A mixture consists of two radioactive materials A 1 and A 2 with half-lives of 20 s and 10 s , respectively. Initially, the mixture has 40 g of A 1 and 160 g of A 2 . The amount of the two in the mixture will become equal after

Question

A mixture consists of two radioactive materials A 1 and A 2 with half-lives of 20 s and 10 s , respectively. Initially, the mixture has 40 g of A 1 and 160 g of A 2 . The amount of the two in the mixture will become equal after, 60 s, 80 s, 20 s, 40 s

MARKS App · Accepted Answer

Let after time t s, A 1 and A 2 will become equal in the mixture. As N ⁡ = N ⁡ 0 1 2 n where n is the number of half-lives For A ⁡ 1, N ⁡ 1 = N ⁡ 0 1 1 2 t / 2 0 For A ⁡ 2, N ⁡ 2 = N ⁡ 0 2 1 2 t / 1 0 According to question, N 1 = N 2 4 0 2 t / 2 0 = 1 6 0 2 t / 1 0 2 t / 1 0 = 4 2 t / 2 0 or 2 t / 1 0 = 2 2 2 t / 2 0 2 t / 1 0 = 2 t 2 0 + 2 t 1 0 = t 2 0 + 2 or t 1 0 - t 2 0 = 2 or t 2 0 = 2 or t = 4 0 s

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