To determine how much Potassium-40 will remain after [tex]$1.022 \times 10^{10}$[/tex] years from an initial sample of 500.3 grams, we need to follow these steps:
1. Identify the half-life and the time elapsed:
- Half-life ([tex]\( t_{1/2} \)[/tex]) of Potassium-40: [tex]\( 1.277 \times 10^9 \)[/tex] years
- Time elapsed: [tex]\( 1.022 \times 10^{10} \)[/tex] years
- Initial sample size: [tex]\( 500.3 \)[/tex] grams
2. Calculate the number of half-lives that have passed:
- Number of half-lives: [tex]\( \frac{\text{time elapsed}}{\text{half-life}} \)[/tex]
- [tex]\[
\frac{1.022 \times 10^{10} \text{ years}}{1.277 \times 10^9 \text{ years}} \approx 8.003
\][/tex]
So, approximately 8.003 half-lives have passed.
3. Use the decay formula to determine the remaining sample:
The remaining amount of a radioactive substance after a certain period is given by:
[tex]\[
\text{Remaining} = \text{Initial} \times \left(\frac{1}{2}\right)^{\text{number of half-lives}}
\][/tex]
- [tex]\[
\text{Remaining} = 500.3 \times \left(\frac{1}{2}\right)^{8.003}
\][/tex]
4. Calculate the remaining sample:
Using the number of half-lives calculated, we find:
- [tex]\[
500.3 \times \left(0.5\right)^{8.003} \approx 1.950
\][/tex]
Therefore, after [tex]\( 1.022 \times 10^{10} \)[/tex] years, approximately 1.950 grams of the initial 500.3 grams of Potassium-40 will remain. Thus, the correct answer is:
- approximately 1.950
