Half-life of a radioactive substance `A` is two times the half-life of another radioactive substance `B`. Initially, the number of `A` and `B` are `N_(A)` and `N_(B)`, respectively . After three half-lives of `A`, number of nuclei of both are equal. Then, the ratio `N_(A)//N_(B)` is .

Options

A. `1//4`
B. `1//8`
C. `1//3
D. `1//6`

Answer

b

Explanation

Three half-lives of `A` is equivalent to six half-lives of `B`. Hence,
`N_(A)((1)/(2))^(3) =N_(B)((1)/(2))^(6)`
or `(N_(A))/(N_(B))=(1)/(8)` .