Nucleus `A` decays to `B` with decay constant `lambda_(1)` and `B` decays to `C` with decay constant `lambda_(2)`. Initially at `t=0` number of nuclei of `A` and `B` are `2N_(0)` and `N_(0)` respectively. At `t=t_(o)`, no. of nuclei of `B` is `(3N_(0))/(2)` and nuclei of `B` stop changing. Find `t_(0)`?A. `(1)/(lambda_(1))ln((2lambda_(1))/(3lambda_(2)))`B. `(1)/(lambda_(1))ln((8lambda_(1))/(3lambda_(2)))`C. `(1)/(lambda_(1))ln((7lambda_(1))/(3lambda_(2)))`D. `(1)/(lambda_(1))ln((4lambda_(1))/(3lambda_(2)))`

Nucleus `A` decays to `B` with decay constant `lambda_(1)` and `B` dec

1.	Nucleus `A` decays to `B` with decay constant `lambda_(1)` and `B` decays to `C` with decay constant `lambda_(2)`. Initially at `t=0` number of nuclei of `A` and `B` are `2N_(0)` and `N_(0)` respectively. At `t=t_(o)`, no. of nuclei of `B` is `(3N_(0))/(2)` and nuclei of `B` stop changing. Find `t_(0)`?A. `(1)/(lambda_(1))ln((2lambda_(1))/(3lambda_(2)))`B. `(1)/(lambda_(1))ln((8lambda_(1))/(3lambda_(2)))`C. `(1)/(lambda_(1))ln((7lambda_(1))/(3lambda_(2)))`D. `(1)/(lambda_(1))ln((4lambda_(1))/(3lambda_(2)))`
Answer» Correct Answer - D `(dN_(B))/(dt)=lambda_(1)N_(A)-lambda_(2)N_(B)` `0= lambda_(1)(2N_(0))e^(-lambda t_(0))-lambda_(2)((3N_(0))/(2)), t_(0)=(1)/(lambda_(1)) ln((4 lambda_(1))/(3 lambda_(2)))`

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