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)/(2)`
B. `(1)/(8)`
C. `(1)/(3)`
D. `(1)/(6)`

Explanation

We know that, the amount remaining after n half lives can be calculated as:
`N = N_(0) ((1)/(2))^(n)`
Remaining amount of `A = N_(A) ((1)/(2))^(3)`
Remaining amount of `B = N_(B) ((1)/(2))^(6)`
`N_(A) ((1)/(2))^(3 = N_(B) ((1)/(2))^(6)`
`(N_(A))/(N_(B)) = (8)/(64) = (1)/(8)`