Question

A small amount of solution containing $N{a}^{24}$ radio nuclide with activity A=2 X${10}^3$ dps was dministered into blood of a patient in a hospital . After 5 hr, a sample of the blood drawn out from lthe patient showed an activity of 16 dpm per c.c $t_{\frac{1}{2}}$ for $N{a}^{24}$ =15 . Find

a. Volume of the blood in the patient
b. Activity of blood sample drawn after a further time of 5 hr.

Answer 1

Let V mL blood is present in patient

a. $r_0$ of $N{a}^{24}$ = 2 X ${10}^3$ dps

= 2 X ${10}^3$ X 60 dpm

= 120 X ${10}^3$ dpm for V mL blood

r of $N{a}^{24}$=16 dpm m$L^{-1}$ at t= 5hr

= 16 X V dpm/V mL

$\frac{r_0}{r}$ =$\frac{N_0}{N}$

$\therefore$ $\frac{N_0}{N}$ =$\frac{120X{10}^3}{16V}$ and t= $\frac{2.303}{K}$log$\frac{N_0}{N}$

5=$\frac{2.303X15}{0.693}$log$\frac{120X{10}^3}{16V}$

$\therefore$ V=5.93 X${10}^3$ mL

b. Activity of blood sample after 5 hr more, i..e.,

t=10 hr

t= $\frac{2.303}{K}$log$\frac{N_0}{N}$

10= $\frac{2.303X15}{0.693}$log$\frac{120X{10}^3}{N}$

$\therefore$ N=75.6X${10}^3$ dpm per 5.95X ${10}^3$ mL

=$\frac{75.6X{10}^3}{5.96X{10}^3}$ dpm per mL

=12.71 dpm per mL

=$\frac{12.71}{60}$=0.2118 dps per mL