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

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).`