InterviewSolution
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InterviewSolution
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Nuclei of radioactive element A are being produced at a constant rate. `alpha`. The element has a decay constant `lambda`. At time `t=0`, there are`N_(0)` nuclei of the element. (a) Calculate the number N of nuclei of A at time t. (b) IF `alpha =2 N_(0) lambda` , calculate the number of nuclei of A after one half-life time of A and also the limiting value of N at `t rarr oo`. |
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Answer» The rate of formation of radioactive nuclei is `alpha` . Rate of decay of radioactive nuclei is `lambda N`. Therefore, `(dN)/(dt) = alpha -lambda N` `rArr (dN)/(alpha -lambda N)=dt` Integrating, `(log e(alpha - lambda N_(0)))/(- lambda) =t +A` Where A is constant of integration. At `t =0,N =N_(0)` ` because (log_(e) (alpha lambda-lambdaN_(0)))/(-lambda)=A` (i) Therefore, Eq. (i) gives `(log_(e)(alpha-lambdaN))/(- lambda)=t+(log_(e)(alpha-lambdaN))/(-lamda)` `rArrlog_(e) ((alpha-lambdaN)/(alpha-lambdaN_(0)))=-lambdat` `rArr (alpha-lambdaN)/(alpha-lambdaN_(0))=e^(lambdat)` ltbtgt `rArr (alpha-lambdaN)/(alpha-lambdaN_(0))=e^(lambdat)` `rArr N=(alpha)/(lambda) (1 -e^(-lambda t)) + N_(0)e^(- lambda t)` (b) Given, `a = 2 N_(0) lambda` `t=T_(1)//2 =(0.693)/(lambda)` ` ` because N =1.5 N_(0)` When `t rarr infty`, Eq(ii) gives `N=(alpha)/(lambda).` |
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