InterviewSolution
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InterviewSolution
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`A` and `B` are two radioactive substances whose half lives are `1` and `2` years respectively. Initially `10 gm` of `A` and `1 gm` of `B` is taken. The time (approximate) after which they will have same quantity remaining is.A. `6.62yr`B. 5yrC. `3.2yr`D. 7yr |
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Answer» Correct Answer - A We have `N=N_(0)((1)/(2))^(t//T)` `thereforeN_(A)=10((1)/(2))^(t//1)impliesN_(B)=1((1)/(2))^(t//2)` Given, `N_(A)=N_(B)` So, `10((1)/(2))=((1)/(2))^(t//2)` So, `10((1)/(2))^(-t//2)or 10=2^(t//2)` `therefore "log"_(10)=t/2log_(10)2` `1=t/2xx0.3010` `therefore` Time, t = 0.62 yr |
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