InterviewSolution
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InterviewSolution
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In a radioactive material the activity at time `t_(1)` is `R_(1)` and at a later time `t_(2)`, it is `R_(2)`. If the decay constant of the material is `lambda`, thenA. `R_(1)=R_(2)e^(-lambda(t_(1)-t_(2)))`B. `R_(1)=R_(2)e^(lambda(t_(1)-t_(2)))`C. `R_(1)=R_(2)(t_(2)//t_(1))`D. `R_(1)=R_(2)` |
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Answer» Correct Answer - A The decay rate R of a radioactive material is the number of decays per second. From radioactive decay law. `-(dN)/(dt)alpha N or -(dN)/(dt)=lambdaN` `"Thus", R=-(dN)/(dt)` `or R=lambdaN or R=lamdaN_(0)E^(-lambdat).....(i)` Where, `R_(0)=lambdaN_(0)` is the activity of the radioactive material at time t=0. At time, `t_(1), R_(1)=R_(0)e^(-lambdat_(1))......(ii)` At time `t_(2)`, `R_(2)=R_(0)e^(-lambdat_(2))......(iii)` Dividing Eq, (ii) by (iii), we have `(R_(1))/(R_(2))=(e^(-lambdat_(1)))/(e^(-lambdat_(2)))=e^(-lambda(t_(1)-t_(2)))` `or R_(1)=R_(2)e^(-lambda(t_(1)-t_(2)))` |
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