Question

A $100\, ml$ solution having activity $50\, dps$ is kept in a beaker. It is now constantly diluted by adding water at a constant rate of $10\, ml / \sec$ and $2\, m l / \sec$ 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:

• A. $10\left[1-\left(\frac{2}{7}\right)^{1 / 4}\right]$
• B. $10\left[1-\left(\frac{5}{7}\right)^{1 / 2}\right]$
• C. $50\left[1-\left(\frac{5}{7}\right)^{1 / 4}\right]$
• D. $50\left[1-\left(\frac{2}{7}\right)^{1 / 2}\right]$

Answer 1

Answer: (C)

The volume of liquid in beaker at any instant of time $t$ is
$V=100+8\, t$
The volume of liquid ejected in $t$ seconds is $2 t$
Number of active atoms being taken out is
$-d N=\frac{N}{V} 2 d t$
$\Rightarrow -\frac{d N}{d t}=\frac{2 N}{V}=\frac{2 N}{100+8 t}$
multiplying both sides with disintegration constant.
$-\lambda d N=\lambda N \frac{2 d t}{V}$
or $-d A=A \cdot \frac{2 d t}{V}$
where $A$ is activity of the solution.
The time taken for $10\, ml$ solution to come out is $5$ second.
$\Rightarrow \int\limits_{A_{0}}^{A} \frac{d A}{A}=\int\limits_{0}^{5} \frac{-2 i}{100+8 t} d t$
$\Rightarrow A=A_{0}\left(\frac{5}{7}\right)^{1 / 4}$
Thus required activity of the ejected solution is
$A-A_{0}=A_{0}\left[1-\left(\frac{5}{7}\right)^{1 / 4}\right]$