To determine which jar will contain the least amount of the isotope of sulfur after 10 seconds, we need to understand how the amount of a radioactive substance decreases over time. The key concept here is the half-life, which is the time required for half of the original amount of the substance to decay. The formula for the remaining fraction of a substance after a given time is:
[tex]\[ \text{remaining\_fraction} = 0.5^{\left(\frac{\text{time\_elapsed}}{\text{half\_life}}\right)} \][/tex]
Let's calculate the remaining fraction for each isotope after 10 seconds:
1. Sulfur-30 (half-life = 1.18 seconds):
[tex]\[
\text{remaining\_fraction} = 0.5^{\left(\frac{10}{1.18}\right)} \approx 0.002811
\][/tex]
2. Sulfur-35 (half-life = 87.5 days = 87.5 \times 24 \times 3600 seconds):
[tex]\[
\text{remaining\_fraction} = 0.5^{\left(\frac{10}{87.5 \times 24 \times 3600}\right)} \approx 0.999999
\][/tex]
3. Sulfur-37 (half-life = 5.05 minutes = 5.05 \times 60 seconds):
[tex]\[
\text{remaining\_fraction} = 0.5^{\left(\frac{10}{5.05 \times 60}\right)} \approx 0.977384
\][/tex]
4. Sulfur-41 (half-life = 1.99 seconds):
[tex]\[
\text{remaining\_fraction} = 0.5^{\left(\frac{10}{1.99}\right)} \approx 0.030710
\][/tex]
The remaining fractions for the isotopes after 10 seconds are approximately:
- Sulfur-30: 0.002811
- Sulfur-35: 0.999999
- Sulfur-37: 0.977384
- Sulfur-41: 0.030710
By comparing these values, we can see that the smallest remaining fraction is for sulfur-30, which means it has decayed the most.
Therefore, the correct answer is:
A. sulfur-30
