In a successive decay chain A → B → C, the decay constant of A is twice the decay constant of B. At the instant when the number of B nuclei is maximum, the number of A nuclei is 400. Find the number of B nuclei at that instant.
Physics · Nuclear physics · radioactivity and law of decay
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Question

In a successive decay chain A → B → C, the decay constant of A is twice the decay constant of B. At the instant when the number of B nuclei is maximum, the number of A nuclei is 400. Find the number of B nuclei at that instant.

Physics · Nuclear physics · radioactivity and law of decay

G1 Physics · Accepted Answer

The rate equation for B is dN(B) ÷ dt = λ(A)N(A) − λ(B)N(B). At the instant when N(B) is maximum: dN(B) ÷ dt = 0. Therefore, λ(A)N(A) = λ(B)N(B). Given, λ(A) = 2λ(B). Hence, 2λ(B)×400 = λ(B)N(B). Cancel λ(B): N(B) = 800. Hence N = 800.