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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 afterA. `60s`B. `80s`C. `20s`D. `40s` |
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Answer» Correct Answer - D Here, `T_(1)=20s, T_(2)=10s, N_(01)=40g`, `N_(02)=160g`. Let the amount of the two radioactive materials become equal after t sec. form `N/(N_(0))=(1/2)^(n)=(1/2)^(t//T)`, we get `N_(1)=N_(01)(1/2)^(t//20)=40(1/2)^(t//20)` `N_(2)=N_(02)(1/2)^(t//10)=160(1/2)^(t//10)` As `N_(1)=N_(2) :. 40(1/2)^(t//20)=160(1/2)^(t//10)` or `(1/2)^(t//20)=4(1/2)^(t//20)=(1/2)^(t/10-2)` `:. t/20=t/10-2 or t=40s` |
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