InterviewSolution
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InterviewSolution
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Half-life of a radioactive substance `A` is two times the half-life of another radioactive substance `B`. Initially, the number of `A` and `B` are `N_(A)` and `N_(B)`, respectively . After three half-lives of `A`, number of nuclei of both are equal. Then, the ratio `N_(A)//N_(B)` is .A. `(1)/(2)`B. `(1)/(8)`C. `(1)/(3)`D. `(1)/(6)` |
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Answer» We know that, the amount remaining after n half lives can be calculated as: `N = N_(0) ((1)/(2))^(n)` Remaining amount of `A = N_(A) ((1)/(2))^(3)` Remaining amount of `B = N_(B) ((1)/(2))^(6)` `N_(A) ((1)/(2))^(3 = N_(B) ((1)/(2))^(6)` `(N_(A))/(N_(B)) = (8)/(64) = (1)/(8)` |
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