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

Given, `N_(A)=N_(B)`
`N_(0)^(A)((1)/(2))^(1//20)=N_(0)^(B)((1)/(2))^(t//10)`
`40((1)/(2))^(1//20)=160((1)/(2))^(t//10)impliest=40s`