Question

A sample of a radioactive nucleus A disintegrates to another radioactive nucleus B, which in turn disintegrates to some other stable nucleus C. Plot of a graph showing the variation of number of atoms of nucleus B versus time is : (Assume that at t=0, there are no B atoms in the sample) 

• A. A
• B. B
• C. C
• D. D

Hint:

In successive disintegration \( A \to B \to C \), the intermediate product population always follows a "rise-then-fall" curve.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This describes a sequential radioactive decay process: \( A \xrightarrow{\lambda_A} B \xrightarrow{\lambda_B} C \).
Initially (\( t=0 \)), there are only atoms of \( A \). Nucleus \( B \) is produced by the decay of \( A \) and consumed by its own decay into \( C \).
Step 2: Key Formula or Approach:
The rate of change of the number of atoms of \( B \) is:
\[ \frac{dN_B}{dt} = \lambda_A N_A - \lambda_B N_B \]
Since \( N_A = N_0 e^{-\lambda_A t} \), the solution for \( N_B(t) \) is:
\[ N_B(t) = \frac{\lambda_A N_0}{\lambda_B - \lambda_A} (e^{-\lambda_A t} - e^{-\lambda_B t}) \]
Step 3: Detailed Explanation:
1. At \( t = 0 \), \( N_B = 0 \) as given in the problem.
2. Initially, because \( N_A \) is large and \( N_B \) is small, the production rate \( \lambda_A N_A \) is higher than the decay rate \( \lambda_B N_B \). Thus, \( N_B \) increases.
3. As time passes, \( N_A \) decreases exponentially and \( N_B \) increases, making the production rate fall and the decay rate rise.
4. Eventually, the decay rate equals the production rate (\( \frac{dN_B}{dt} = 0 \)), and \( N_B \) reaches its maximum.
5. After this peak, the production from the nearly exhausted supply of \( A \) cannot keep up with the decay of \( B \), so \( N_B \) starts decreasing toward zero as \( t \to \infty \).
Step 4: Final Answer:
The graph must start at zero, reach a peak, and then decay back to zero.
Graph (C) correctly depicts this variation.