A 100ml solution having activity 50 dps is kept in a beaker It is now constantly diluted by adding water at a constant rate of `10ml//sec` and `2 ml//sec` of solution is constantly being taken out. Find the activity of 10 ml solution which is taken out, assuming half life to be effectively very large :

Answer

`A_(0)[1-((5)/(7))^(1//4)]` where `A_(0)=50dps`

Explanation

The volume of liquid in beaker at any instant of time is
`V=100+8t`
The volume of liquid ejected in `t` seconds is `2 t`
`:.` Number of active atoms being taken out is
`-(dN)/(V)2dt` or `-(dN)/(dt)=(2N)/(V)=(2N)/(100+8t)`
multiplying both sides with disintergration constant.
`-lambda dN=lambdaN(2dt) " " or -dA=A.(2dt)/(V)`
where `A` is activity of the solution. The time taken for `10 ml` solution to come out is `5` seconds.
or `underset(A_(0))overset(A)int(dA)/(A)=underset(0)overset(5)int(-2t)/(100+8t)dt" " or A=A_(0)((5)/(7))^(1//4)`
`:.` required activity of the ejected solution is `A_(0)-A=A_(0)[1-((5)/(7))^(1//4)]`
Ans. `A_(0)[1-((5)/(7))^(1//4)]`where `A_(0)=50dps`