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)`?

Options

A. the value of `t_(0)"is"(1)/(lambda_(1))ln.(4)/(3)(lambda_(1))/(lambda_(2))`
B. the value of `t_(0)"is"(1)/(lambda_(2))ln.(4)/(3)(lambda_(1))/(lambda_(2))`
C. the value of `N_(A)` at `t_(0)"is""(3N_(0))/(2)(lambda_(2))/(lambda_(1))`
D. the value of `N_(A)` at `t_(0)"is""(2N_(0))/(3)(lambda_(2))/(lambda_(1))`

Answer

A, C