Question

Consider radioactive decays \(A \to B\) with half-life \((T_{1/2})_A\), and \(B \to C\) with half-life \((T_{1/2})_B\). At any time t, the number of nuclides of B is given by
\( (N_B)_t = \frac{\lambda_A{\lambda_B - \lambda_A}(N_A)_0 (e^{-\lambda_A t} - e^{-\lambda_B t}) \),
where \((N_A)_0\) is the number of nuclides of A at \(t = 0\). The decay constants of A and B are \(\lambda_A\) and \(\lambda_B\), respectively.
If \((T_{1/2})_B<(T_{1/2})_A\), then the ratio \(\frac{(N_B)_t}{(N_A)_t}\) at time \(t \gg (T_{1/2})_A\) is}
\((N_A)_t\) is the number of nuclides of A at time t

• A. \(\frac{\lambda_A}{\lambda_B - \lambda_A}\)
• B. \(\frac{\lambda_B}{\lambda_A}\)
• C. \(\frac{\lambda_A}{\lambda_B}\)
• D. \(\frac{\lambda_B}{\lambda_B - \lambda_A}\)

Hint:

In serial decay \(A \to B \to C\): - If \(\lambda_A \ll \lambda_B\) (\(T_{1/2,A} \gg T_{1/2,B}\)), we reach secular equilibrium, and for large t, \(\lambda_A N_A \approx \lambda_B N_B\). - If \(\lambda_A<\lambda_B\) (\(T_{1/2,A}>T_{1/2,B}\)), we reach transient equilibrium, and for large t, \(\frac{N_B}{N_A} \to \frac{\lambda_A}{\lambda_B - \lambda_A}\). - If \(\lambda_A>\lambda_B\), no equilibrium is reached. Remembering these conditions can help you quickly identify the expected answer.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem deals with serial radioactive decay, specifically the case of transient equilibrium. We are given the number of nuclides of the daughter nucleus B, \((N_B)_t\), and we need to find the ratio of the number of B nuclides to the number of A nuclides, \(\frac{(N_B)_t}{(N_A)_t}\), at a very long time \(t\).
Step 2: Key Formula or Approach:
We are given the formula for \((N_B)_t\). We also know the formula for the decay of the parent nucleus A: \[ (N_A)_t = (N_A)_0 e^{-\lambda_A t} \] The relationship between half-life \(T_{1/2}\) and decay constant \(\lambda\) is \(\lambda = \frac{\ln 2}{T_{1/2}}\). The condition \((T_{1/2})_B<(T_{1/2})_A\) implies \(\frac{\ln 2}{\lambda_B}<\frac{\ln 2}{\lambda_A}\), which means \(\lambda_B>\lambda_A\). This is the condition for transient equilibrium. We need to evaluate the ratio \(\frac{(N_B)_t}{(N_A)_t}\) in the limit of large \(t\).
Step 3: Detailed Explanation:
1. Write the expression for the ratio: \[ \frac{(N_B)_t}{(N_A)_t} = \frac{\frac{\lambda_A}{\lambda_B - \lambda_A}(N_A)_0 (e^{-\lambda_A t} - e^{-\lambda_B t})}{(N_A)_0 e^{-\lambda_A t}} \] The \((N_A)_0\) terms cancel out. \[ \frac{(N_B)_t}{(N_A)_t} = \frac{\lambda_A}{\lambda_B - \lambda_A} \frac{e^{-\lambda_A t} - e^{-\lambda_B t}}{e^{-\lambda_A t}} \] 2. Simplify the expression: Distribute the denominator \(e^{-\lambda_A t}\) into the parenthesis in the numerator: \[ \frac{(N_B)_t}{(N_A)_t} = \frac{\lambda_A}{\lambda_B - \lambda_A} \left( \frac{e^{-\lambda_A t}}{e^{-\lambda_A t}} - \frac{e^{-\lambda_B t}}{e^{-\lambda_A t}} \right) \] \[ \frac{(N_B)_t}{(N_A)_t} = \frac{\lambda_A}{\lambda_B - \lambda_A} \left( 1 - e^{-(\lambda_B - \lambda_A)t} \right) \] 3. Apply the long time limit: We are given the condition \((T_{1/2})_B<(T_{1/2})_A\), which implies \(\lambda_B>\lambda_A\). Therefore, the term \((\lambda_B - \lambda_A)\) is positive. We need to evaluate the ratio for a time \(t\) that is much larger than the half-life of A, i.e., \(t \gg (T_{1/2})_A\). In this limit (\(t \to \infty\)), the exponential term \(e^{-(\lambda_B - \lambda_A)t}\) will approach zero because its exponent is large and negative. \[ \lim_{t \to \infty} e^{-(\lambda_B - \lambda_A)t} = 0 \quad (\text{since } \lambda_B - \lambda_A>0) \] 4. Calculate the final ratio: Substituting this limit into our simplified expression for the ratio: \[ \lim_{t \gg (T_{1/2})_A} \frac{(N_B)_t}{(N_A)_t} = \frac{\lambda_A}{\lambda_B - \lambda_A} (1 - 0) \] \[ \frac{(N_B)_t}{(N_A)_t} = \frac{\lambda_A}{\lambda_B - \lambda_A} \] This state is known as transient equilibrium, where the ratio of the activities (and hence the number of atoms) of the daughter to the parent becomes constant. Step 4: Final Answer:
In the long time limit, the ratio \(\frac{(N_B)_t}{(N_A)_t}\) approaches \(\frac{\lambda_A}{\lambda_B - \lambda_A}\). This matches option (A).