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

Answer

A

Explanation

Amount of A left in `n_(1)` half-lives `=(1/2)^(n_(1)) [A_(0)]`
Amount of B left in `n_(2)` half-lives `=(1/2)^(n_(2)) [B_(0)]`
At the end `([A_(0)])/(2^(n_(1)))=([B_(0))]/(2^(n_(2))) rArr 4/(2^(n_(1)))=(1)/(2^(n_(2))), [A_(0)]=4[B_(0)]`
`therefore 2^(n_(1))/(2^(n_(2))=4 rArr 2^(n_(1)-n_(2)) =(2)^(2) therefore n_(1)-n_(2)=2`
Also, `t=n_(1) xx t_(1//2(A)) t=n_(2) xx t_(1//2(B))` (Let, concentration of both become equal after time t)
`therefore n_(1) xx t_(1//2(A))/(n_(2) xx t_(1//2(B))=1 rArr (n_(1) xx5)/(n_(2) xx 15) =1 rArr n_(1)/n_(2)=3`
From Eqs (1) and (2), we get
`n_(1)=3, n_(2)=1`
`t=3 xx 5=15" minutes"`