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Half-lives of two radioactive substances A and B are respectively 20 minutes and 40 minutes. Initially, the samples of A and B have equal number of nuclei. After 80 minutes the ratio of remaining numbers of A and B nuclei is

1. 1 : 16
2. 4 : 1
3. 1 : 4
4. 1 : 1

Correct answer: 1 : 16

Solution

The number of remaining nuclei after time t is given by N = N₀(1/2)^(t/T), where T is the half-life. For A, after 80 minutes (4 half-lives), the fraction remaining is (1/2)^4 = 1/16. For B, after 80 minutes (2 half-lives), the fraction remaining is (1/2)^2 = 1/4. The ratio of remaining nuclei of A to B is (1/16) : (1/4) = 1 : 16.

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