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A radioisotope has half life of 10 years. What percentage of the original amount of it would you expect to remain after 20 years? |
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Answer» Correct Answer - C `t_(1//2) = 10` years, `T = 20` year We know `T = nt_(1//2) implies n = (20)/(10) = 2` And `N = ((1)/(2))^(n) N_(0)` `(N)/(N_(0)) = ((1)/(2))^(2) = (1)/(4)` % of `(N)/(N_(0)) = (1)/(4) xx 100 = 25%` |
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