InterviewSolution
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InterviewSolution
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As a given instant there are `25%` undercayed radio - active nucles in az sample . After `10` second the number of undecayed nucles reduces to `12.5%` Calculate (i() mean - like of the nucleus, and (ii) the time in which the number of undecayed nuclei will further to `6.25 %` of the reducted number . |
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Answer» Correct Answer - A::C::D From the given in information , it is clear that half life of the radioactive nuclie is `10 sec` (since half the amount is consumed in `10 sec 12.5% ` is half of `25 %` pis note) Mean life ` tau = (1)/(lambda) = (1)/(0.693//t_(1//2)) = (t_(1//2))/(0.693) = (10)/(0.693) = 14.43 sec ` (ii) N = N_(0) e ^(- lambda t) rArr (N)/(N_(0)) = (6.25)/(100)` `lambda = 0.693 s^(-1)` `(6.25)/(100) = e^(-0.0693t) rArr e^(-0.0693t) = (100)/(6.25) = 16` `0.0693 t = in 16 = 2.773 or t =(2.733)/(0.0693) = 40 sec` |
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