A small amount of solution containing $$ Na^{24} $$ radionuclide with activity $$ A= 2.0.10^3 $$ disintegration per second was injected into the bloodstream of a man. The activity of 1 $$ cm^3 $$ of the blood sample taken t= 5.0 hours later turned out to be A'= 16 disintegration per minute per $$ cm^3 $$ . The half-life of the radionuclide is T= 15 hours. Find the volume of the man's blood.
Let $$V$$ = volume of blood in the body of the human being. Then the total activity of the blood is $$A'V $$.
Assuming all this activity is due to the injected $$Na^{24} $$ and taking account of the decay of this radionuclide, we get
$$ V A' = A\ e^{-\lambda\ t} $$
Now $$ \lambda = \dfrac{ln\ 2}{15} $$ per hour ,
$$ t = 15 $$ hour
Thus $$ V = \dfrac{A}{A'} e^{-ln\ 2/3} = \dfrac{2.0 \times 10^{3}}{(16 / 60)} e^{-ln\ 2/3} cc = 5.95 $$ litre