A solution containing active cobalt $_{27}^{60} C o$ having activity of $0.8 μ C i$ and decay constant $λ$ is injected in an animal's body. If $1 c m^{3}$ 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? ( $1 C i = 3.7 \times 10^{10}$ decay per second and at $t = 10 h r s$ $e^{- λ t} = 0.84$ )

Question

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Answer 1

The activity equation can be written as
$- \frac{d N}{d t} = λ N_{o} e^{- λ t}$
given that
$λ N_{o} = 0.8 μ C_{i}$
Putting the values,
$λ N_{o} = 2.96 \times 1 0^{4}$
Let the volume of the blood flowing be V,
the activity would reduce by a factor of $\frac{1 0 ^{- 3}}{V}$
Hence
$\frac{λ N _{o} 1 0 ^{- 3}}{V} e^{- λ t} = 300 / 60$ (Both R.H.S. and L.H.S. are decay/s)
Putting the values of $e^{- λ t}$ and $λ N_{o}$ we get
$V = 5 l i t r e$

Answer 2

The activity equation can be written as

- \frac{d N}{d t} = λ N_{o} e^{- λ t}

given that

λ N_{o} = 0.8 μ C_{i}

Putting the values,

λ N_{o} = 2.96 \times 1 0^{4}

Let the volume of the blood flowing be V,

the activity would reduce by a factor of

\frac{1 0 ^{- 3}}{V}

Hence

\frac{λ N _{o} 1 0 ^{- 3}}{V} e^{- λ t} = 300 / 60

(Both R.H.S. and L.H.S. are decay/s)

Putting the values of

e^{- λ t}

and

λ N_{o}

we get

V = 5 l i t r e