Two substances `A` and `B` are present such that `[A_(0)] = 4[B_(0)]` and half life `A` is `5 min` and that of `B` is `15 min`. If they start decaying at the same time following first order kinetics now much time later will take if the concentration of both of them would be same?

Options

A. `15` minute
B. `10` minute
C. `5` minute
D. `12` minute

Answer

a

Explanation

Amount of `A` left in `n_(1)` halves=`([A_(0)])/(2^(n_(1)))`
Amount of `B` left in `n_(2)` halves=`([B_(0)])/(2^(n_(2)))`
If `([A_(0)])/(2^(n_(1)))=([B_(0)])/(2^(n_(2)))` after `A` decays for `n_(1)` halves and `B` decays for `n_(2)` halves.
`:' [A_(0)]=4[B_(0)]`
`:. 4=2^(n_(1))/2^(n_(2))` or `2^(2)=2^(n_(1)n_(2))`
`:. n_(1)-n_(2)=2`
`:. n_(2)=n_(1)-2`
Also `T=n_(1)xxt_(1//2) A` and `T=n_(2)xxt_(1//2) B`
`:. (n_(1)xxt_(1//2)xxA)/(n_(2)xxt_(1//2)xxB)=1` or `n_(1)/n_(2)=15/5`
or `n_(1)=3n_(2)`
Thus `n_(1) =3` and `n_(2) =1`.
Thus `T=3xx5=15 "minute"`.