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A mixture consists of two radioactive ma...

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

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Let after t s, quantity of `A_1` and `A_2` are equal .
We know , `N=N_0(1/2)^(t//T)` [where, T=half -life]
`therefore ` For `A_1, N_1 = N_(01)(1/2)^(t//20)` and for `A_2, N_2 = N_(02)(1/2)^(t//10)`
`because N_1=N_2`
`therefore 40xx(1/2)^(t//20) = 160(1/2)^(t//10)`
or `2^(2-(t//20))=2^(4-(t//10))` or, `2-t/20 = 4-t/10` or, t=40 s.
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