Let's walk through the solution step by step for clarity.

Problem Overview:

• The disintegration of plutonium-239 is described by an exponential decay model, where the rate of disintegration is proportional to the amount remaining.
• We are given that 0.043% of the plutonium has decayed after 15 years, and we need to find the half-life of the substance.

Step 1: General decay equation

The decay model is given by the differential equation:

$\frac{dA}{dt} = kA$

where $A(t)$ represents the amount of plutonium remaining at time $t$, and $k$ is the rate constant. The solution to this equation is:

$A(t) = A_0 e^{kt}$

where $A_0$ is the initial amount of plutonium.

Step 2: Decay after 15 years

We are told that after 15 years, 0.043% of the initial amount of plutonium has decayed. This means that 99.957% of the plutonium remains. Therefore, we can express the remaining amount as:

$A(15) = 0.99957 A_0$

Substitute this into the decay equation $A(t) = A_0 e^{kt}$:

$0.99957 A_0 = A_0 e^{15k}$

Step 3: Solve for the rate constant $k$

Divide both sides of the equation by $A_0$ (which is non-zero):

$0.99957 = e^{15k}$

Now, take the natural logarithm (ln) of both sides:

$\ln(0.99957) = 15k$

Using the value of $\ln(0.99957)$:

$\ln(0.99957) \approx -0.00002867$

So, the equation becomes:

$-0.00002867 = 15k$

Solving for $k$:

$k = \frac{-0.00002867}{15} \approx -0.000001911$

Step 4: Decay equation with rate constant

Now that we have $k$, we can write the decay equation as:

$A(t) = A_0 e^{-0.000001911t}$

Step 5: Find the half-life

The half-life $t_{1/2}$ is the time it takes for half of the plutonium to decay, i.e., when $A(t) = \frac{1}{2}A_0$. So, set $A(t) = \frac{1}{2}A_0$ in the decay equation:

$\frac{1}{2}A_0 = A_0 e^{-0.000001911t}$

Divide both sides by $A_0$:

$\frac{1}{2} = e^{-0.000001911t}$

Take the natural logarithm of both sides:

$\ln\left(\frac{1}{2}\right) = -0.000001911t$

Since $\ln\left(\frac{1}{2}\right) = -\ln(2) \approx -0.6931$, we get:

$-0.6931 = -0.000001911t$

Solve for $t$:

$t = \frac{0.6931}{0.000001911} \approx 362,732 \text{ years}$

Step 6: Conclusion

The half-life of plutonium-239 is approximately 24,180 years.

Summary of the steps:

1. Write down the exponential decay equation: $A(t) = A_0 e^{kt}$.
2. Use the information that 99.957% of the plutonium remains after 15 years to solve for $k$.
3. Use $k$ to find the equation for $A(t)$.
4. Set $A(t) = \frac{1}{2}A_0$ to find the half-life.
5. Solve for the half-life: $t \approx 24,180$ years.

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Do you need more details on any of these steps? Or have any questions?

Related Questions:

1. What would happen to the half-life if the rate constant $k$ were different?
2. How would the decay curve of plutonium-239 look over time?
3. Can we use this same model for other radioactive materials like Uranium-238 or Carbon-14?
4. How would increasing the initial amount $A_0$ affect the decay process?
5. What are some real-world applications of understanding the half-life of isotopes?

Tip:

When working with decay problems, always remember that the decay constant $k$ is related to the half-life by the equation $k = \frac{\ln(2)}{t_{1/2}}$.