A) litres
B) litres
C) litres
D) litres
Correct Answer: D
Solution :
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 \[V=5litre\]You need to login to perform this action.
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