A stable nuclei C is formed from two radioactive nuclei A and B with decay constant of `lambda_1` and `lambda_2` respectively. Initially, the number of nuclei of A is `N_0` and that of B is zero. Nuclei B are produced at a constant rate of P. Find the number of the nuclei of C after time t.

Answer

`N_(0)(1-e^(-lambda_(1)t))+P(t+(e^(-lambda_(2)t-1))/(lambda))`

Explanation

`(dN_(B))/(dt) =P-lambda_2 N_B`
`int_(0)^(N_2) (dN_(B))/(P)-lambda_(2) (N_(B))/(P)=int_(0)^(t)dt`
`ln((P-lambda_(2) (N_(B)))/(P))-lambda_2t`
`N_B=P(1 -e^(-lambda_(2)t)`
The number of nuclei of A after time `t` is `N_(A)=N_(0) e^(-lambda t)`. Thus,
`(dN_(C))/(dt) =lambda_(1)N_(A) +lambda_(2)N_(B)`
`=+lambda_(1)N_(0) e^(-lambda t)+P(1-e^(-lmabda_(2) t)`
`N_(C)+N_(0)(1-e^(- lmabda_1 t)) + P(t+(e^(-lambda_2 t-1))/(lambda_2))`