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

Answer

A

Explanation

Since `(N)/(N_(0)) = ((1)/(2))^((t)/(T_(1//2)))`
`rArr (1)/(e)= ((1)/(2))^((5)/(T_(1//2)))`
`rArr-ln e= (-5)/(T_(1//2)) ln 2`
`rArr T_(1//2)= 5ln 2= 5 log_(e)2`