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

Answer

A

Explanation

We have, `N=N_(0)e^(-lamdat)implies(N_(0))/(e)=N_(0)e^(-lamda(5))implieslamda=1/5`
Now, `(N_(0))/(2)=N_(0)e^(-lamdat)impliest=(1)/(lamda)1n2 = 5 1n2`