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. 60s
B. 80s
C. 50s
D. 40s

Answer

D

Explanation

Let after t s amount of the `A _(1) and A _(2)` will become equal in the mixturer.
As `N = N _(0) ((1)/(2)) ^(n)`
where n is the number of half-lives
For `A _(1), N_(1) = N _(01) ((1)/(2))`
For `A _(2), N _(2) = N _(02) ((1)/2)) ^(t //10)`
According to question, `N _(1) = N _(2)`
`(40)/(2 ^(t //20)) = (160)/(2 ^(t //10))`
`2 ^(t //10) = 4 ( 2 ^(t //20))or 2 ^(t) //10 = 2 ^(2) 2 ^(t //20)`
`2 ^(t //10) =2 ((t )/(20) + 2)`
`(t)/(10) = (t)/(20) + 2 or (t)/(10) - (t)/(20) =2`
`or (t)/(20)=2 or t =40 s`