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Radioactivity of a sample at `T_(1)` time is `R_(1)` and at time `T_(2)` is `R_(2).` If half-life of sample is T, then in time `(T_(2)-T_(1)),` the number of decayed atoms is proportional toA. `R_(1)T_(2)-R_(2)T_(2)`B. `(R_(1)-R_(2))T`C. `((R_(1)-R_(2)))/(T)`D. `R_(1)-R_(2)` |
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Answer» Correct Answer - B Let `R_(1)=N_(1)lamdaandR_(2)=N_(2)lamda` Mean life, `T=(1)/(lamda)` `R_(1)-R_(2)=(N_(1)-N_(2))lamda=(N_(1)-N_(2))(1)/(T)` So, number of atoms disintegrated in `(T_(2)-T_(1))` is `=N_(1)-N_(2)=(R_(1)-R_(2))T` |
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