InterviewSolution
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InterviewSolution
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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. |
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Answer» Correct Answer - `N_(0)(1-e^(-lambda_(1)t))+P(t+(e^(-lambda_(2)t-1))/(lambda))` `(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))` |
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