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Chapter 19: Problem 72

A proposed system for storing nuclear wastes involves storing the radioactive material in caves or deep mine shafts. One of the most toxic nuclides that must be disposed of is plutonium-239, which is produced in breeder reactors and has a half-life of \(24,100\) years. A suitable storage place must be geologically stable long enough for the activity of plutonium- 239 to decrease to 0.1\(\%\) of its original value. How long is this for plutonium-239?

Short Answer

Expert verified
It would take approximately 180,609 years for the activity of plutonium-239 to decrease to 0.1% of its original value. Therefore, a geologically stable storage place must be capable of securing the nuclear waste for at least 180,609 years.

Step by step solution

01

Understand the concept of half-life

Half-life is the time required for a quantity to reduce to half of its initial value. In the context of nuclear decay, it means that half of the radioactive nuclei will have decayed after one half-life period. In this problem, we are given the half-life of plutonium-239 as 24,100 years.
02

Set up the decay equation

The general equation to describe radioactive decay is given by: \[A(t) = A_0 \cdot (1/2)^{\frac{t}{T}}\] Where: - \(A(t)\) is the activity of the radioactive substance at time \(t\) - \(A_0\) is the initial activity of the radioactive substance - \(T\) is the half-life of the radioactive substance - \(t\) is the time elapsed In our problem, we want to find the time \(t\) when the activity of plutonium-239 decreases to 0.1% (0.001) of its initial activity \(A_0\). Hence, \[0.001 A_0 = A_0 \cdot (1/2)^{\frac{t}{24100}}\]
03

Solve for the time \(t\)

To find the time \(t\) when the activity reaches 0.1% of its initial value, we divide both sides of the equation by \(A_0\) and solve for \(t\). We have: \[0.001 = (1/2)^{\frac{t}{24100}}\] Now, we can take the natural logarithm of both sides of the equation: \[\ln(0.001) = \ln((1/2)^{\frac{t}{24100}})\] Using the logarithm property \(\ln(a^b) = b \ln(a)\), we get: \[\ln(0.001) = \frac{t}{24100} \ln(1/2)\] Finally, we can solve for \(t\): \[t = \frac{24100 \cdot \ln(0.001)}{\ln(1/2)}\] An approximation gives: \[t \approx 180609 \text{ years}\]
04

Interpret the result

It would take approximately 180,609 years for the activity of plutonium-239 to decrease to 0.1% of its original value. Therefore, a geologically stable storage place must be capable of securing the nuclear waste for at least 180,609 years.

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