A radioactive isotope A decays into radioactive isotope B, which then decays into stable isotope C. Initially the sample contains 2000 nuclei of A and no B or C. The half-life of A is 2 h and the half-life of B is 4 h. Find the number of B nuclei present after 4 h.
Physics · Nuclear physics · radioactivity and law of decay
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Question

A radioactive isotope A decays into radioactive isotope B, which then decays into stable isotope C. Initially the sample contains 2000 nuclei of A and no B or C. The half-life of A is 2 h and the half-life of B is 4 h. Find the number of B nuclei present after 4 h.

Physics · Nuclear physics · radioactivity and law of decay

G1 Physics · Accepted Answer

Let λ(A) = ln2 ÷ 2 and λ(B) = ln2 ÷ 4. For a successive decay chain A → B → C with N(B,0)=0, N(B,t) = [λ(A)N₀ ÷ (λ(B)−λ(A))] × [exp(−λ(A)t) − exp(−λ(B)t)]. Here, λ(A) = 2λ(B). At t = 4 h: exp(−λ(A)t) = exp(−2ln2) = 1 ÷ 4, exp(−λ(B)t) = exp(−ln2) = 1 ÷ 2. Also, λ(A) ÷ [λ(B)−λ(A)] = 2λ(B) ÷ [−λ(B)] = −2. Therefore, N(B) = (−2)(2000)[1 ÷ 4 − 1 ÷ 2] = (−4000)(−1 ÷ 4) = 1000. Hence N = 1000.