4 litres
5 litres
6 litres
7 litres
The activity equation can be written as -dN/dt = λN0e-λt. Given that λN0 = 0.8 µCi. Putting the values, λN0 = 2.96 × 104. Let the volume of the blood flowing be V, the activity would reduce by a factor of 10-6V. Hence λN010-6Ve-λt = 300/60 (Both R.H.S. and L.H.S. are decay/s). Putting the values of e-λt and λN0 we get V = 5 litres