Question

A solution containing active cobalt ${}_{27}^{60}Co$ 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? ($1Ci=3.7\times {10}^{10}$ decay per second and at $t=10hrs$ $e^{-\lambda t}=0.84$)

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

Answer 1

Answer: (D)

The activity equation can be written as

$-\frac{dN}{dt}=\lambda N_o{e}^{-\lambda t}$

given that

$\lambda N_o=0.8\mu C_i$

Putting the values,

$\lambda N_o=2.96\times {10}^4$

Let the volume of the blood flowing be V,

the activity would reduce by a factor of $\frac{{10}^{-3}}{V}$

Hence

$\frac{\lambda N_o{10}^{-3}}{V}{e}^{-\lambda t}=300/60$ (Both R.H.S. and L.H.S. are decay/s)

Putting the values of $e^{-\lambda t}$ and $\lambda N_o$ we get

$V=5litre$