Question

At time t = 0, a material is composed of two radioactive atoms A and B, where N$_A$(0) = 2N$_B$(0). The decay constant of both kind of radioactive atoms is \(\lambda\). However, A disintegrates to B and B disintegrates to C. Which of the following figures represents the evolution of N$_B$(t)/N$_B$(0) with respect to time t? 
[ $N_A(0)$ = No. of A atoms at t = 0 ] 
[ $N_B(0)$ = No. of B atoms at t = 0 ] 

• A. A
• B. B
• C. C
• D. D

Hint:

This is a standard problem of serial radioactive decay. The key is to set up the correct differential equation for the intermediate nuclide (B). Remember that its population changes due to both formation and decay. Analyzing the function's behavior at \(t=0\), \(t \to \infty\), and finding its maxima/minima is a reliable way to identify the correct graph.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We have a radioactive decay chain \( A \rightarrow B \rightarrow C \). We are given the initial conditions and the decay constant. We need to find the correct graph for the ratio of the number of B atoms at time t to the initial number of B atoms, i.e., \( N_B(t)/N_B(0) \).
Step 2: Key Formula or Approach:
The rate of change of the number of atoms of B, \(N_B\), is determined by its rate of formation from A and its own rate of decay. This can be expressed as a differential equation:
\[ \frac{dN_B}{dt} = (\text{Rate of formation of B}) - (\text{Rate of decay of B}) \] \[ \frac{dN_B}{dt} = \lambda N_A(t) - \lambda N_B(t) \] We need to solve this differential equation.
Step 3: Detailed Explanation:
First, find the number of A atoms at time t, \(N_A(t)\). This is a simple decay process:
\[ N_A(t) = N_A(0) e^{-\lambda t} \] Substitute this into the rate equation for \(N_B\):
\[ \frac{dN_B}{dt} = \lambda (N_A(0) e^{-\lambda t}) - \lambda N_B \] \[ \frac{dN_B}{dt} + \lambda N_B = \lambda N_A(0) e^{-\lambda t} \] This is a first-order linear differential equation. The general solution is:
\[ N_B(t) = (N_B(0) + \lambda N_A(0) t) e^{-\lambda t} \] We are given the initial condition \(N_A(0) = 2N_B(0)\). Substituting this into the solution:
\[ N_B(t) = (N_B(0) + \lambda (2N_B(0)) t) e^{-\lambda t} \] \[ N_B(t) = N_B(0) (1 + 2\lambda t) e^{-\lambda t} \] We need to plot the ratio \( \frac{N_B(t)}{N_B(0)} \):
\[ \frac{N_B(t)}{N_B(0)} = (1 + 2\lambda t) e^{-\lambda t} \] Let's analyze the behavior of this function, let's call it \(f(t)\):
1. At t = 0: \( f(0) = (1 + 0)e^0 = 1 \). The graph starts at 1 on the y-axis.
2. As t \(\rightarrow \infty\): The exponential term \(e^{-\lambda t}\) goes to zero faster than the linear term \((1 + 2\lambda t)\) grows, so \(f(t) \rightarrow 0\).
3. Maximum value: To find if there's a peak, we take the derivative with respect to t and set it to zero.
\[ \frac{d}{dt} \left( \frac{N_B(t)}{N_B(0)} \right) = \frac{d}{dt} \left[ (1 + 2\lambda t) e^{-\lambda t} \right] \] Using the product rule:
\[ = (2\lambda)e^{-\lambda t} + (1 + 2\lambda t)(-\lambda e^{-\lambda t}) \] \[ = e^{-\lambda t} [2\lambda - \lambda(1 + 2\lambda t)] = e^{-\lambda t} [\lambda - 2\lambda^2 t] \] Set the derivative to zero:
\[ \lambda - 2\lambda^2 t = 0 \implies t = \frac{\lambda}{2\lambda^2} = \frac{1}{2\lambda} \] Since the second derivative is negative at this point, it is a maximum.
Step 4: Final Answer:
The function \( \frac{N_B(t)}{N_B(0)} \) starts at 1, increases to a maximum at \( t = \frac{1}{2\lambda} \), and then decays to zero as \( t \rightarrow \infty \). The graph in option (B) correctly depicts this behavior.