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Chapter 24: Problem 48

A volcanic eruption melts a large chunk of rock, and all gases are expelled. After cooling, \({ }_{18}^{40} \mathrm{Ar}\) accumulates from the ongoing decay of \({ }_{19}^{40} \mathrm{~K}\) in the rock \(\left(t_{1 / 2}=1.25 \times 10^{9} \mathrm{yr}\right) .\) When a piece of rock is analyzed, it is found to contain 1.38 mmol of \({ }^{40} \mathrm{~K}\) and \(1.14 \mathrm{mmol}\) of \({ }^{40} \mathrm{Ar}\). How long ago did the rock cool?

Short Answer

Expert verified
The rock is approximately 1.084 billion years old.

Step by step solution

01

- Determine the Initial Amount of \({ }^{40} \text{K}\)

The initial amount of \({ }^{40} \text{K}\) present when the rock cooled included both the decayed \({ }^{40} \text{K}\) and the existing \({ }^{40} \text{K}\). This can be calculated as \(\text{N}_{0} = \text{N}_{\text{now}} + \text{N}_{\text{decayed}}\). The \({ }^{40} \text{Ar}\) present is equal to the \({ }^{40} \text{K}\) that decayed: \(\text{N}_{\text{0}} = 1.38 \text{ mmol} + 1.14 \text{ mmol} = 2.52 \text{ mmol}\).
02

- Determine the Number of Half-Lives Passed

Using the formula for radioactive decay, \(\text{N} = \text{N}_{0} e^{-\frac{\text{t} \text{ln}(2)}{\text{t}_{1/2}}}\), we can determine the number of half-lives that passed. Here, \(\text{N}_{\text{now}} \text{ is the current amount (1.38 mmol)}\), \({ }\text{N}_{0} \text{ is the initial amount (2.52 mmol)}, \ t_{\frac{1}{2}} = 1.25 \times 10^9 \text{years}.\).
03

- Calculate the Age of the Rock

First, solve for \ t \ in the decay formula. \(\text{1.38} = \text{2.52} e^{-\frac{\text{t} \text{ln}(2)}{1.25 \times 10^9}}\). Isolate \ t \ as follows: \(\text{t} = \frac{-1.25 \times 10^9 \times \text{ln}(\frac{1.38}{2.52})}{\text{ln}(2)} \). Simplifying this expression gives the result for \ t \.
04

- Perform the Calculation

Calculate the natural logarithm and the constant values: \[ \text{t} = \frac{-1.25 \times 10^9 \times \text{ln}(0.5476)}{\text{ln}(2)} = \frac{-1.25 \times 10^9 \times (-0.602)}{0.693} = 1.084 \times 10^9 \text{ years} \]. The rock is approximately 1.084 billion years old.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life Calculation
Half-life is a crucial concept in radioactive decay. It’s the time taken for half of the radioactive isotopes in a sample to decay. In the given exercise, the half-life of \({}^{40} \text{K}\) is provided as 1.25 billion years.
To calculate the half-life, you can use the formula: \[ N(t) = N_0 e^{-\frac{t \ln(2)}{t_{1/2}}} \] Here:
  • \( N(t) \) is the amount of radioactive isotope remaining after time \( t \)
  • \( N_0 \) is the initial amount of the isotope.
  • \( t \) is the time that has passed.
  • \( t_{1/2} \) is the half-life of the isotope.
This formula helps to establish how much of the substance has decayed after a specific period. Understanding this decay process is essential for radiometric dating methods like the one used in the exercise.
Argon-Potassium Dating
Argon-Potassium (K-Ar) dating is a method used to date volcanic rocks and ash layers. It takes advantage of the radioactive decay of \({}^{40} \text{K}\) to \({}^{40} \text{Ar}\). This technique is immensely valuable in geology.
Here’s how it works:
  • When volcanic rocks cool, they expel any argon gas, resetting the radiometric clock to zero.
  • Over time, \({}^{40} \text{K}\) decays to \({}^{40} \text{Ar}\).
  • By measuring the ratios of \({}^{40} \text{K}\) and \({}^{40} \text{Ar}\), the elapsed time since the rock cooled can be calculated.
This method is effective for dating rocks that are millions to billions of years old. In recent exercise, we used K-Ar dating to determine that the rock is approximately 1.084 billion years old.
Radioactive Isotopes in Geochemistry
In geochemistry, radioactive isotopes serve as natural clocks. They allow scientists to date rocks, understand geological processes, and study the history of the Earth. Isotopes like \({}^{40} \text{K}\) play a significant role.
Here are key points about radioactive isotopes in geology:
  • Different isotopes have different half-lives, making them suitable for dating varied age ranges.
  • They can trace the formation, metamorphism, and movement of rocks.
  • Analyzing the ratios of parent and daughter isotopes helps to determine the age and history of geological samples.
They are critical in establishing chronological frameworks for geological events and understanding Earth's ancient history.
Natural Logarithm in Decay Equations
The natural logarithm (\( \ln \)) is fundamental in the mathematical description of radioactive decay. In the decay equation \[ N(t) = N_0 e^{-\frac{t \ln(2)}{t_{1/2}}} \], \( \ln(2) \approx 0.693 \).
Here’s why \( \ln \) is important:
  • It helps to linearize the exponential decay process.
  • Makes it easier to solve for the time \( t \) when measuring the age of rocks and other materials.
  • \( \ln(2) \) is used because we’re dealing with decay that follows a half-life formula.
In our example:
Given \( N(t) = 1.38 \text{ mmol} \), \( N_0 = 2.52 \text{ mmol} \), and \( t_{1/2} = 1.25 \times 10^9 \text{ years} \), we isolate \( t \) as follows:
\[ 1.38 = 2.52 e^{-\frac{t \ln(2)}{1.25 \times 10^9}} \] Solving this equation using natural logs enables us to determine that the rock’s age is approximately 1.084 billion years.

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