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 ?

Options

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

Answer

A

Explanation

Amount of A left in `n_(1)` half-lives `=((1)/(2))^(n_(1))[A_(0)]`
Amount of B left `n_(2)` half-lives `=((1)/(2))^(n_(2))[B_(0)]`
At the end, `([A_(0)])/(2^(n_(1)))=([B_(0)])/(2^(n_(2)))`
`implies(4)/(2^(n_(1)))=(1)/(2^(n_(2))){[A_(0)]=4[B_(0)]}`
`:.(2^(n_(1)))/(2^(n_(2)))=4implies2^(n_(1)-n_(2))=(2)^(2)`
`:.n_(1)-n_(2)=2`
`:.n_(2)=(n_(1)-2)`
Also, `t=n_(1)xxt_(1//2(B))t=n_(2)xxt_(1//2)(B))`
(Let, concentration of both become equal after time `t`)
`:.(n_(1)xxt_(1//2(A)))/(n_(2)xxt_(1//2(B)))=1implies(n_(1)xx5)/(n_(2)xx15)=1implies(n_(1))/(n_(2))=3`
From Eqs. `(1)` and `(2)`, we get
`n_(1)=3`, `n_(2)=1`
`t=3xx5=15` minutes