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: $$_{19}^{40} \mathrm{K}+_{-1}^{0} \mathrm{e} \longrightarrow_{\mathrm{i} 8}^{40} \mathrm{Ar}(10.7 \%)$$ $$_{19}^{40} \mathrm{K} \longrightarrow_{20}^{40} \mathrm{Ca}+_{-1}^{0} \mathrm{e}(89.3 \%)$$ $$t_{1 / 2}=1.27 \times 10^{9}$$ 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 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?

In order to date geological materials, we need a stable and reliable end product of decay that can be isolated and measured from the original material to estimate the age. In this case, choosing the Argon-40 to Potassium-40 ratio is preferred because Argon is a noble gas, and it can be easily trapped in the rock as it gets formed. Contrary to this, Calcium, being a reactive solid element, is a part of several minerals and forms various compounds, making it difficult to isolate and measure the Calcium-40 produced from Potassium-40 decay. Therefore, the Argon to Potassium ratio is more reliable and suitable for dating materials.

Using the Argon-40/ Potassium-40 ratio in dating materials involves a few major assumptions: 1. No Argon-40 was initially trapped in the rock when it was formed. All Argon-40 found in the rock now has been produced only by the decay of Potassium-40. 2. The decay rate of Potassium-40 remains constant over time. 3. No Argon-40 has been lost from the rock since it was formed until now. If any of these assumptions are not met, the calculation of the age of the rock may result in inaccuracies.

Given the Argon-40 to Potassium-40 ratio, we can calculate the age of the rock using the following equation: \[ N = N_0e^{\lambda t} \] Where \( N \) is the amount of remaining Potassium-40, \( N_0 \) is the initial amount of Potassium-40, \( \lambda \) is the decay constant, and \( t \) is the age of the rock. Since we are given the ratio, we can write: \[ \frac{\mathrm{Ar}^{40}}{\mathrm{K}^{40}} = \frac{N_0 - N}{N} = \frac{0.95}{1} \] Now, we can find N using this relation: \[ N = \frac{N_0}{1.95} \] We can now apply the decay equation to find the age of the rock: \[ \frac{1}{1.95} = e^{\lambda t} \] The decay constant (\( \lambda \)) can be calculated using the half-life given (\( t_{1/2} = 1.27 \times 10^9 \)): \[ \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{1.27*10^9} \] Now, we can solve the equation for \( t \): \[ t = \frac{ln(1.95)}{\lambda} = \frac{ln(1.95)}{\frac{ln(2)}{1.27*10^9}} \] Calculating the value for \(t\): \[ t \approx 3.06 \times 10^8 \] The age of the rock is approximately \( 3.06 \times 10^8 \) years.

If some Argon-40 has escaped from the rock sample, the measured Argon-40 to Potassium-40 ratio will be lower than the actual ratio. As a result, the calculated age will be less than the actual age of the rock, since a lower ratio suggests that less Potassium-40 has decayed into Argon-40. The actual age of the rock will always be more than the measured age if some Argon-40 has escaped from the sample.