We have two radioactive nuclei `A` and `B`. Both convert into a stable nucleus `C`. Nucleus `A` converts into `C` after emitting two two `alpha`-particles and three `beta`-particles. Nucleus `B` converts into `C` after emitting one.
`alpha`-particle anf five `beta`-particles. A time `t = 0`, nuclei of `A` are `4N_(0)` and that of `B` are `N_(0)`. Half-life of `A`(into the conversion of `C`) is `1min` and that of `B` is `2min`. Initially number of nuclei of `C` are zero.
What are number of nuclei of `C`, when number of nuclei of `A` and `B` are equal ?

Options

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

Answer

C

Explanation

The no. of nuclei of A and B are equal apart `t=4` min and the number of nuclei of A and B are `(N_(0))/(4)` each.
Hence number of nuclei of C formed from `A=4N_(0)-(N_(0))/(4)=(15N_(0))/(4)` and number of nuclei of C formed from `B=N_(0)-(N_(0))/(4)=(3N_(0))/(4)`. Hence
total number of nuclei of C formed `(18N_(0))/(4)=(9N_(0))/(2)`