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

Explanation

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