Let's solve the problem discussing each step carefully.
1. Understanding the problem:
- We have a radioactive substance, cesium-137, with a half-life of 30 years.
- The amount of the substance left after a certain period can be calculated using the exponential decay formula:
[tex]\[
A(t) = 458 \left(\frac{1}{2}\right)^{\frac{t}{30}}
\][/tex]
- We need to find two things:
1. Initial amount in the sample (at t = 0 years).
2. The amount remaining after 80 years.
2. Finding the Initial Amount:
- The initial amount of cesium-137 is represented by [tex]\( A(0) \)[/tex]. Substitute [tex]\( t = 0 \)[/tex] into the equation:
[tex]\[
A(0) = 458 \left(\frac{1}{2}\right)^{\frac{0}{30}} = 458 \left(\frac{1}{2}\right)^0
\][/tex]
- Any number raised to the power of 0 is 1:
[tex]\[
A(0) = 458 \times 1 = 458 \text{ grams}
\][/tex]
- Therefore, the initial amount of cesium-137 in the sample is 458 grams.
3. Finding the Amount After 80 Years:
- To find the amount of cesium-137 remaining after 80 years, we need to evaluate [tex]\( A(80) \)[/tex]. Substitute [tex]\( t = 80 \)[/tex] into the equation:
[tex]\[
A(80) = 458 \left(\frac{1}{2}\right)^{\frac{80}{30}}
\][/tex]
- First, calculate the exponent:
[tex]\[
\frac{80}{30} \approx 2.6667
\][/tex]
- Then, compute the powers of 0.5:
[tex]\[
A(80) = 458 \left(\frac{1}{2}\right)^{2.6667}
\][/tex]
- Taking the power of [tex]\(\left(\frac{1}{2}\right)^{2.6667}\)[/tex] gives us a value which, when multiplied by 458 grams, gives an approximate value:
[tex]\[
A(80) \approx 72 \text{ grams}
\][/tex]
- Therefore, the amount of cesium-137 remaining after 80 years is approximately 72 grams.
The results are:
- Initial amount: 458 grams
- Amount after 80 years: 72 grams
These values are rounded to the nearest gram as requested.
