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.

Options

A. `(5N_(0))/(2)`
B. `2N_(0)`
C. `3N_(0)`
D. `(9N_(0))/(2)`

Answer

D

Explanation

Initeally `Pto4N_(0),QtoN_(0)`
Half-life, `T_(P)=1min,T_(Q)=2min`
Let after time t, number of nuclei of P and Q equal
`i.e.,(4N_(0))/(2^(t//1))=(N_(0))/(2^(t//2))" "(becauseN=(N_(0))/(2^(t//T_(1)//2)))`
`or(4)/(2^(t//2))=1or t=4min`
So, at t= 4 min
`N_(P)=((4N_(0)))/(2^(4//1))=(N_(0))/(4)`
and at `t=4 min,N_(Q)=(N_(0))/(2^(4//2))=(N_(0))/(4)`
So, number of nuclei of R in teh sample
`=(4N_(0)-(N_(0))/(4))+(N_(0)-(N_(0))/(4))=(9N_(0))/(2)`