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Chapter 18: Problem 64

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 \(285,119\) years for the activity of plutonium-239 to decrease to \(0.1\%\) of its original value.

Step by step solution

01

Recall the radioactive decay formula

The decay of a radioactive substance can be modeled by the equation: \[A(t) = A_0 e^{-\lambda t}\] Where: \(A(t)\) is the activity of the substance at time t, \(A_0\) is the initial activity of the substance, \(\lambda\) is the decay constant, and \(t\) is the time elapsed.
02

Convert half-life to decay constant

We are given the half-life of plutonium-239, which is 24,100 years. We can convert this to the decay constant using the formula: \[\lambda = \frac{\ln 2}{t_{1/2}}\] Where: \(t_{1/2}\) is the half-life of the substance. Substitute the given half-life value: \[\lambda = \frac{\ln 2}{24,100}\]
03

Set up the equation for the given problem

We know that the suitable storage place must be geologically stable long enough for the activity of plutonium-239 to decrease to 0.1% of its original value. In terms of the decay equation, this means: \[A(t) = 0.001 A_0\] Now, we substitute the decay equation and the decay constant into this equation: \[0.001 A_0 = A_0 e^{-\frac{\ln 2}{24,100} t}\]
04

Solve for the time t

Now we need to solve for \(t\). First, divide both sides of the equation by \(A_0\): \[0.001 = e^{-\frac{\ln 2}{24,100} t}\] Next, take the natural logarithm of both sides: \[\ln 0.001 = -\frac{\ln 2}{24,100} t\] Now, solve for \(t\): \[t = \frac{24,100\ln 0.001}{-\ln 2}\] Finally, calculate the value for \(t\): \[t \approx 285,119\, \text{years}\] Therefore, it would take approximately 285,119 years for the activity of plutonium-239 to decrease to 0.1% of its original value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plutonium-239
Plutonium-239 is a radioactive isotope commonly encountered in nuclear science. It's one of the dangerous products found in nuclear waste due to its high radiotoxicity and long half-life. Plutonium-239 is mainly used in nuclear reactors and weapons. It鈥檚 primarily created in breeder reactors, which convert uranium-238 into plutonium-239 through neutron absorption.
This isotope is notorious for its potential to remain hazardous for thousands of years, which poses significant storage challenges for nuclear waste management. Understanding its properties is crucial in dealing with long-term storage solutions so that we can safely contain it and prevent contamination of the environment.
Half-life
The concept of half-life is central to understanding radioactive decay. A half-life is the time required for half of the radioactive atoms in a sample to decay. For plutonium-239, the half-life is a daunting 24,100 years.
This long half-life means that plutonium-239 remains active and hazardous for many generations, requiring careful consideration in storage and containment. To manage nuclear wastes effectively, we need to understand how long plutonium-239 remains dangerous. By calculating its half-life, we can estimate how long these materials need to be securely stored before reaching safer levels of radiation.
Decay constant
The decay constant (\( \lambda \)) represents the probability of a single atom decaying per unit time. You can find this by using the relationship with half-life:
  • The formula \( \lambda = \frac{\ln 2}{t_{1/2}} \)
  • For plutonium-239, \( \lambda = \frac{\ln 2}{24,100} \)
The decay constant helps us model the radioactive decay process quantitatively. Knowing this, we can calculate how quickly or slowly the sample decays over time. For long-lived isotopes like plutonium-239, this calculation becomes vital for determining necessary precautions and storage durations to ensure safety.
Nuclear waste storage
Nuclear waste storage involves securing radioactive materials in a way that they pose no threat to current and future generations. Plutonium-239, due to its long half-life, presents a unique challenge that requires meticulous planning.
Suitable storage options include deep geological repositories or specially designed storage facilities that prevent the release of radiation. Geological stability is critical; the sites must remain undisturbed for the thousands of years it takes for substances like plutonium-239 to decay to safe levels. Proper storage solutions take into account potential natural disasters, human interference, and other factors that could jeopardize the safety of the stored materials.
By understanding the science behind radioactive decay and the specific characteristics of plutonium-239, we can devise strategies to protect both the environment and public health.

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Most popular questions from this chapter

The rate constant for a certain radioactive nuclide is \(1.0 \times\) \(10^{-3} \mathrm{h}^{-1} .\) What is the half-life of this nuclide?

Natural uranium is mostly nonfissionable \(^{238} \mathrm{U} ;\) it contains only about \(0.7 \%\) of fissionable \(^{235}\) U. For uranium to be useful as a nuclear fuel, the relative amount of \(^{235}\) U must be increased to about \(3 \% .\) This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of \(^{238} \mathrm{UF}_{6}(g)\) and \(^{235} \mathrm{UF}_{6}(g) .\) The fluoride mixture is then enriched through a multistage diffusion process to produce a \(3 \%^{235} \mathrm{U}\) nuclear fuel. The diffusion process utilizes Graham's law of effusion (see Chapter \(8,\) Section \(8-7\) ). Explain how Graham's law of effusion allows natural uranium to be enriched by the gaseous diffusion process.

When nuclei undergo nuclear transformations, \(\gamma\) rays of characteristic frequencies are observed. How does this fact, along with other information in the chapter on nuclear stability, suggest that a quantum mechanical model may apply to the nucleus?

The most significant source of natural radiation is radon-222. \(^{222} \mathrm{Rn},\) a decay product of \(^{238} \mathrm{U},\) is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of 3.82 days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \(^{238} \mathrm{U}\) decays to \(^{222} \mathrm{Rn} ?\) What nuclei are produced when \(^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \(^{222} \mathrm{Rn} ?\) c. Another problem associated with \(^{222} \mathrm{Rn}\) is that the decay of \(^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) 3.11 min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \(^{222}\) Rn levels not exceed 4 pCi per liter of air (1 \(\mathrm{Ci}=1\) curie \(=3.7 \times 10^{10}\) decay events per second; \(1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\). Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(^{222}\) Rn per liter of air.

Using the kinetic molecular theory (see Section 8.6), 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 50 for the appropriate mass values.)
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