To solve this problem, we need to determine how long it will take for 36 grams of plutonium-240 to decay to 12 grams, given the decay constant [tex]\(k = 0.00011\)[/tex]. The decay follows the exponential decay function [tex]\(Q(t) = Q_0 e^{-kt}\)[/tex], where [tex]\(Q_0\)[/tex] is the initial quantity and [tex]\(Q(t)\)[/tex] is the quantity remaining after [tex]\(t\)[/tex] years.
Here are the steps to find the time [tex]\(t\)[/tex]:
1. Initial and Final Quantities:
- Initial quantity, [tex]\(Q_0 = 36\)[/tex] grams.
- Final quantity, [tex]\(Q(t) = 12\)[/tex] grams.
2. Set Up the Equation:
Plugging in the values into the decay formula:
[tex]\[
12 = 36 \cdot e^{-0.00011t}
\][/tex]
3. Isolate the Exponential Term:
Divide both sides by 36 to simplify:
[tex]\[
\frac{12}{36} = e^{-0.00011t}
\][/tex]
[tex]\[
\frac{1}{3} = e^{-0.00011t}
\][/tex]
4. Apply the Natural Logarithm:
To solve for [tex]\(t\)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[
\ln\left(\frac{1}{3}\right) = \ln\left(e^{-0.00011t}\right)
\][/tex]
Using the property of logarithms [tex]\(\ln(e^x) = x\)[/tex], we get:
[tex]\[
\ln\left(\frac{1}{3}\right) = -0.00011t
\][/tex]
5. Solve for [tex]\(t\)[/tex]:
Divide both sides by [tex]\(-0.00011\)[/tex]:
[tex]\[
t = \frac{\ln\left(\frac{1}{3}\right)}{-0.00011}
\][/tex]
6. Calculate the Time:
[tex]\[
t \approx 9987.384442437362 \text{ years}
\][/tex]
7. Round to the Nearest 10 Years:
The final result rounded to the nearest 10 years is approximately:
[tex]\[
t_{\text{rounded}} \approx 9990 \text{ years}
\][/tex]
Hence, the answer is B. 9,990 years.
