Question

${}^{131}I$ is an isotope of Iodine that $\beta -$ decays to an isotope of Xenon with a half-life of $8\text{days}$. A small amount of a serum labelled with ${}^{131}I$ is injected into the blood of a person. The activity of the amount of ${}^{131}I$ injected was $2.4\times {10}^5\text{Becquerel (Bq)}$. It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After $11.5\text{hours},2.5\text{ml}$ of blood is drawn from the person's body, and gives an activity of $115\text{Bq}$. The total volume of blood in the person's body, in $\text{liters}$ is approximately _______.

Answer 1

Given:
Half life, $t_{1/2}=8\text{days}=8\times 24\text{hr}$

$R_0=2.4\times {10}^5\text{Bq}$

We know that for half life,

$\begin{array}{}{t}_{1/2}=\frac{\ln 2}{\lambda }=\frac{0.693}{\lambda }& \text{(1)}\end{array}$

Also, $\frac{R}{R_0}=e^{-\lambda t}$

On rearranging it can be written as

$\begin{array}{}\ln \frac{R_0}{R}=\lambda t& \text{(2)}\end{array}$

Using (1) and (2), we get

$\frac{0.693}{t_{1/2}}\times t=\ln \frac{R_0}{R}$

$\frac{0.692}{8\times 24}\times 11.5=\ln \frac{2.4\times {10}^5}{R}$

$\frac{2.4\times {10}^5}{R}=e^{0.041}=1.042$

$R=\frac{2.4\times {10}^5}{1.041}=2.3\times {10}^5\text{Bq}$

Given that,
$115\text{Bq}$ is in volume = $2.5\text{ml}$

$\therefore 2.3\times {10}^5\text{Bq}$ is in volume

$=\frac{2.5}{115}\times 2.3\times {10}^5=5000\text{ml}$

$\therefore \text{Volume of blood}=5\text{liters}$

Hence, $5$ is the correct answer.