Question

A radioactive nucleus A can decay to P or Q with decay constant $2\lambda$ and $\lambda$ respectively. Another radioactive nucleus B decays in Q with decay constant $3\lambda$. Initially number of active nuclei of A and that of B are equal ($N_0$). Select correct relation for any later moment t.

• A. $N_A = N_B = N_0e^{-3\lambda t}$
• B. $N_P = N_Q = \frac{N_0}{3}(1-e^{-3\lambda t})$
• C. $N_P = 2N_Q = \frac{N_0}{3}(1-e^{-3\lambda t})$
• D. $2N_P = N_Q = \frac{4N_0}{3}(1-e^{-3\lambda t})$

Answer 1

Answer: (D)

Option D

Answer 2

1. For nucleus A, the total decay constant is 2λ + λ = 3λ, so the number remaining is
     $N_A = N_0 e^{-3\lambda t}.$
2. In its decay, P is produced with probability 2/3 and Q with probability 1/3. Thus,
     $N_P = \frac{2}{3}[N_0 - N_A] = \frac{2N_0}{3}(1 - e^{-3\lambda t}).$   $\text{Q from A } = \frac{1}{3}[N_0 - N_A] = \frac{N_0}{3}(1 - e^{-3\lambda t}).$
3. Nucleus B decays to Q with decay constant 3λ, so
     $N_B = N_0 e^{-3\lambda t} \quad \text{and} \quad \text{Q from B } = N_0 - N_B = N_0(1 - e^{-3\lambda t}).$
4. Total Q produced is:
     $N_Q = \frac{N_0}{3}(1 - e^{-3\lambda t}) + N_0(1 - e^{-3\lambda t}) = \frac{4N_0}{3}(1 - e^{-3\lambda t}).$
5. Notice that
     $2N_P = 2 \times \frac{2N_0}{3}(1 - e^{-3\lambda t}) = \frac{4N_0}{3}(1 - e^{-3\lambda t}) = N_Q.$

Thus, the relation is $2N_P = N_Q = \frac{4N_0}{3}(1 - e^{-3\lambda t})$.