Question

A solution containing active cobalt $\_{27}^{60}Co$ having an activity of $0.8 \, μCi$ and decay constant $\lambda $ is injected in an animal's body. When $1 \, cm^{3}$ blood is drawn from the animal's body after $10 \, h$ of injection, the activity was found to be $300 \, dpm$ (decays per minute), then the total volume of blood in the animal's body is close to $\left[\right.1 \, Ci=3.7\times 10^{10} \, dps$ (decays per second) and at $t=10 \, h$ the value of $e^{- \lambda t}=0.84\left]\right.$

• A. $4L$
• B. $6L$
• C. $5L$
• D. $7L$

Answer 1

Answer: (C)

Let the total volume of blood is $\textit{v}$ , initial activity $A_{0}=0.Q \, \mu C_{i}$ its activity at time $\textit{t}$ , $A = A_{o} e^{- \lambda t}$ , the activity of $\text{x}$ volume, $A^{1}=\left(\frac{A}{V}\right)X=X\left(\frac{A_{o}}{V}\right)e^{- \lambda t}$
$\Rightarrow V=X\left(\frac{A_{0}}{A^{1}}\right) e^{-\lambda t} \Rightarrow V=\left(1 cm ^{3}\right)\left(\frac{8 \times 10^{-7} \times 3.7 \times 10^{10}}{\frac{300}{60}}\right)(0.84)$
$=4.97 \times 10^{3} \quad cm ^{2}=4.97 \quad$ liter $=5$ L.