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 ?

Explanation

Suppose their concentration become equal after `n_(1)` half-lives of A and `n_(2)` half lives of B.
Amount of A left after `n_(1)` half lives `=(1)/(2^(n_(1)))[A_(0)]`
Amount of B left `n_(2)` half lives `=(1)/(2^(n_(2)))[B_(0)]`
Now, as the amounts become equal,
`(1)/(2^(n_(1)))[A_(0)]=(1)/(2^(n_(2)))" or "(1)/(2^(n_(1)))4[B]_(0)=(1)/(2^(n_(2)))[B_(0)]" or "(4)/(2^(n_(1)))=(1)/(2^(n_(2)))`
or `(2^(n_(1)))/(2^(n_(2)))=4" or "2^(n_(1)-n_(2))=2^(2)" or "n_(1)-n_(2)=2" "...(i)`
Suppose concentrations become equal after time t, then
`t=n_(1)xxt_(1//2(A))=n_(2)xxt_(1//2(B))`
or `(n_(1))/(n_(2))=(t_(1//2(B)))/(t_(1//2(A)))=(15)/(5)=3" or "n_(1)=3n_(2)" "...(ii)`
Substituting in (i),
`3n_(2)-n_(2)=2" or "2n_(2)=2" or "n_(2)=1`
`:." "t=n_(2)xxt_(1//2(B))=1xx15" min "=15 min`