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 `4N_0` and that of Q are `N_0`. Half-life of P (for conversion to R) is 1 min whereas that of Q is 2min. 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 -

(A) `2N_0`
(B) `3N_0`
(C) `(9N_0)/2`
(D) `(5N_0)/2`

Options

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

Explanation

`{:(,P,Q),("No. of nuclei, at t = 0",4N_(0),N_(0)),("Half - life","1 min","2 min"),("No. of nuclei after",N_(p),N_(Q)):}`
time t
Let 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 mintues, 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)`