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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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Most popular questions from this chapter

A recent study concluded that any amount of radiation exposure can cause biological damage. Explain the differences between the two models of radiation damage, the linear model and the threshold model.

The earth receives \(1.8 \times 10^{14} \mathrm{kJ} / \mathrm{s}\) of solar energy. What mass of solar material is converted to energy over a 24 -h period to provide the daily amount of solar energy to the earth? What mass of coal would have to be burned to provide the same amount of energy? (Coal releases \(32 \mathrm{kJ}\) of energy per gram when burned.)

Consider the following information: i. The layer of dead skin on our bodies is sufficient to protect us from most \(\alpha\) -particle radiation. ii. Plutonium is an \(\alpha\) -particle producer. iii. The chemistry of \(\mathrm{Pu}^{4+}\) is similar to that of \(\mathrm{Fe}^{3+}\). iv. Pu oxidizes readily to \(\mathrm{Pu}^{4+}\) Why is plutonium one of the most toxic substances known?

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