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

Answer

B

Explanation

Initially, number of nuclei `P=4N_(0)`
number of nuclei`Q=N_(0)`
Half life of P, `T_(P)=1mi n`
Half life of Q, `T_(Q)=2mi n`
No. of nuclei of P after time t,
`n_(P)=4N_(0)(1/2)^(t//T_(P))=4N_(0)(1/2)^(t//1)`
No. of nuclei of Q after time t,
`n_(Q)=N_(0)(1/2)^(t//T_(Q))=N_(0)(1/2)^(t//2)`
Suppose number of nuclei P and Q in given sample are equal after time t, then `n_(P)=n_(Q)`
`4N_(0)(1/2)^(t//1)=N_(0)(1/2)^(t//2)`
`(1/2)^(t//2)=1/4=(1/2)^(2)`
or `t/2=2 or t=4 mi n`
`:. n_(P)=4N_(0)(1/2)^(4//1)=(N_(0))/4`
` and n_(Q)=N_(0)(1/2)^(4//2)=(N_(0))/4`
`:.` Population of R in the sample is
`=(4N_(0)-(N_(0))/4)+(N_(0)-(N_(0))/4)=(9N_(0))/2`