Problem 21 - Master Resource Book in JEE Main Physics

Question

Two radioactive materials \(X_{1}\) and \(X_{2}\) have decay constants \(10 \lambda\) and \(\lambda\) respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of \(X_{1}\) to that of \(X_{2}\) will be \(1 / e\) after a time (a) \(\frac{1}{10 \lambda}\) (b) \(\frac{1}{11 \lambda}\) (c) \(\frac{11}{10 \lambda}\) (d) \(\frac{1}{9 \lambda}\)

Step-by-Step Solution

Answer

(d) \(\frac{1}{9 \lambda}\)

Step 1: Understand the Decay Process

Radioactive decay follows an exponential decay law. For a radioactive material with decay constant \(k\), the number of undecayed nuclei at time \(t\) is given by \(N(t) = N_0 e^{-kt}\), where \(N_0\) is the initial number of nuclei.

Step 2: Apply Decay Formula to Both Materials

For material \(X_1\) with decay constant \(10 \lambda\), the number of nuclei at time \(t\) becomes \(N_{X_1}(t) = N_0 e^{-10 \lambda t}\). For material \(X_2\) with decay constant \(\lambda\), the number of nuclei at time \(t\) is \(N_{X_2}(t) = N_0 e^{-\lambda t}\).

Step 3: Set Up the Ratio Condition

The condition given is \(\frac{N_{X_1}(t)}{N_{X_2}(t)} = \frac{1}{e}\). Substitute the expressions for \(N_{X_1}(t)\) and \(N_{X_2}(t)\) into this condition:\[\frac{N_0 e^{-10 \lambda t}}{N_0 e^{-\lambda t}} = \frac{1}{e}\]

Step 4: Simplify the Ratio

Cancel \(N_0\) from the numerator and denominator:\[e^{-10 \lambda t + \lambda t} = \frac{1}{e}\]This simplifies to:\[e^{-9 \lambda t} = \frac{1}{e}\]

Step 5: Solve for Time \(t\)

Equate the exponents to find \(t\):\[-9 \lambda t = -1\]Solve for \(t\) by dividing both sides by \(-9 \lambda\):\[t = \frac{1}{9 \lambda}\]

Step 6: Choose the Correct Option

The calculated time \(t = \frac{1}{9 \lambda}\) corresponds to option (d).

Key Concepts

Decay Constant

The decay constant is a crucial part of understanding radioactive decay. It represents the probability that a given nucleus will decay per unit time. In other words, it dictates the rate at which a radioactive substance undergoes decay. The larger the decay constant, the faster the material decays. In our problem, the decay constants are given as \(10 \lambda\) for material \(X_1\) and \(\lambda\) for \(X_2\). This indicates that \(X_1\) decays much faster than \(X_2\).

Here, the decay constant influences the exponential decay of nuclei. For radioactive decay, time is always linked with an exponential function characterized by the decay constant. This tells us how rapidly the number of undecayed nuclei will decrease over time, providing insights into the longevity and behavior of the radioactive materials involved.

Exponential Decay

Exponential decay describes the way in which the quantity of a radioactive substance diminishes over time. The process is governed by the equation \(N(t) = N_0 e^{-kt}\), where \(N(t)\) is the number of remaining nuclei at time \(t\), \(N_0\) is the initial number of nuclei, and \(k\) is the decay constant.

In this specific exercise, exponential decay applies to both materials \(X_1\) and \(X_2\). For \(X_1\), the formula becomes \(N_{X_1}(t) = N_0 e^{-10 \lambda t}\), suggesting a rapid decline due to its significantly large decay constant. Whereas for \(X_2\), with \(N_{X_2}(t) = N_0 e^{-\lambda t}\), the decrease happens more gradually.

Understanding exponential decay helps us predict how quickly or slowly a substance will reduce, which is key in fields like nuclear physics and medicine.

Radioactive Nuclei Ratio

The radioactive nuclei ratio compares the number of remaining nuclei of one substance to another at any given time. This ratio helps to understand the comparative decay rates of different materials. In this exercise, the ratio \( \frac{N_{X_1}(t)}{N_{X_2}(t)} \) is set to \( \frac{1}{e} \).

By substituting the exponential decay formulas for each material, the ratio is expressed as \( \frac{N_0 e^{-10 \lambda t}}{N_0 e^{-\lambda t}} \). After simplification, it becomes \( e^{-9 \lambda t} \), which equals \( \frac{1}{e} \). This relation allows us to solve for the time \(t\).

This comparison demonstrates how much faster one substance decays than another under similar conditions, by analyzing the time it takes for their nuclei ratio to reach a specific value.