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. `4 : 1`
B. `1 : 2`
C. `8 : 1`
D. `16 : 1`

Answer

A

Explanation

`T_(1//2(A))= 40min, T_(1//2(B))= 20 min`
t = 80 min
`n_(A) = t/T_(1//2(A))=80/40 = 2`
`n_(B) = t/T_(1//2(B))=80/20 = 4`
`N_(A)/N_(B)=(N_(0)//2^(2))/(N_(0)//2^(4))=16/4 = 4 : 1`