To solve this, we need to use the formula for exponential decay, which is based on the concept of half-life:

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

Where:

• $A$ is the amount remaining after time $t$,
• $A_0$ is the initial amount,
• $T_{1/2}$ is the half-life,
• $t$ is the time passed.

We are asked to find the time it takes for 70% of the plutonium-239 to decay. This means we want 30% of the original amount to remain, so:

$A = 0.30 \times A_0$

Given:

• $A_0 = 15 \, \text{g}$,
• $T_{1/2} = 24,360 \, \text{years}$.

Plug these values into the equation:

$0.30 \times 15 = 15 \times \left( \frac{1}{2} \right)^{\frac{t}{24,360}}$

Simplify:

$0.30 = \left( \frac{1}{2} \right)^{\frac{t}{24,360}}$

Now, take the natural logarithm of both sides:

$\ln(0.30) = \ln \left( \left( \frac{1}{2} \right)^{\frac{t}{24,360}} \right)$

Using the logarithmic property $\ln(a^b) = b \ln(a)$, we get:

$\ln(0.30) = \frac{t}{24,360} \times \ln \left( \frac{1}{2} \right)$

Since $\ln \left( \frac{1}{2} \right) \approx -0.6931$, we substitute this in:

$\ln(0.30) = \frac{t}{24,360} \times (-0.6931)$

$-1.2040 = \frac{t}{24,360} \times (-0.6931)$

Solve for $t$:

$t = \frac{-1.2040 \times 24,360}{-0.6931}$

$t \approx 41,968 \, \text{years}$

So, it will take approximately 41,968 years for 70% of the plutonium-239 to decay.

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Would you like any further clarification or more details on this process?

Related questions:

1. How would the result change if the half-life was longer?
2. What would happen if only 50% of the isotope decayed?
3. How do you calculate the remaining amount of a substance after a certain time has passed?
4. What is the concept of half-life in radioactive decay?
5. How do exponential decay functions relate to real-world phenomena like drug metabolism or population decline?

Tip: When using logarithms to solve exponential decay problems, always remember that the natural logarithm ($\ln$) is often the easiest tool to apply, especially when dealing with base-2 decay models.