Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

Potassium \(-40\) decays to argon \(-40\) with a half life of \(1.27 \times 10^{9} \mathrm{yr}\). What is the age of a rock in which the mass ratio of \({ }^{40} \mathrm{Ar}\) to \({ }^{40} \mathrm{~K}\) is \(3.6\) ? (a) \(2.8 \times 10^{10} \mathrm{yr}\) (b) \(1.4 \times 10^{9} \mathrm{yr}\) (c) \(1.4 \times 10^{10} \mathrm{yr}\) (d) \(2.8 \times 10^{9} \mathrm{yr}\)

Step-by-Step Solution

Answer

The age of the rock is approximately \(2.8 \times 10^9 \text{ years}\), corresponding to option (d).

Step 1: Understanding the Decay Process

Potassium-40 decays into argon-40 over time. The given ratio is the mass of argon-40 to potassium-40 in the rock. The stated half-life is the time it takes for half of a sample of potassium-40 to transform into argon-40.

Step 2: Define Variables and Equation

Let \( N_0 \) be the initial amount of potassium-40, \( N \) its remaining amount, and \( D \) the amount of argon-40 formed, with \( D = 3.6N \). The decay equation is \( N = N_0 e^{-\lambda t} \), where \( \lambda \) is the decay constant, given by \( \lambda = \frac{\ln 2}{T_{1/2}} \), and \( T_{1/2} = 1.27 \times 10^9 \text{ years} \).

Step 3: Decay Constant and Rearrange Equation

Calculate the decay constant \( \lambda = \frac{\ln 2}{1.27 \times 10^9} \approx 5.464 \times 10^{-10} \text{ yr}^{-1} \). The total initial amount is \( N_0 = N + D = N + 3.6N = 4.6N \). Substitute \( N = N_0 e^{-\lambda t} \) into \( 4.6N = N e^{\lambda t} \).

Step 4: Solve for Time Fusing Ratio

From \( 4.6 = e^{\lambda t} \), take the natural logarithm on both sides to get \( \ln 4.6 = \lambda t \). Then, solve for \( t \): \( t = \frac{\ln 4.6}{\lambda} \).

Step 5: Substitute Values and Calculate

Substitute \( \lambda \approx 5.464 \times 10^{-10} \text{ yr}^{-1} \) into \( t = \frac{\ln 4.6}{5.464 \times 10^{-10}} \). This results in \( t \approx 2.8 \times 10^9 \text{ years} \).

Key Concepts

Potassium-Argon Dating

Potassium-argon dating is a method used to determine the age of rocks and minerals based on the radioactive decay of potassium-40 to argon-40. Scientists use this technique to date geological formations and archaeological artifacts. Here's how it works: a sample's potassium content begins to decay into argon over time. Since argon is a gas and escapes freely from molten rock, only when the rock cools and solidifies does the clock start.

This process is particularly useful for dating ancient rocks, as the half-life of potassium-40 is significant enough (about 1.27 billion years).

• Therefore, it is ideal for materials that are millions of years old.
• The more argon relative to potassium, the older the rock formation.

By measuring this ratio, scientists can calculate the time that has passed since the rock solidified.

Half-Life Calculation

The concept of half-life is crucial to understanding how radioactive decay works. It represents the time required for half of the radioactive substance to transform into another element. For potassium-40, its half-life is approximately 1.27 billion years. Understanding half-life allows us to predict the behavior of decaying elements over time.

To calculate the age of a sample, scientists use the known half-life to understand how much of the original potassium-40 remains compared to what has decayed into argon-40.

• If only a small fraction of potassium-40 remains, and more argon-40 is detected, it signifies an ancient sample.
• Half-life calculations involve exponential decay equations for accurate results.

This chronological technique is a powerful tool in paleontology, archaeology, and geology.

Exponential Decay Equation

The exponential decay equation is central to calculating the amount of a radioactive substance that remains over time. This equation is expressed as \( N = N_0 e^{-\lambda t} \). Here

• \( N_0 \) stands for the initial quantity of the substance,
• \( N \) is the remaining quantity after time \( t \),
• \( \lambda \) is the decay constant, and
• \( t \) is the elapsed time.

The decay constant \( \lambda \) is derived from the half-life \( T_{1/2} \) using the relation \( \lambda = \frac{\ln 2}{T_{1/2}} \). This formula instantly ties into half-life calculations.

In potassium-argon dating, knowing the decay constant allows us to substitute values to find the age of rocks. Once the exponential equation is solved for \( e^{\lambda t} \), the result demonstrates how much the sample has transformed, reflecting its age.

Natural Logarithm

The natural logarithm (ln) plays a significant role in solving exponential decay equations. It helps transform equations involving exponential functions into linear forms, which are easier to handle and interpret.

In the context of potassium-argon dating, after setting up the equation \( 4.6 = e^{\lambda t} \), taking the natural logarithm of both sides simplifies finding the elapsed time \( t \). Converting the equation via natural logarithms rephrases it as \( \ln 4.6 = \lambda t \).

• This approach allows for straightforward manipulation of variables.
• It also facilitates a comparison between the numerical relationships of known values and unknown time \( t \).

Using natural logarithms is essential for unraveling time-dependent processes in decay equations and achieving accurate dating results.