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. 20s
D. 40s

Answer

D

Explanation

`(N_(1))/(N_(01)) = ((1)/(2))^((t)/(20)), (N_(2))/(N_(02)) = ((1)/(2))^(t//10)`
`N_(1)= N_(2)`
`rArr (40)/((2)^(t//20))= (160)/((2)^(t//10))`
`2^(t//20) = 2^(((t)/(10)-2))`
`rArr (t)/(20) = (t)/(10)-2 rArr t = 40`