Chapter 18: Problem 40
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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