Nuclei of a 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=2N_0lambda`, calculate the number of nuclei of A after one half-life of A, and also the limiting value of N as `trarroo`.

Answer

(i) `1/(lambda) {alpha - (alpha - lambdaN_(0))e^(-lambdat)]` (ii) (A) `3/2N_(0)` B `2N_(0)`

Explanation

(i) Let at time t, number of radioactive nuclei are N. Net rate of formation of nuclel of A
`(dN)/(dt) = alpha - lambda N rArr (dN)/(alpha - lambda N) = dt rArr underset(N_(0))overset(N) int (dN)/(alpha - lambdaN) = underset(0)overset(t)int dt`
Solving this equation, we get
`N = (1)/(lambda) [alpha - (alpha - lambda N_(0))e^(-lambda t)]` .....(i)
(ii) (i) Substituting `alpha = 2 lambda N_(0)` and `t - t_(1//2) = (ln(2))/(lambda) 1n`
equation (i) we get, `N = (3)/(2) N_(0)`
(ii) Substituting `alpha = 2 lambdaN_(0)` and `t rarr oo` in equation (i), we get
`N = (alpha)/(lambda) = 2N_(0) rArr N = 2N_(0)`