A radioactive sample containing approximately 6.52×10¹3 radioactive nuclides has a half-life of 5730 years. Calculate the number of radioactive nuclides present in the sample after 8000 years.

icon
Related questions
Question
**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.
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.
Expert Solution
steps

Step by step

Solved in 3 steps

Blurred answer