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![**Problem Statement:**
A radioactive sample containing approximately \(6.52 \times 10^{13}\) radioactive nuclides has a half-life of 5730 years. Calculate the number of radioactive nuclides present in the sample after 8000 years.
**Solution:**
To determine the number of nuclides remaining after a given period, we can use the formula for exponential decay based on half-life:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
where:
- \( N(t) \) is the remaining quantity of nuclides at time \( t \),
- \( N_0 \) is the initial quantity of nuclides (\(6.52 \times 10^{13}\)),
- \( t \) is the time elapsed (8000 years),
- \( T_{1/2} \) is the half-life (5730 years).
**Steps for Calculation:**
1. Plug the known values into the exponential decay formula:
\[
N(8000) = 6.52 \times 10^{13} \left( \frac{1}{2} \right)^{\frac{8000}{5730}}
\]
2. Calculate the exponent:
\[
\frac{8000}{5730} \approx 1.396
\]
3. Evaluate the decay factor:
\[
\left( \frac{1}{2} \right)^{1.396} \approx 0.341
\]
4. Multiply by the initial quantity:
\[
N(8000) \approx 6.52 \times 10^{13} \times 0.341 \approx 2.22 \times 10^{13}
\]
**Conclusion:**
After 8000 years, approximately \(2.22 \times 10^{13}\) radioactive nuclides will remain in the sample.](https://content.bartleby.com/qna-images/question/fc44715e-e997-4081-aac6-58f75fd336c7/471b0eb4-43b4-47fc-964f-4216e5655f1b/ekfgj8c_processed.png)
Transcribed Image Text:**Problem Statement:**
A radioactive sample containing approximately \(6.52 \times 10^{13}\) radioactive nuclides has a half-life of 5730 years. Calculate the number of radioactive nuclides present in the sample after 8000 years.
**Solution:**
To determine the number of nuclides remaining after a given period, we can use the formula for exponential decay based on half-life:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
where:
- \( N(t) \) is the remaining quantity of nuclides at time \( t \),
- \( N_0 \) is the initial quantity of nuclides (\(6.52 \times 10^{13}\)),
- \( t \) is the time elapsed (8000 years),
- \( T_{1/2} \) is the half-life (5730 years).
**Steps for Calculation:**
1. Plug the known values into the exponential decay formula:
\[
N(8000) = 6.52 \times 10^{13} \left( \frac{1}{2} \right)^{\frac{8000}{5730}}
\]
2. Calculate the exponent:
\[
\frac{8000}{5730} \approx 1.396
\]
3. Evaluate the decay factor:
\[
\left( \frac{1}{2} \right)^{1.396} \approx 0.341
\]
4. Multiply by the initial quantity:
\[
N(8000) \approx 6.52 \times 10^{13} \times 0.341 \approx 2.22 \times 10^{13}
\]
**Conclusion:**
After 8000 years, approximately \(2.22 \times 10^{13}\) radioactive nuclides will remain in the sample.
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