The mass ratios of \({ }^{40}\) Ar to \({ }^{40} \mathrm{~K}\) also can be used to date geologic materials. Potassium- 40 decays by two processes: $$\begin{array}{l}{ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{18}^{40} \mathrm{Ar}(10.7 \%) \quad t_{1 / 2}=1.27 \times 10^{9} \text { years } \\\\{ }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{-1}^{0} \mathrm{e}(89.3 \%) & \end{array}$$ 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?

The reason \({ }^{40} \mathrm{Ar}/{ }^{40}\mathrm{~K}\) ratios are used instead of \({ }^{40} \mathrm{Ca}/{ }^{40} \mathrm{~K}\) ratios is because argon is a noble gas and does not react chemically with other elements in the rock. This means that any \({ }^{40}\mathrm{Ar}\) present in the rock is most likely a result of radioactive decay of \({ }^{40}\mathrm{~K}\). On the other hand, calcium has higher reactivity and can be introduced or removed from the rock by multiple processes making its ratio with potassium not as reliable in dating the materials. #b. What assumptions must be made using this technique?#

When using the potassium-argon dating technique, we make the following assumptions: 1. The rock had negligible initial argon-40 at the time of its formation. 2. The decay rate and half-life of potassium-40 have remained constant over time. 3. There have been no gains or losses of potassium-40 and argon-40 due to environmental factors or other chemical reactions. #c. A sedimentary rock has an \({ }^{40} \mathrm{Ar} /{ }^{40} \mathrm{~K}\) ratio of \(0.95\). Calculate the age of the rock.#

To calculate the age of the rock, we will use the decay equation: \(N(t) = N_0 e^{-\lambda t}\), where \(N(t)\) is the amount of the remaining radioactive element at time \(t\), \(N_0\) is the initial amount of the element, \(\lambda\) is the decay constant and \(t\) is the time. According to the given decay process, 10.7% of potassium-40 decays into argon-40, so we need to adjust our equations accordingly: \(N_{\mathrm{Ar}} = 0.95 N_{\mathrm{K}}\), \(N_{\mathrm{K}} = N_0 e^{-\lambda t}\), and \(N_{\mathrm{Ar}} = 0.107\cdot N_0(1 - e^{-\lambda t})\). Substitute the first equation into the third equation: \(0.95\cdot N_{\mathrm{K}} = 0.107\cdot N_0(1 - e^{-\lambda t})\). Next, substitute the second equation into this new equation: \(0.95 e^{-\lambda t} = 0.107(1 - e^{-\lambda t})\). Now, solve for \(t\): \(e^{-\lambda t} = \frac{0.107}{0.107 + 0.95}\) \(-\lambda t = \ln\left(\frac{0.107}{1.057}\right)\) \(t = -\frac{1}{\lambda}\ln\left(\frac{0.107}{1.057}\right)\) The decay constant \(\lambda\) is related to the half-life \(t_{1/2}\) by \(\lambda = \frac{\ln(2)}{t_{1/2}}\), and we are given that \(t_{1/2} = 1.27\times10^9\) years. We can now find the age of the rock: \(t = -\frac{1.27\times10^9 \,\text{years}}{\ln(2)}\ln\left(\frac{0.107}{1.057}\right)\) \(t \approx 1.14\times10^9\) years. #d. How will the measured age of a rock compare to the actual age if some \({ }^{40}\) Ar escaped from the sample?#

If some \({ }^{40}\mathrm{Ar}\) escaped from the sample, the measured ratio of \({ }^{40}\mathrm{Ar}/{ }^{40}\mathrm{~K}\) would be lower than it should be. This would lead to an underestimation of the rock's age because it would suggest that less \({ }^{40}\mathrm{~K}\) had decayed into \({ }^{40}\mathrm{Ar}\) than it actually did.