Let `T` be the mean life of a radioactive sample. `75%` of the active nuclei present in th sample initially will deacy in time

Options

A. `2T`
B. `1//2(ln2)T`
C. `4T`
D. `2(ln2)T`

Answer

D

Explanation

`N = (N_(0))/(4) rArr (N)/(N_(0)) = (1)/(4) = ((1)/(2))^(2) = ((1)/(2))^(n)`
Number of half lives `n = 2`
`t = nT_(1//2) = 2T_(1//2) = 2 ((ln2)/(lambda)) = 2ln2 ((1)/(lambda)) = 2(ln2)T`