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. `5log_(10) 2`
D. `5 log_(e) 2`

Answer

D

Explanation

form `N=N_(0)e^(-lambdat)`
`(N_(0))/e=N_(0)e^(-lambdaxx5) :. 5lambda=1, lambda=1/5`
The activity reduces to half its initial value in half
life periode, `T=(log_(e)2)/lambda=(log_(e) 2)/(1//5) 5log_(e) 2`