Question

A solution containing active cobalt $^{60}_{27}Co$ having activity of $0.8 \, \mu Ci$ and decay constant $\lambda$ is injected in an animal?s body. If $1 \, cm^3$ of blood is drawn from the animal?s body after 10 hrs of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body ? ($1 \, Ci=3.7 \times 10^{10}$ decays per second and at $t=10 hrs e^{- \lambda t} =0.84$)

• A. 6 liters
• B. 7 liters
• C. 4 liters
• D. 5 liters

Answer 1

The Correct Option is D

We know that
$\frac{d N}{d t}=-N_{0} \lambda e^{-\lambda t}$
It is given that activity $=0.8 \mu Ci$. Therefore, $\lambda N_{0}=0.8 \mu Ci$.
Given: If $1\, cm ^{3}$ of blood is drawn from the animal's body after 10 hours of injection then activity was 300 decays per minute.
Let $V$ be the volume of blood flowing, then activity reduces as $\frac{1}{V}$. Thus,
$\frac{1}{V} \times \lambda N_{0} e^{-\lambda}=\frac{300}{60}$
Put $\lambda N_{0}=0.8 \mu Ci , e^{-\lambda t}=0.84, Ci =3.7 \times 10^{10}$ in above equation, we obtain
$\frac{1}{V} \times 0.8 \times 10^{-6} \times 3.7 \times 10^{10} \times 0.84=\frac{300}{60}$
$\Rightarrow \frac{1}{V} \times 24.86=5$
$\Rightarrow V=\frac{24.86}{5}=4.97 \sim 5$
Therefore, volume of blood flowing $=5$ litres