Question

A small amount of solution containing $N{a}^{24}$ radionuclide with activity $A={\mathrm{2.0.10}}^3$ disintegration per second was injected into the bloodstream of a man. The activity of 1 $c{m}^3$ of the blood sample taken t= 5.0 hours later turned out to be A'= 16 disintegration per minute per $c{m}^3$ . The half-life of the radionuclide is T= 15 hours. Find the volume of the man's blood.

Answer 1

Let $V$ = volume of blood in the body of the human being. Then the total activity of the blood is $A^{\prime }V$.

Assuming all this activity is due to the injected $N{a}^{24}$ and taking account of the decay of this radionuclide, we get
$V{A}^{\prime }=A{e}^{-\lambda t}$
Now $\lambda =\frac{ln2}{15}$ per hour ,

$t=15$ hour
Thus $V=\frac{A}{A^{\prime }}{e}^{-ln2/3}=\frac{2.0\times {10}^3}{\left(16/60\right)}{e}^{-ln2/3}cc=5.95$ litre