InterviewSolution
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InterviewSolution
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The half-life of a radioactive sample is `T`. If the activities of the sample at time `t_(1)` and `t_(2)` `(t_(1) lt t_(2))` and `R_(1)` and `R_(2)` respectively, then the number of atoms disintergrated in time `t_(2)-t_(1)` is proportional toA. `(R_(1)-R_(2))T`B. `(R_(1)+R_(2))T`C. `(R_(1)R_(2))/(R_(1)+R_(2))T`D. `(R_(1)R_(2))/(T)` |
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Answer» Correct Answer - A Activity `R= lambda N` so that `R_(1)=lambda N_(1)` and `R_(2)=lambda N_(2)` `R_(1)-R_(2)=lambda(N_(1)-N_(2))=(0.6931)/(T)(N_(1)-N_(2))` `N_(1)-N_(2)=((R_(1)-R_(2))T)/(0.6931)` `N_(1)-N_(2) prop (R_(1)-R_(2))T, :. t_(2)-t_(1) alpha(R_(1)-R_(2))T` |
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