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.

Options

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

Answer

D

Explanation

`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`