Two radioactive nuclei `P` and `Q`, in a given sample decay into a stable nucleus `R`. At time `t = 0`, number of `P` species are `4 N_0` and that of `Q` are `N_0`. Half-life of `P` (for conversation to `R`) is `1mm` whereas that of `Q` is `2 min`. Initially there are no nuclei of `R` present in the sample. When number of nuclei of `P` and `Q` are equal, the number of nuclei of `R` present in the sample would be :

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)`