Radioactive nuclei P and Q both decay into stable R. At t = 0, Nₚ = 4N₀ and number of Q nuclei = N₀. The half-lives of P and Q are 1 min and 2 min, respectively. Initially no R is present. When Nₚ = number of Q nuclei, how many R nuclei are present?
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Question

Radioactive nuclei P and Q both decay into stable R. At t = 0, Nₚ = 4N₀ and number of Q nuclei = N₀. The half-lives of P and Q are 1 min and 2 min, respectively. Initially no R is present. When Nₚ = number of Q nuclei, how many R nuclei are present?

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G1 Physics · Accepted Answer

Nₚ=4N₀(1/2)^t and number of Q nuclei=N₀(1/2)^{t/2}, with t in minutes. Equality gives t=4 min. Then Nₚ=number of Q nuclei=N₀/4, so remaining P+Q=N₀/2. From initial total 5N₀, R nuclei = 5N₀−N₀/2 = 9N₀/2. Correct answer: C.