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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