A small quantity of solution containing  $N{a}^{24}$ radionuclide (with a half life of 15 hours) of activity 1.0$\mu$  curie is injected into the blood of a person. A 1 cm³ blood sample shows 296 disintegrations per minute after 5 hours. Assuming uniform spreading of the radionuclide in the person’s blood, estimate the total volume of his blood. (in litres)

Consider a 1 cm³ sample of the blood. If its initial activity was A₀, then the activity A  after time t will be  $A=A_0{e}^{-\lambda t}$. In terms of half life  $T_{1/2}$, the expression can be  written as  $A=A_0{\left(\frac{1}{2}\right)}^{t/T_{1/2}}$, i.e.,  $\frac{296}{60s}=A_0{\left(\frac{1}{2}\right)}^{5hr/15hr}$ or   $A_0={\left(\frac{296}{60s}\right)}^{2^{1/3}}=6.22s^{-1}.$
Since the total initial activity is given to be  $A_0^{\text{'}}=1.0\times {10}^{-6}$  curie   $=1.0\times {10}^{-6}\times (3.7\times {10}^{10}\text{\hspace{0.17em}disintegration/second)=3.7}\times {\text{10}}^{\text{4}}{\text{s}}^{\text{-1}},$
 $\therefore$ The total volume of the blood is $V=\frac{\text{Total activity}}{{\text{activity\hspace{0.17em}per\hspace{0.17em}cm}}^{\text{3}}}=\frac{3.7\times {10}^4{s}^{-1}}{6.22s^{-1}{\text{\hspace{0.17em}cm}}^{\text{3}}}$

$=5.95\times {10}^{3}c{m}^3=5.95litres.$