Problem 38 - Master Resource Book in JEE Main Physics

Question

The radioactivity isotope \(X\) with a half-life of \(10^{9}\) year decays to \(Y\) which is stable. A sample of rocks were found to contain both the elements \(X\) and \(Y\) in the ratio \(1: 7\). What is the age of the rocks? (a) \(2 \times 10^{9} \mathrm{yr}\) (b) \(3 \times 10^{9} \mathrm{yr}\) (c) \(6 \times 10^{9} \mathrm{yr}\) (d) \(7 \times 1^{9} \mathrm{yr}\)

Step-by-Step Solution

Answer

The age of the rocks is \(3 \times 10^9\) years.

Step 1: Understanding the Problem

We need to determine the age of the rocks based on the given ratio of radioactive isotope \(X\) to the stable isotope \(Y\). Since the ratio is \(1 : 7\), it means for every 1 part of \(X\), there are 7 parts of \(Y\). First, we'll find out how much of \(X\) has decayed initially.

Step 2: Initial Quantities

Given that the ratio of \(X\) to \(Y\) is \(1 : 7\), the total parts can be expressed as \(X + Y = 1 + 7 = 8\). If \(N_0\) is the initial amount of \(X\), then after decay, \(N\) will be the amount of \(X\) remaining. Hence, \(N = N_0/8\).

Step 3: Exponential Decay Formula

The formula for radioactive decay is given by \(N = N_0 e^{-\lambda t}\), where \(\lambda\) is the decay constant and \(t\) is the time. Since we have \(N = N_0/8\), the equation becomes \(N_0/8 = N_0 e^{-\lambda t}\).

Step 4: Finding the Decay Constant

The decay constant \(\lambda\) is related to the half-life \(t_{1/2}\) by \(\lambda = \frac{\ln 2}{t_{1/2}}\). Given the half-life \(t_{1/2} = 10^{9}\) years, we have \(\lambda = \frac{\ln 2}{10^9}\).

Step 5: Substitute and Solve for Time

Substitute \(\lambda\) back into the exponential decay equation: \(\frac{1}{8} = e^{-\left(\frac{\ln 2}{10^9}\right) t}\). Take the natural logarithm of both sides to solve for \(t\): \(-\ln(8) = -\frac{\ln 2}{10^9} t\). Simplifying, we find \(t = \frac{\ln 8}{\ln 2} \times 10^9\).

Step 6: Calculate the Result

Since \(\ln 8 = 3 \ln 2\), the calculation becomes \(t = \frac{3 \ln 2}{\ln 2} \times 10^9\) years, which simplifies to \(t = 3 \times 10^9\) years. Hence, the age of the rocks is \(3 \times 10^9\) years.

Key Concepts

The Concept of Half-life

Half-life is a term that describes the time it takes for half of a given sample of a radioactive substance to decay. Think of it as a clock ticking down to a point where only half of the original substance remains. This concept is crucial in understanding radioactive decay and helps predict the rate at which a radioactive isotope will decay over time.

For the isotope in the original exercise, the half-life is given as \(10^9\) years. This means that in one billion years, only half of the original isotope \(X\) would remain, with the other half having decayed into a stable form \(Y\).

Understanding the half-life allows us to calculate the age of the rocks by using the provided ratio of elements \(X\) and \(Y\). The original ratio \(1:7\) indicates that a significant amount of \(X\) has already decayed into \(Y\), giving us the information we need to calculate the elapsed time.

Exponential Decay

Exponential decay is the process by which the quantity of a radioactive substance decreases over time at a rate proportional to its current amount. This type of decay is characterized by a gradual reduction, where the rate of decrease slows down as the quantity lessens.

The mathematical formula for radioactive decay is: \[ N = N_0 e^{-\lambda t} \]Where:

• \(N\) is the remaining quantity of the substance at time \(t\)
• \(N_0\) is the initial quantity of the substance
• \(\lambda\) is the decay constant
• \(t\) is the time elapsed

In the context of our exercise, as the decay occurs, the amount of \(X\) steadily declines according to this exponential pattern, with \(Y\) increasing as a result of \(X\)'s decay. Thus, the exponential decay function not only tells us how quickly the substance decays but also allows us to calculate the time when we know the current and initial ratios of the isotopes.

Decay Constant

The decay constant \(\lambda\) is a crucial component in the formula for radioactive decay. It is a measure of the probability per unit time that a single atom will decay. A higher decay constant means that the substance decays more quickly.

In our radioactive decay formula, the decay constant is related to the half-life through the relationship:\[ \lambda = \frac{\ln 2}{t_{1/2}} \]This indicates that the decay constant is inversely proportional to the half-life. A longer half-life results in a smaller \(\lambda\), signifying a slower decay rate.

For our isotope \(X\), given the half-life \(10^9\) years, the decay constant can be calculated as:\[ \lambda = \frac{\ln 2}{10^9} \]This value is then used in the exponential decay formula to determine how long it takes for a given amount of isotope \(X\) to decay to the measured ratio of \(1:7\) with \(Y\)..