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?

Step by step solution

01

a. Why Argon-40/Potassium-40 ratios are used to date materials instead of Calcium-40/Potassium-40 ratios

Argon-40 is a noble gas, and it does not bond with other elements, allowing it to accumulate after the decay of Potassium-40. This allows the Argon-40 to be easily distinguished and measured compared to the materials surrounding it. Calcium-40, on the other hand, is a common element in the Earth's crust and can easily bond with other elements. This makes it harder to determine which calcium ions resulted from Potassium-40 decay, and therefore the Calcium-40/Potassium-40 ratio would not provide accurate dating information.

02

b. Assumptions made using this technique

Some important assumptions made while using the Argon-40/Potassium-40 dating technique are: 1. The initial amount of Argon-40 was zero when the rock sample formed. 2. The amount of Argon-40 in the sample has only increased due to the decay of Potassium-40. 3. No Argon-40 has been lost or added to the rock sample since its formation. 4. The decay rates of Potassium-40 remained constant over time.

03

c. Calculating the age of the rock

We are given the Argon-40/Potassium-40 ratio as 0.95. Let's denote the amount of Argon-40 as A and the amount of Potassium-40 as K, so A/K = 0.95. To calculate the age of the rock, we will use the decay equation \( t = \frac{1}{\lambda} * \ln(1 + \frac{A_{40}}{K_{40}}) \), where \(t\) is the age of the rock, and \(\lambda\) is the decay constant related to half-life by the equation: \(\lambda = \frac{\ln(2)}{t_{1/2}}\). We have half-life \(t_{1/2} = 1.27 \times 10^9\) years. Now, we will calculate the decay constant, \(\lambda\). \(\lambda = \frac{\ln(2)}{1.27 \times 10^9} \approx 5.46 \times 10^{-10}\mathrm{~} \text{year}^{-1}\) Now we can calculate the age of the rock using the Argon-40/Potassium-40 ratio: \(t = \frac{1}{5.46 \times 10^{-10}} * \ln(1 + 0.95) = 4.01 \times 10^8\) years. The age of the rock is approximately 401 million years.

04

d. Comparing the measured age of a rock with its actual age if some Argon-40 escaped from the sample

If some Argon-40 escaped from the sample, the measured Argon-40/Potassium-40 ratio would be smaller than the actual ratio. As a result, the calculated age of the rock using the above equation would be less than the real age. The lower Argon-40-to-Potassium-40 ratio would lead to a younger estimated age, as it implies less Potassium-40 decayed into Argon-40 than what actually occurred.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!