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A solution containing active cobalt ${ }_{27}^{60} \mathrm{Co}$ having activity of $0.8 \mu \mathrm{Ci}$ and decay constant $\lambda$ is injected in an animal's body. If $1 \mathrm{~cm}^3$ of blood is drawn from the animal's body after $10 \mathrm{hrs}$ of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body? $\left(1 \mathrm{Ci}=3.7 \times 10^{10}\right.$ decay per second and at $\left.\mathrm{t}=10 \mathrm{hrs}^{-\lambda \mathrm{t}}=0.84\right)$
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5 litres
5 litres
Let initial activity $=\mathrm{No}=0.8 \mu \mathrm{ci}$ $0.8 \times 3.7 \times 10^4 \mathrm{dps}$
Activity in $1 \mathrm{~cm}^3$ of blood at $\mathrm{t}=10 \mathrm{hr}$, $\mathrm{n}=\frac{300}{60} \mathrm{dps}=5 \mathrm{dps}$
$\mathrm{N}=$ Activity of whole blood at time $\mathrm{t}=10 \mathrm{hr}$. Total volume of the blood in the person, V $=\frac{N}{n}$
$=\frac{N_0 e-\lambda t}{n}=\frac{0.8 \times 3.7 \times 10^4 \times 0.7927}{5} \cong 5$ litre
Activity in $1 \mathrm{~cm}^3$ of blood at $\mathrm{t}=10 \mathrm{hr}$, $\mathrm{n}=\frac{300}{60} \mathrm{dps}=5 \mathrm{dps}$
$\mathrm{N}=$ Activity of whole blood at time $\mathrm{t}=10 \mathrm{hr}$. Total volume of the blood in the person, V $=\frac{N}{n}$
$=\frac{N_0 e-\lambda t}{n}=\frac{0.8 \times 3.7 \times 10^4 \times 0.7927}{5} \cong 5$ litre
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