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

Answer

C

Explanation

From the law of radioactive decay,
`((N)/(N_(0))) = ((1)/(2))^(n) "but n" = (t)/(T) = (5)/(T)`
`therefore" "((N)/(N_(0)))= ((1)/(2))^(5//T)" "...(i)`
But it is given that `(N)/(N_(0)) = (1)/(e)` after 5 minutes …(ii)
`therefore` From eq. (i) and (ii), `(1)/(e) = ((1)/(2))^(5//T)`
Taking logarithms on both sides, we get
`log 1 - log_(e) e = (5)/(T) log_(e) ((1)/(2))`
`therefore" "-1 = (5)/(T) log_(e) ((1)/(2)) rArr 1 = (5)/(T) log_(e) 2`
`rArr" "T = 5 log_(e)2`
But T is the half life time or time in which the activity is reduced to half.
`therefore t = T = 5 log_(e)2`