To solve this problem, we will use the formula for exponential decay, which is:

$A(t) = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$

Where:

• $A(t)$ is the amount of plutonium remaining after time $t$,
• $A_0$ is the initial amount of plutonium,
• $T_{1/2}$ is the half-life of plutonium (24,360 years in this case),
• $t$ is the time that has passed.

Step 1: Find the remaining amount of plutonium

Since 81% of the plutonium decays, only 19% remains. Therefore, $A(t)$ will be 19% of the initial amount, so:

$A(t) = 0.19 \times A_0$

Step 2: Set up the equation

Using the decay formula:

$0.19 A_0 = A_0 \left( \frac{1}{2} \right)^{\frac{t}{24,360}}$

We can cancel out $A_0$ from both sides:

$0.19 = \left( \frac{1}{2} \right)^{\frac{t}{24,360}}$

Step 3: Solve for $t$

To isolate $t$, take the natural logarithm (ln) of both sides:

$\ln(0.19) = \frac{t}{24,360} \ln \left( \frac{1}{2} \right)$

Since $\ln \left( \frac{1}{2} \right) = -\ln(2)$, we get:

$\ln(0.19) = \frac{t}{24,360} (-\ln(2))$

Now solve for $t$:

$t = \frac{\ln(0.19)}{-\ln(2)} \times 24,360$

Step 4: Calculate the result

Using approximate values for the natural logarithms:

• $\ln(0.19) \approx -1.6605$
• $\ln(2) \approx 0.6931$

Substitute these into the equation:

$t = \frac{-1.6605}{-0.6931} \times 24,360$

$t \approx 2.396 \times 24,360 \approx 58,310 \text{ years}$

Final Answer:

It will take approximately 58,310 years for 81% of the plutonium-239 to decay.

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Would you like more details on any step, or do you have further questions on this topic?

Follow-up questions:

1. How does the half-life formula work in general?
2. Can the same formula be used for other isotopes with different half-lives?
3. What other factors influence radioactive decay besides time and half-life?
4. How do we calculate the time it takes for a certain percentage of an isotope to remain instead of decay?
5. Why is the natural logarithm used in decay equations?

Tip: Always double-check the units when working with time-based decay problems to ensure consistency!