A) litres
B) litres
C) litres
D) litres
Correct Answer: D
Solution :
[d] 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 |
\[V=5litre\] |
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