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 = $$3.7 \times 10^{10}$$ decay per second and at t = 10 hrs the value of $$e^{-\lambda t}$$ = 0.84)

We begin with the activity that was actually injected into the animal. It is stated to be $$0.8\;\mu\text{Ci}$$. The symbol $$\mu$$ means “micro”, i.e. $$10^{-6}$$, so in curies this is

$$A_0 = 0.8\times10^{-6}\;\text{Ci}$$

The activity falls exponentially with time. The law of radioactive decay is

$$A = A_0\,e^{-\lambda t}$$

where $$A$$ is the activity after the elapsed time $$t$$ and $$\lambda$$ is the decay constant. For the given interval, the numerical value of the exponential term is supplied:

$$e^{-\lambda t}=0.84 \quad\text{at}\; t = 10\;\text{h}$$

Hence the total activity present in the body after ten hours becomes

$$A = A_0\,e^{-\lambda t}=0.8\times10^{-6}\;\text{Ci}\times0.84 =0.672\times10^{-6}\;\text{Ci}$$

The activity that we can measure experimentally is expressed in “decays per minute” (dpm), therefore we now convert the curie value into that unit. The fundamental relation is

$$1\;\text{Ci}=3.7\times10^{10}\;\text{decays per second (dps)}$$

and one minute contains $$60$$ s, so

$$1\;\text{Ci}=3.7\times10^{10}\times60=2.22\times10^{12}\;\text{decays per minute (dpm)}$$

Because a micro-curie is $$10^{-6}$$ of a curie, the activity that we just found corresponds to

$$A = 0.672\times10^{-6}\;\text{Ci}\times2.22\times10^{12}\;\dfrac{\text{dpm}}{\text{Ci}} =0.672\times2.22\times10^{6}\;\text{dpm} =1.492\times10^{6}\;\text{dpm}$$

This entire activity of $$1.492\times10^{6}$$ decays per minute is now assumed to be uniformly distributed throughout the animal’s blood. Let the blood volume be $$V\;\text{cm}^3$$ (remember that $$1\;\text{L}=1000\;\text{cm}^3$$).

The activity contained in each cubic centimetre of blood would therefore be

$$\dfrac{A}{V}=\dfrac{1.492\times10^{6}\;\text{dpm}}{V}$$

After ten hours, a 1 cm$$^3$$ sample of blood was taken and its activity was measured to be

$$300\;\text{dpm}$$

Equating the theoretical activity per cm$$^3$$ to the observed value, we write

$$\dfrac{1.492\times10^{6}}{V}=300$$

Solving for $$V$$ gives

$$V=\dfrac{1.492\times10^{6}}{300} =4.973\times10^{3}\;\text{cm}^3$$

Converting cubic centimetres into litres,

$$V= \dfrac{4.973\times10^{3}}{1000}\;\text{L} =4.973\;\text{L}\approx5\;\text{L}$$

Hence, the correct answer is Option D.