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

Let number of atoms becomes equal after time `t`
`N_(P) = N_(Q)`
`4N_(0)((1)/(2))^((t)/(1)) = N_(0)((1)/(2))^((t)/(2))`
`((1)/(2))^(t-(t)/(2)) = (1)/(4) = ((1)/(2))^(2)`
`((1)/(2))^((t)/(2)) = ((1)/(2))^(2) rArr (t)/(2) = 2 rArr t = 4min`
`N_(P) = 4N_(0)((1)/(2))^((4)/(1)) = (N_(0))/(4)`,
NUmber of `R` formed `= 4N_(0) - (N_(0))/(4) = (15N_(0))/(4)`
`N_(Q) = N_(0) ((1)/(2))^((4)/(2)) = (N_(0))/(4)`,
Number of `R` formed `= N_(0)-(N_(0))/(4) = (3N_(0))/(4)`
Number of nuclei of `R = (15N_(0))/(4) +(3N_(0))/(4) = (9N_(0))/(2)`