Nuceli `A` and `B` convert into a stable nucleus `C`. Nucleus A is converted into `C` by emitting `2 alpha`particels and `3 beta`-particles. Nucleus `B` is converted into `C` by emitting one `alpha`-particle and `5 beta`-particles. At time `t=0`, nuclei of A are `4N_(0)` and nuceli of `B` are `N_(0)`. Initially, number of nuclei of `C` are zero. Half-life of `A` (into conservation of `C`) is `1 min` and that of `B` is `2 min`. Find the time (in minutes) at which rate of disintegration of `A` and `B` are equal.

Answer

6

Explanation

We have to find the time at which
`lambda_(A)N_(A)=lambda_(B) N_(B)`
`((ln 2)/(T_(A)))(4N_(0) e^(-lambda_(A) t))=(N_(0))(ln 2)/(T_(B))(e^(-lambda_(B) t))`
`e^((lambda_(A)-lambda_(B))t)=8`
`(lambda_(A)-lambda_(B))t=ln 8=9(ln 2)`
`((ln2)/(1)-(ln2)/(2))t=3 ln(2) rArr t=6` min`.