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If a radioactive substance reduces to `(1)/(16)` of its original mass in `40` days, what is its half-life ?A. 10B. 5C. 2.5D. 20 |
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Answer» Correct Answer - A (a) From Rutherford and Soddy law for radioactive decay `N = N_0 e^(-lamda t)` where `N_0` and `N` are number of atom in a radioactive substance at time` t = 0` and `t = t` and `lamda` is decay constant. Also, half-life `T_(1//2) = (0.693)/(lamda)` `:. (N)/(N_0) = e^((0.683)/(-T^(1//2)) t)` Given, `t = 40 days, (N)/(N_0) = (1)/(16)` `:. (1)/(16) = e^((-(0.693 xx 40))/(T^(1//2)))` `rArr (0.693 xx 40)/(T^(1//2)) = log 16` `T_(1//2) = (0.693 xx 40)/(log 16)` =`9.99 ~~ 10 days`. |
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