InterviewSolution
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InterviewSolution
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A radioisotope has `t_(1//2) = 5` years. After a given amount decays for 15 years, what fraction of the original isotope remains? |
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Answer» Half life `(t_(1//2) = 5` years Time for decay `(T) = 15` years We know that, `T = n xx t_(1//2)` So, `15 = n xx 5` or n = 3 We know that, `N = ((1)/(2))^(n) N_(0)` or `(N)/(N_(0)) = ((1)/(2))^(n) = ((1)/(2))^(3) = (1)/(8)` Thus, after 15 years `(1)/(8)th` of the original amount remains. |
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