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 _(e) 2//5`B. `(5)/(log_(e)2)`C. `5 log _(10)2`D. `5 log _(e) 2`

The activity of a radioactive sample is measures as `N_0` counts per m

1.	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 _(e) 2//5`B. `(5)/(log_(e)2)`C. `5 log _(10)2`D. `5 log _(e) 2`
Answer» Correct Answer - D `N=N_(0)e^(-lamdat) implies (N_(0))/(e )=N_(0)e^(-lamda(5))implies lamda =(1)/(5)` `Now (N_(0))/(2)=N_(0)e^(-lamda(t)) implies t=(1)/(lamda)ln2=5ln2`

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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 _(e) 2//5`B. `(5)/(log_(e)2)`C. `5 log _(10)2`D. `5 log _(e) 2`

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