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

(d) Fraction remains after `n` half-lives
`(N)/(N_0) = ((1)/(2))^n = ((1)/(2))^(t//T)`
Given `N = (N_0)/(e) rArr (N_0)/(e N_0) = ((1)/(2))^(5//T)`
or `(1)/(e) = ((1)/(2))^(5//T)`
Taking log on both sides, we get
`log 1 - log e = (5)/(T) log (1)/(2)`
`-1 = (5)/(T)(-log 2) rArr T = 5 log_e 2`
Now, let `t'` be the time which activity reduces to half.
`((1)/(2)) = ((1)/(2))^(t'//5 log_e 2) rArr t' = 5 log_e 2`.