Chapter 19: 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: \({ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{18}^{40} \mathrm{Ar}(10.7 \%) \quad t_{1 / 2}=1.27 \times 10^{9}\) years \({ }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{-1}^{0} \mathrm{e}(89.3 \%)\) 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 \({ }^{40} \mathrm{Ar}\) escaped from the sample?
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