Potassium-40 has a half-life of [tex]$1.277 \times 10^9$[/tex] years. After [tex]$1.022 \times 10^{10}$[/tex] years, how much potassium-40 will remain from a 500.3 g sample?

A. approximately 1.950 g
B. approximately 3.010 g
C. approximately 62.540 g
D. approximately 71.470 g



Answer :

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