To determine the percentage of plutonium-238 that remains radioactive after a certain period, we use the concept of half-lives. The formula to find the remaining quantity of a radioactive substance after a certain period is:
\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
where:
- \( N \) is the remaining quantity of the substance,
- \( N_0 \) is the initial quantity of the substance,
- \( t \) is the elapsed time,
- \( t_{1/2} \) is the half-life of the substance.
In this case, the half-life (\( t_{1/2} \)) of plutonium-238 is 87.7 years, and the elapsed time (\( t \)) is 263.1 years. First, we calculate the number of half-lives that have passed:
\[ \text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{263.1}{87.7} \approx 3 \]
Now, we can calculate the remaining percentage of the original sample:
\[ N = N_0 \left( \frac{1}{2} \right)^3 = N_0 \left( \frac{1}{8} \right) \]
This means that after 263.1 years, \( \frac{1}{8} \) of the original plutonium-238 remains:
\[ \text{Percentage remaining} = \left( \frac{1}{8} \right) \times 100% = 12.5% \]
Therefore, after 263.1 years, approximately 12.5% of the plutonium-238 atoms will remain radioactive.
To find the percentage of atoms that have changed to another isotope, we subtract the remaining percentage from 100%:
\[ \text{Percentage changed} = 100% - 12.5% = 87.5% \]
Thus, after 263.1 years, approximately 87.5% of the atoms in the sample of plutonium-238 will have changed to another isotope.
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