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 \(2.8 \times 10^9\) years.

Step 1: Understanding the Problem

We need to determine the age of the rock given the half-life of potassium-40 and the mass ratio of argon-40 to potassium-40 in the rock. The ratio provides insight into how much of the original potassium-40 has decayed into argon-40.

Step 2: Set Up the Decay Equation

The decay of potassium-40 to argon-40 can be modeled using the equation for radioactive decay: \[ N_t = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \] where \( N_t \) is the remaining quantity of potassium-40, \( N_0 \) is the original quantity, \( t \) is time, and \( T_{1/2} \) is the half-life.

Step 3: Express Quantities in Terms of the Mass Ratio

Let \( N_0 = 1 \) be the total initial amount of \( ^{40}\text{K} \). Then \( N_t = \frac{1}{4.6} = \frac{1}{1 + 3.6} \), as \( 3.6 \) is the ratio of argon to remaining potassium, thus \( 1 \) part of the \( 4.6 \) total is still the potassium-40.

Step 4: Substitute Values into the Decay Equation

We substitute \( N_t = \frac{1}{4.6} \), \( N_0 = 1 \), and \( T_{1/2} = 1.27 \times 10^9 \) into the decay equation:\[ \frac{1}{4.6} = \left(\frac{1}{2}\right)^{\frac{t}{1.27 \times 10^9}} \]

Step 5: Solve for the Age of the Rock \( t \)

To solve for \( t \), take the logarithm of both sides:\[ \ln\left(\frac{1}{4.6}\right) = \frac{t}{1.27 \times 10^9} \ln\left(\frac{1}{2}\right) \]Rearrange to solve for \( t \):\[ t = \frac{\ln\left(\frac{1}{4.6}\right)}{\ln\left(\frac{1}{2}\right)} \times 1.27 \times 10^9 \]

Step 6: Calculate the Result

Perform the calculation:\[ \ln\left(\frac{1}{4.6}\right) \approx -1.526 \]\[ \ln\left(\frac{1}{2}\right) \approx -0.693 \]Substitute these into the equation for \( t \):\[ t \approx \frac{-1.526}{-0.693} \times 1.27 \times 10^9 \]\[ t \approx 2.8 \times 10^9 \mathrm{yr} \]

Step 7: Interpret the Result

The calculated age of the rock, \( t \), is \( 2.8 \times 10^9 \) years. Given the options, the answer is (d).

Key Concepts

Half-life

The concept of half-life is essential when studying radioactive decay. A half-life refers to the time it takes for half of a radioactive substance to decay. For potassium-40 (\(^40\mathrm{K}\)), the half-life is approximately \(1.27 \times 10^{9}\) years. This means every \(1.27 \times 10^{9}\) years, half of the \(^40\mathrm{K}\) in a given sample will have decayed into another element, such as argon-40 (\(^40\mathrm{Ar}\)).

Understanding half-life helps scientists determine the ages of rocks and fossils. Because the process of radioactive decay is predictable, measuring how much \(^40\mathrm{K}\) remains in a rock, as opposed to how much has transformed into \(^40\mathrm{Ar}\), allows us to approximate the time that has passed since the rock was formed. This method of dating is called radiometric dating.

Potassium-40 Decay

Potassium-40 decay is a natural process that occurs due to the instability of the potassium-40 isotope. It undergoes decay in two different ways:

• Beta decay into calcium-40.
• Eletron capture, leading to argon-40.

In the context of potassium-argon dating, we are primarily concerned with the transformation into argon-40. This specific decay mode allows geologists to date ancient rocks as \(^40\mathrm{K}\) is ubiquitously found in minerals.

The decay process is characterized by the reduction of the original \(^40\mathrm{K}\) atoms over time, and it is governed by a probabilistic model of decay. Even though the occurrence of decay on a per atom basis is random, over large numbers, the decay follows an exponential decay pattern represented by:\[N_t = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\]where \(N_t\) is the amount of \(^40\mathrm{K}\) remaining, \(N_0\) is the initial amount, \(t\) is time, and \(T_{1/2}\) is the half-life.

Argon-40 Accumulation

As potassium-40 decays, argon-40 accumulates within the rock or mineral sample. The accumulation is significant because argon is a noble gas, which does not bond with other elements and gets trapped in the mineral structure.

This trapped argon-40 thus becomes a reliable indicator of the time elapsed since the rock was last reset or formed. By comparing the ratio of \(^40\mathrm{Ar}\) to \(^40\mathrm{K}\), geologists can ascertain the geological age of the sample.

The exercise provided a mass ratio of \(3.6\), indicating much of the original \(^40\mathrm{K}\) was converted into \(^40\mathrm{Ar}\) over time. Plugging this ratio into the decay equation and calculating provides the information needed to derive the age accurately. Hence, the measuring and understanding of this accumulation process enable the effective use of radiometric dating in geological studies.