In a successive decay chain A → B → C, initially only A is present. The half-life of A is 2 h and the half-life of B is 4 h. After how many hours is the number of B nuclei maximum?
Physics · Nuclear physics · radioactivity and law of decay
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Question

In a successive decay chain A → B → C, initially only A is present. The half-life of A is 2 h and the half-life of B is 4 h. After how many hours is the number of B nuclei maximum?

G1 Physics · Accepted Answer

For a successive chain with no initial B, the daughter population becomes maximum at t(max) = ln[λ(B) ÷ λ(A)] ÷ [λ(B)−λ(A)]. Here, λ(A) = ln2 ÷ 2, λ(B) = ln2 ÷ 4. Therefore, λ(B) ÷ λ(A) = 1 ÷ 2. Numerator: ln(1 ÷ 2) = −ln2. Denominator: λ(B)−λ(A) = ln2 ÷ 4 − ln2 ÷ 2 = −ln2 ÷ 4. Thus, t(max) = (−ln2) ÷ (−ln2 ÷ 4) = 4 h. Hence N = 4.