Nucleus A decays to B with decay constant `lamda_(1)` and B decuys to C with decay constants `lamda_(2)`. Initially at t=0, number of nuclei of A and B are `2N_(0)` and `N_(0)` respectively. At some instant `t=t_(0)`, number of nuclei of B stop changing. Find `t_(0)` if at this intant number of nuclei of B is `(3N_(0))/(2)`

Options

A. `t_(0)=(1)/(lamda_(1))ln""(4lamda_(1))/(3lamda_(2))`
B. `t_(0)=(1)/(lamda_(2))ln""(4lamda_(1))/(3lamda_(2))`
C. `N_(A)=(3N_(0))/(2)(lamda_(2))/(lamda_(1))" to "t=t_(0)`
D. `N_(A)=(2N_(0))/(3)(lamda_(2))/(lamda_(1))" at "t=t_(0)`

Answer

option 4