InterviewSolution
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InterviewSolution
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The activity of a radioactive sample is measures as `N_0` counts per minute at `t = 0` and `N_0//e` counts per minute at `t = 5 min`. The time (in minute) at which the activity reduces to half its value is.A. `log_e2/5`B. `5/(log_e2)`C. `5log_10 2`D. `5log_e 2` |
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Answer» Correct Answer - d According to activity law `R=R_0e^(-lambdat)` …(i) where, `R_0`=initial activity at t=0 R=activity at time t `lambda`=decay constant According to given problem, `R_0=N_0` counts per minute `R=N_0/e` counts per minute t=5 minutes Substituting these values in equation (i), we get `N_0/e=N_0e^(-5lambda)` `e^(-1)=e^(-5lambda)` `5lambda=1` or `lambda=1/5` per minute At `t=T_(1//2)` , the activity R reduces to `R_0/2`. where `T_(1//2)` = half life of a radioactive sample From equation (i), we get `R_0/2=R_0 e^(-lambdaT_(1//2))` `e^(lambdaT_(1//2))=2` Taking natural logarithms on both sides of above equation, we get `lambdaT_(1//2) = log_e2` or `T_(1//2)=(log_e 2)/lambda=(log_e 2) /((1/5))=5 log_e 2` minutes |
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