Sure, let's solve this step by step.
1. Identify what is given:
- The half-life of Strontium-90 (Sr-90), which is 28 years.
- The desired remaining percentage of the sample, which is [tex]\(64\%\)[/tex] decomposed. Therefore, the remaining sample is [tex]\(100\% - 64\% = 36\%\)[/tex] of the original amount.
2. Understand the concept of half-life:
- Half-life is the time it takes for half of a radioactive substance to decay.
3. Use the exponential decay formula:
[tex]\[
N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\][/tex]
where:
- [tex]\( N(t) \)[/tex] is the remaining quantity of the substance after time [tex]\( t \)[/tex].
- [tex]\( N_0 \)[/tex] is the initial quantity of the substance.
- [tex]\( t \)[/tex] is the time that has passed.
- [tex]\( t_{1/2} \)[/tex] is the half-life of the substance.
For this question, we want to find the time [tex]\( t \)[/tex] it takes for the substance to decompose 64%, meaning that 36% remains. We can set up the equation as follows:
[tex]\[
0.36 = \left( \frac{1}{2} \right)^{\frac{t}{28}}
\][/tex]
4. Solve for [tex]\( t \)[/tex]:
Rearrange the equation to isolate [tex]\( t \)[/tex]:
[tex]\[
\left( \frac{1}{2} \right)^{\frac{t}{28}} = 0.36
\][/tex]
To solve for [tex]\( t \)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[
\ln \left( \left( \frac{1}{2} \right)^{\frac{t}{28}} \right) = \ln(0.36)
\][/tex]
Using the property of logarithms [tex]\( \ln(a^b) = b \cdot \ln(a) \)[/tex]:
[tex]\[
\frac{t}{28} \cdot \ln \left( \frac{1}{2} \right) = \ln(0.36)
\][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[
t = 28 \cdot \frac{\ln(0.36)}{\ln(0.5)}
\][/tex]
5. Substitute the values of the natural logarithms:
[tex]\[
t = 28 \cdot \frac{\ln(0.36)}{\ln(0.5)}
\][/tex]
From the calculation, we find that:
[tex]\[
t \approx 18.03
\][/tex]
6. Conclusion:
It will take approximately 18.03 years for a given sample of Strontium-90 (Sr-90) to be 64% decomposed.
