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

Answer

C

Explanation

Number of nuclei, at `t = o4N_(0)N_(0)` Half-life 1 min 2 min
Number of nuclei after time t `N_(p) N_(g)` Suppose, after t min the number of nuclei of P and Q are equal.
`therefore N_(p) = 4N_(0)((1)/(2))^(t//1)`
and `N_(Q) = N_(0)((1)/(2))^(t//2)`
As `N_(p) = N_(Q)`
`therefore 4N_(0)((1)/(2))^(t//1) = N_(0)((1)/(2))^(t//2)`
`(4)/(2^(t//1))= (1)/(2^(t//2)) or 4 = (2^(t))/(2^(t//2))`
or `4 = 2^(t//2) or 2^(2) = 2^(t//2)`
or `(t)/(2) = 2 or t = 4 min`
After 4 minutes, both P and Q have equal number of nuclei
`therefore` Number of nuclei of R
` = (4N_(0) -(N_(0))/(4)) +(N_(0) -(N_(0))/(4))`
` = (15N_(0))/(4) + (3N_(0))/(4) = (9N_(0))/(2)`