InterviewSolution
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InterviewSolution
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The radioactivity of a sample is `R_(1)` at a time `T_(1)` and `R_(2)` at time `T_(2)`. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time `(T_(2) -T_(1))` is proporational toA. `R_(1) T_(1)=R_(2) T_(2)`B. `R_(1) - R_(2)`C. `(R_(1)-R_(2))/(T)`D. `(R_(1) -R_(2)) T` |
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Answer» Correct Answer - d `R_(1) =N_(1) lambda , R_(2)=N_(2) lambda` Also, `T=log_(e).(2)/(lambda)` or `lambda=log_(e). (2)/(T)` `:. R_(1) -R_(2) =(N_(1) -N_(2)) lambda` `=(N_(1) -N_(2))log_(e).(2)/(T)` `:. (N_(1) -N_(0))=((R_(1) -R_(2))T)/(log_(e) 2)` i.e., `(N_(1) -N_(2)) prop (R_(1) -R_(2))T`. |
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