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Chapter 21: Problem 44

The half-life for the process \(^{238} \mathrm{U} \longrightarrow^{206} \mathrm{Pb}\) is \(4.5 \times 10^{9} \mathrm{yr}.\) A mineral sample contains 75.0 \(\mathrm{mg}\) of \(^{238} \mathrm{U}\) and 18.0 \(\mathrm{mg}\) of \(^{206} \mathrm{pb} .\) What is the age of the minineral?

Short Answer

Expert verified
The age of the mineral sample containing ^238U and ^206Pb is approximately \(1.959 \times 10^9\ \text{years}\).

Step by step solution

01

Understand the Half-Life Formula

The formula to calculate the number of half-lives that have occurred for a radioactive process is: \[N(t) = N_{0} \times (1/2)^{t/T}\] Where: - \(N(t)\) is the remaining amount of the substance at time \(t\) - \(N_{0}\) is the initial amount of the substance - \(t\) is the time elapsed - \(T\) is the half-life of the process
02

Calculate the Amount of Original ^238U

We know that at time t, the remaining ^238U in the sample is 75.0 mg, and the amount of ^206Pb is 18.0 mg. Since ^238U decays into ^206Pb, we can find the original amount of ^238U by summing the remaining ^238U and the amount of ^206Pb produced. \[N_0 = 75.0 + 18.0 = 93.0\ \text{mg}\]
03

Rewrite the Half-Life Formula for Time

Now we need to solve for the time \(t\), so we must rearrange the half-life formula to isolate \(t\). \[\frac{N(t)}{N_0} = (1/2)^{t/T}\] Taking the logarithm of both sides will eliminate the exponent and give us the age of the mineral sample: \[t = T \times \frac{\log(\frac{N(t)}{N_0})}{\log(1/2)}\]
04

Substitute Known Values and Calculate Time

Now, substitute the known values into the equation: \[t = 4.5 \times 10^{9}\ \text{yr} \times \frac{\log(\frac{75}{93})}{\log(1/2)}\] Using a calculator, we can find the age of the mineral: \[t \approx 1.959 \times 10^9\ \text{yr}\]
05

Present the Answer

The age of the mineral sample containing ^238U and ^206Pb is approximately \(1.959 \times 10^9\ \text{years}\).

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Most popular questions from this chapter

One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a})_{20}^{40} \mathrm{Ca}\) and \(_{20}^{45} \mathrm{Ca}\) , \((\mathbf{b})^{12} \mathrm{C}\) and \(^{14} \mathrm{C},\) \((\mathbf{c})\) lead-206 and thorium-230. Explain your choice in each case.

Which type or types of nuclear reactors have these characteristics? \(\begin{array}{l}{\text { (a) Does not use a secondary coolant }} \\ {\text { (b) Creates more fissionable material than it consumes }} \\ {\text { (c) Uses a gas, such as He or } \mathrm{CO}_{2}, \text { as the primary coolant }}\end{array}\)

The spent fuel elements from a fission reactor are much more intensely radioactive than the original fuel elements. (a) What does this tell you about the products of the fission process in relationship to the belt of stability, Figure 21.2? (b) Given that only two or three neutrons are released per fission event and knowing that the nucleus undergoing fission has a neutron- to-proton ratio characteristic of a heavy nucleus, what sorts of decay would you expect to be dominant among the fission products?

The atomic masses of hydrogen-2 (deuterium), helium-4, and lithium-6 are 2.014102 amu, 4.002602 amu, and 6.0151228 amu, respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon. (d) Which of these three isotopes has the largest nuclear binding energy per nucleon? Does this agree with the trends plotted in Figure 21.12\(?\)

Which of the following statements best explains why alpha emission is relatively common, but proton emission is extremely rare? \begin{equation}\begin{array}{l}{\text { (a) Alpha particles are very stable because of magic numbers }} \\ \quad {\text { of protons and neutrons. }} \\\ {\text { (b) Alpha particles occur in the nucleus. }} \\ {\text { (c) Alpha particles are the nuclei of an inert gas. }} \\ {\text { (d) An alpha particle has a higher charge than a proton. }}\end{array}\end{equation}
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