Sure, let's break down this problem step by step.
1. Understand the problem:
We need to determine how much of a 16.8-gram sample of Leutium-176 remains after [tex]\( 1.155 \times 10^{11} \)[/tex] years given that its half-life is [tex]\( 3.85 \times 10^{10} \)[/tex] years.
2. Identify the data given:
- Initial mass of Leutium-176: 16.8 grams
- Half-life of Leutium-176: [tex]\( 3.85 \times 10^{10} \)[/tex] years
- Time elapsed: [tex]\( 1.155 \times 10^{11} \)[/tex] years
3. Calculate the number of half-lives that have passed:
We use the formula:
[tex]\[
\text{Number of half-lives} = \frac{\text{Time elapsed}}{\text{Half-life}}
\][/tex]
Plugging in the provided values:
[tex]\[
\text{Number of half-lives} = \frac{1.155 \times 10^{11} \text{ years}}{3.85 \times 10^{10} \text{ years}} = 3
\][/tex]
4. Calculate the remaining mass:
The remaining mass of a substance after a certain number of half-lives can be calculated using the following formula:
[tex]\[
\text{Remaining mass} = \text{Initial mass} \times \left( \frac{1}{2} \right)^{\text{Number of half-lives}}
\][/tex]
Applying the values we have:
[tex]\[
\text{Remaining mass} = 16.8 \text{ grams} \times \left( \frac{1}{2} \right)^3
\][/tex]
Simplify the expression [tex]\( \left( \frac{1}{2} \right)^3 = \frac{1}{8} \)[/tex]:
[tex]\[
\text{Remaining mass} = 16.8 \text{ grams} \times \frac{1}{8} = 2.1 \text{ grams}
\][/tex]
5. Conclusion:
After [tex]\( 1.155 \times 10^{11} \)[/tex] years, 2.1 grams of the original 16.8-gram sample of Leutium-176 will remain.
So, the correct answer is:
[tex]\[ 2.10 \text{ g} \][/tex]
