A mixture consists of two radioactive materials `A_1` and `A_2` with half-lives of `20 s` and `10 s` respectively. Initially the mixture has `40 g` of `A_1` and `160 g` of `a_2`. The amount the two in the mixture will become equal after

Options

A. 60 s
B. 80 s
C. 20 s
D. 40 s

Answer

D

Explanation

Let the amount of the two in the mixture will become equal after t years .
The amount of `A_(1)` , which remains after t years `N_(1) = (N_(0) 1)/((2)^(t//20))`
The amount of `A_(2)` , which remains , after t years `N_(2) = (N_(0) 2)/((2)^(t//10))`
According to the problem `N_(1) = N_(2)`
`(40)/((2)^(t//20)) = (160)/((2)^(t //10)) implies 2^(t//20) = 2^(((t)/(10) - 2))`
`(t)/(20) - (t)/(10) = 2 implies (t)/(20) = 2 , t = 40` s