Nucleus A decays into B with a decay constant `lamda_(1)` and B further decays into C with a decay constant `lamda_(2)` . Initially, at t = 0, the number of nuclei of A and B were `3N_(0)` and `N_(0)` respectively. If at t = `t_(0)` number of nuclei of B becomes constant and equal to `2N_(0)`, then

Options

A. `(N_(0))/(4)`
B. `(lamda_(A))/(lamda_(B))N_(0)e^(-lamda_(B)t`
C. `(lamda_(A))/(lamda_(B)) (N_(0))/(4)`
D. None of these

Answer

C

Explanation

We have, `lamda_(B)N_(B)=lamda_(A)N_(A)`
`therefore" "N_(B)=(lamda_(A))/(lamda_(B)).N_(A)`
The give time is equivalent to two half-lives of A. Hence,
`N_(A)=(N_(0))/(4)" "(thereforeN=N_(0)((1)/(2))^(2)`
`N_(B)=(lamda_(A))/(lamda_(B))((N_(0))/(4))`