[ A(152.88) = 16.352 \times \left( \frac{1}{2} \right)^{2.148} \]
- Using values from the calculation:
\[ \left( \frac{1}{2} \right)^{2.148} \approx 0.221 \]
- Therefore:
\[ A(152.88) \approx 16.352 \times 0.221 \approx 3.614 \, \text{mCi} \]
Final Answer:
\[ \approx 3.614 \, \text{mCi} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F703b43e5-ae23-420c-b3a8-d626b9b6023e%2F34f780f0-e166-4948-9f51-d2fdebe9e2c7%2Fc4hdk2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A radioactive sample decays with a half-life of 71.19 minutes. If the original activity of the sample was 16.352 mCi, what is the activity of the sample after 2.548 hours have passed?
Express your answer in terms of mCi using at least three significant figures.
**Solution Template:**
- Calculate total elapsed time in minutes: \[ \text{Elapsed time} = 2.548 \, \text{hours} \times 60 \, \text{minutes/hour} \]
- Total elapsed time: \[ 2.548 \times 60 = 152.88 \, \text{minutes} \]
- Use the decay formula: \[ A(t) = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
- Where:
- \( A(t) \) = Activity at time \( t \)
- \( A_0 \) = Original activity (16.352 mCi)
- \( t \) = Elapsed time (152.88 minutes)
- \( T_{1/2} \) = Half-life of the sample (71.19 minutes)
- Substitute the values:
\[ A(152.88) = 16.352 \times \left( \frac{1}{2} \right)^{\frac{152.88}{71.19}} \]
- Calculate:
\[ A(152.88) = 16.352 \times \left( \frac{1}{2} \right)^{2.148} \]
- Using values from the calculation:
\[ \left( \frac{1}{2} \right)^{2.148} \approx 0.221 \]
- Therefore:
\[ A(152.88) \approx 16.352 \times 0.221 \approx 3.614 \, \text{mCi} \]
Final Answer:
\[ \approx 3.614 \, \text{mCi} \]
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