InterviewSolution
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InterviewSolution
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The rate of decay of a radioactive species is given by `R_(1)` at time `t_(1)` and `R_(2)` at later time `t_(2)`. The mean life of this radioactive species is:A. `T = ((t_(1) - t_(2)))/("In" (R_(2)//R_(1)))`B. `T = ((t_(2) - t_(1)))/("In" (R_(2)//R_(1)))`C. `T = ((t_(2) - t_(1)))/("In" (R_(1)//R_(2)))`D. `T = ("In" (R_(2))//R_(1))/((t_(2)- t_(1)))` |
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Answer» Correct Answer - `(a)` `R_(1) = R_(0) e^(-lambda t_(1))` and `R_(2) = R_(0) e^(-lambda t_(2))` Now `(R_(2))/(R_(1)) = e^(lambda(t_(1)-t_(2)))` or In `((R_(2))/(R_(1))) = lambda (t_(1) - t_(2))` `(1)/(lambda) = ((t_(1) - t_(2)))/("In"(R_(2)//R_(1)))` or `T = (t_(1) - t_(2))/("In" (R_(2)//R_(1)))` |
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