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`.

Explanation

(a) `{:(N_(0),0,t=0),(Ararr,B,),(N,,t=t):}`
`(dN)/(dt) = alpha - lambdaN`
`underset(N_(0))overset(N)int (dN)/((alpha - lambdaN)) = underset(0)overset(t)int dt`
`(|ln(alpha - lambdaN)|N_(0)^(N))/(-lambda)=t`
`ln(alpha - lambdaN)-ln(alpha - lambdaN_(0)) =- lambdat`
`ln((alpha - lambdaN)/(alpha - lambdaN_(0))) =- lambdat`
`(alpha - lambdaN)/(alpha - lambdaN_(0))=e^(-lambdat)`
`alpha - lambdaN = (alpha - lambdaN_(0))e^(-lambdat)`
`alpha = (1)/(lambda) [alpha - (alpha - lambdaN_(0))e^(-lambdat)]`
(b) `alpha = 2N_(0)lambda, t = (ln2)/(lambda) rArr e^(-lambdat) = (1)/(2)`
`N = (1)/(lambda) [2N_(0)lambda-(2N_(0)lambda - N_(0)lambda).(1)/(2)]`
`=(1)/(lambda)[2N_(0)lambda-(N_(0)lambda)/(2)] = (3N_(0))/(2)`
`t rarr oo, e^(-lambdat) = 0`
`= (1)/(lambda)[2N_(0)lambda] = 2N_(0)`