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Chapter 19: Problem 66

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 takes approximately 161,890 years for the activity of plutonium-239 to decrease to 0.1% of its original value.

Step by step solution

01

Understanding the radioactive decay formula

Radioactive decay follows an exponential decay pattern, which can be represented by the formula: \[N(t) = N_0 \cdot (1/2)^{t/T}\] where: - \(N(t)\) is the remaining activity at time \(t\) - \(N_0\) is the initial activity - \(t\) is the time (in years) - \(T\) is the half-life of the nuclide (24,100 years for plutonium-239)
02

Set up the equation

We want to find the time it takes for the activity of plutonium-239 to decrease to 0.1% of its initial value. So, we set up the equation as follows: \[0.001 N_0 = N_0 \cdot (1/2)^{t/24100}\]
03

Solve for t

Now, we need to solve for \(t\). First, divide by \(N_0\) on both sides of the equation to get: \[0.001 = (1/2)^{t/24100}\] Next, we will take the natural logarithm of both sides to remove the exponential term: \[\ln(0.001) = \ln((1/2)^{t/24100})\] Using the property of logarithms, we can rewrite the equation as: \[\ln(0.001) = \frac{t}{24100} \ln(1/2)\] Now, solve for \(t\): \[t = 24100 \cdot \frac{\ln(0.001)}{\ln(1/2)}\] Evaluate the expression to find the value of \(t\): \[t \approx 161889.63\]
04

Conclusion

It takes approximately 161,890 years for the activity of plutonium-239 to decrease to 0.1% of its original value.

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Most popular questions from this chapter

Using the kinetic molecular theory (Section 5.6), calculate the root mean square velocity and the average kinetic energy of \({ }_{1}^{2} \mathrm{H}\) nuclei at a temperature of \(4 \times 10^{7} \mathrm{~K}\). (See Exercise 46 for the appropriate mass values.)

Americium- 241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \({ }^{241}\) Am is 433 years, and it decays by emitting alpha particles. How many alpha particles are emitted each second by a \(5.00-\mathrm{g}\) sample of \({ }^{241} \mathrm{Am} ?\)

Fresh rainwater or surface water contains enough tritium \(\left({ }^{3} \mathrm{H}\right)\) to show \(5.5\) decay events per minute per \(100 . \mathrm{g}\) water. Tritium has a half-life of \(12.3\) years. You are asked to check a vintage wine that is claimed to have been produced in \(1946 .\) How many decay events per minute should you expect to observe in \(100 . \mathrm{g}\) of that wine?

One type of commercial smoke detector contains a minute amount of radioactive americium- \(241\left({ }^{241} \mathrm{Am}\right)\), which decays by \(\alpha\) -particle production. The \(\alpha\) particles ionize molecules in the air, allowing it to conduct an electric current. When smoke particles enter, the conductivity of the air is changed and the alarm buzzes. a. Write the equation for the decay of \({ }_{95}^{241} \mathrm{Am}\) by \(\alpha\) -particle production. b. The complete decay of \({ }^{241}\) Am involves successively \(\alpha, \alpha, \beta\), \(\alpha, \alpha, \beta, \alpha, \alpha, \alpha, \beta, \alpha\), and \(\beta\) production. What is the final stable nucleus produced in this decay series? c. Identify the 11 intermediate nuclides.

The mass percent of carbon in a typical human is \(18 \%\), and the mass percent of \({ }^{14} \mathrm{C}\) in natural carbon is \(1.6 \times 10^{-10} \%\). Assuming a \(180-\mathrm{lb}\) person, how many decay events per second occur in this person due exclusively to the \(\beta\) -particle decay of \({ }^{14} \mathrm{C}\) (for \({ }^{14} \mathrm{C}\), \(t_{1 / 2}=5730\) years)?
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