Question

A small amount of solution containing ${}^{24}\text{Na}$ radio nuclide with activity $A=2\times {10}^3$ dps was administered into blood of a patient in hospital. After 5 hour, a sample of the blood drawn out from the patient showed an activity of 16 dpm per cc. $t_{1/2}$ for ${}^{24}\text{Na}=15hr$. The activity of blood sample drawn after a further time of 5 hr is :

• A. $0.2118dpmpermL$
• B. $0.4232dpspermL$
• C. $0.4232dpmpermL$
• D. $0.2118dpspermL$

Answer 1

The correct option is D $0.2118dpspermL$

$Letvolumeofbloodofpatient=VmL$
(a) $Rateofdecay{}^{24}\text{Na}=2\times {10}^{3}dps=2\times 60\times {10}^{3}dpminjectedinVmLblood$
$r_{0}of{}^{24}\text{Na}afterT=5hr=16dpm/mL$
$=16\times VdpmforVmLblood$
$\because r\propto N;\therefore \frac{r_0}{r}=\frac{N_0}{N}$
Now, $t=\frac{2.303}{\lambda }\log \frac{N_0}{N}$
$5=\frac{2.303\times 5}{0.693}\log \frac{120\times {10}^3}{16\times V}$
$\therefore V=5.95\times {10}^{3}mL$
(b) Let activity $\left(A\right)$ of blood after 5 hour more, i.e., $t=10hrs$
$\because t=\frac{2.303}{\lambda }\log \frac{r_0}{r}$
$10=\frac{2.303\times 5}{0.693}\log \frac{120\times {10}^3}{A}$
$\therefore A=75.6\times {10}^{3}dpmforVmL$
$A=\frac{75.6\times {10}^3}{5.95\times {10}^3}dpmfor1mL$
$A=12.71dpmpermL=0.2118dpspermL$