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Chapter 18: Problem 79

Naturally occurring uranium is composed mostly of \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U},\) with relative abundances of \(99.28 \%\) and \(0.72 \%,\) respectively. The half-life for \(^{238} \mathrm{U}\) is \(4.5 \times 10^{9}\) years, and the half-life for \(^{235} \mathrm{U}\) is \(7.1 \times 10^{8}\) years. Assuming that the earth was formed 4.5 billion years ago, calculate the relative abundances of the \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U}\) isotopes when the earth was formed.

Short Answer

Expert verified
The initial relative abundances of \(^{238}\mathrm{U}\) and \(^{235}\mathrm{U}\) isotopes when the earth was formed were approximately \(RA_{238} \approx 98.12\%\) and \(RA_{235} \approx 1.88\%\).

Step by step solution

01

Write the half-life formula

We will use the half-life formula to determine the initial amount of each isotope when the earth was formed. The formula is given by: \(N_t = N_0 e^{(-\lambda t)}\) Where \(N_t\) is the amount remaining after time \(t\), \(N_0\) is the initial amount, \(\lambda\) is the decay constant, and \(t\) is the time elapsed.
02

Find the decay constants for both isotopes

The decay constant can be found using the formula: \(\lambda = \frac{\ln{2}}{t_{1/2}}\) Where \(t_{1/2}\) is the half-life of the isotope. For \(^{238}\mathrm{U}\): \(\lambda_{238} = \frac{\ln{2}}{4.5 \times 10^9 \; \text{years}}\) For \(^{235}\mathrm{U}\): \(\lambda_{235} = \frac{\ln{2}}{7.1 \times 10^8 \; \text{years}}\)
03

Calculate the initial amounts of both isotopes

We want to find the initial amounts of both isotopes, so we rewrite the half-life formula for \(N_0\). The current abundances being 99.28% and 0.72% represent \(N_t\). Let \(N_{0_{238}}\) be the initial amount of \(^{238}\mathrm{U}\), and \(N_{0_{235}}\) be the initial amount of \(^{235}\mathrm{U}\). For \(^{238}\mathrm{U}\): \(0.9928 = N_{0_{238}}e^{(-\lambda_{238} \times 4.5 \times 10^9)}\) For \(^{235}\mathrm{U}\): \(0.0072 = N_{0_{235}}e^{(-\lambda_{235} \times 4.5 \times 10^9)}\) Solve for \(N_{0_{238}}\) and \(N_{0_{235}}\) using the decay constants from Step 2.
04

Calculate the initial relative abundances

With the initial amounts of both isotopes, we can now calculate their initial relative abundances. Let \(RA_{238}\) be the initial relative abundance of \(^{238}\mathrm{U}\), and \(RA_{235}\) be the initial relative abundance of \(^{235}\mathrm{U}\). \(RA_{238} = \frac{N_{0_{238}}}{N_{0_{238}} + N_{0_{235}}} \times 100\) \(RA_{235} = \frac{N_{0_{235}}}{N_{0_{238}} + N_{0_{235}}} \times 100\) Calculate \(RA_{238}\) and \(RA_{235}\) using the initial amounts from Step 3. These will be the initial relative abundances of \(^{238}\mathrm{U}\) and \(^{235}\mathrm{U}\) isotopes when the earth was formed.

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Most popular questions from this chapter

The mass ratios of \(^{40} \mathrm{Ar}\) to \(^{40} \mathrm{K}\) also can be used to date geologic materials. Potassium-40 decays by two processes: $$\begin{aligned} &_{19}^{40} \mathrm{K}+_{-1}^{0} \mathrm{e} \longrightarrow_{18}^{40} \mathrm{Ar}(10.7 \%) \quad t_{1 / 2}=1.27 \times 10^{9} \text { years }\\\ &_{19}^{40} \mathrm{K} \longrightarrow_{20}^{40} \mathrm{Ca}+_{-1}^{0} \mathrm{e}(89.3 \%) \end{aligned}$$a. Why are \(^{40} \mathrm{Ar} /^{40} \mathrm{K}\) ratios used to date materials rather than \(^{40} \mathrm{Ca} /^{40} \mathrm{K}\) ratios? b. What assumptions must be made using this technique? c. A sedimentary rock has an \(^{40} \mathrm{Ar} /^{40} \mathrm{K}\) ratio of 0.95. Calculate the age of the rock. d. How will the measured age of a rock compare to the actual age if some \(^{40}\) Ar escaped from the sample?

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