Half-lives of two radioactive substances `A` and `B` are respectively `20` minutes and `40` minutes. Initially, he sample of `A` and `B` have equal number of nuclei. After `80` minutes the ratio of the remaining number of `A` and `B` nuclei is :

Options

A. `1:16`
B. `4:1`
C. `1:4`
D. `1:1`

Answer

C

Explanation

`(N_(A))/(N_(B)) = (N_(0)((1)/(2))^(80//20))/(N_(0)((1)/(2))^(80//40)) =(((1)/(2))^(4))/(((1)/(2))^(2))=((1)/(2))^(2) = (1)/(4)`