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^(-lambda t)`
Here , t = 5 minutes
`(N_(0))/(e) = N_(0) * e^(-5 lambda) implies 5 lambda = 1 , lambda = (1)/(5)`
Now , `T_(1//2) = (l n 2)/(lambda) = 5 l n 2 `