The activity equation can be written as

\( \frac{dN}{dt} = \lambda N_{o}e^{-\lambda t}\)

Given:

\( \lambda N_{o} = 0.8\mu C_{i}\)

Substituting the values, we get:

\( \lambda N_{o} = 2.96 * 10^{4}\)

Let the volume of the blood flowing be V

The activity would reduce by a factor of\( \frac{10^{-3}}{V}\)

Therefore,

\( \frac{\lambda N_{o}10^{-3}}{V}e^{-\lambda t} = 300/60.\)

Now substituting the values of \( e^{-\lambda t} and \lambda N_{o},\) we get

V = 5 litre

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