InterviewSolution
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InterviewSolution
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A radioactive nuclide is produced at the constant rate of n per second (say, by bombarding a target with neutrons). The expected number `N` of nuclei in existence `t` s after the number is `N_(0)` is given byA. `N=N_(0)e^(-lambda t)`B. `N=N_(0)e^(-lambda t)`C. `N=n/lambda+(N_0+n/lambda)e^(-lambda t)`D. `N=n/lambda+(N_0+n/lambda)e^(-lambda t)` |
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Answer» Correct Answer - c `(dN)/(dt) =n -lambda N` because the population `N` is simultaneously incereasing at rate n and decreasing due to decay at rate `lambda N`. ` underset(N_(0))overset(N) (int)(dN)/(n-lambdaN)=underset(0) overset(t)(int)dt` `(1)/(lambda)In ((n-lambdaN_(0))/(n-lambdaN))=t` `N=(n)/(lambda)+(N_(0)-(n)/(lambda))e^(-lambdat)`. |
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