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 λ. 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]

Solution

Physics sir · 4 steps

B is fed by A and drained by its own decay, so it first rises and peaks when the two rates balance.

1. Rate of change. Let N_B(0) = N₀, so N_A(0) = 2N₀ and N_A = 2N₀ e^(-λ t). B is formed at the rate λ N_A and decays at the rate λ N_B, so dN_B/dt = λ N_A - λ N_B. At t = 0 this is 2λ N₀ - λ N₀ = +λ N₀, so the graph must start by rising from 1.
2. Solve it. Try N_B = N₀ e^(-λ t) f(t). Substituting gives f' = 2λ, and f(0) = 1, so f = 1 + 2λ t.

   N_B(t)/N_B(0) = (1 + 2λ t) e^(-λ t)

3. Find the peak. The peak comes when formation equals decay, that is N_A = N_B. This gives 2e^(-λ t) = (1 + 2λ t)e^(-λ t), so t = 1/2λ. The peak height is 2e^(-1/2) ≈ 1.21. After that the factor e^(-λ t) wins and the ratio falls slowly towards zero without ever reaching it.
4. So. The curve starts at 1, rises to a maximum at t = 1/2λ and then decays towards zero. That is graph (c).

The idea to keep. For a chain A → B → C, the middle member grows while its supply rate beats its own decay rate. The maximum of N_B is exactly where λ_A N_A = λ_B N_B.

Answer from the NTA final answer key. Solution by PrepWiser.