Question
The half-life of plutonium-238 is 87.7 years.
What percentage of the atoms in a sample of plutonium-238 will remain radioactive after 263.1 years?
%
What percentage of the atoms in the same sample of plutonium-238 will have changed to another isotope after 263.1 years?
What percentage of the atoms in a sample of plutonium-238 will remain radioactive after 263.1 years?
%
What percentage of the atoms in the same sample of plutonium-238 will have changed to another isotope after 263.1 years?
Answers
GPT-4o mini
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.**
\[
N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\]
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.**