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 $5hour$, a sample of the blood drawn out from the patient showed an activity of $16dpm$ per $cc$. $t_{1/2}$ for ${}^{24}\text{Na}=15hr$. The activity of blood sample drawn after a further time of $5hr$ is :

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

Answer 1

The correct option is D $0.2118dpspermL$

Let volume of blood of patient be V mL.
Rate of decay of ${}^{24}\text{Na}=2\times {10}^{3}dps=2\times 60\times {10}^3$ dpm injected in V mL of blood
The rate $r_0$ of decay of ${}^{24}\text{Na}$ after $T=5hr=16dpm/mL$
$=16\times Vdpm$ for V mL blood
$\because r\propto N;\therefore \frac{r_0}{r}=\frac{N_0}{N}$
Now, the integrated rate law expression for the first order decay process is $t=\frac{2.303}{\lambda }\log \frac{N_0}{N}$
At the end of 5 hours $5=\frac{2.303\times 15}{0.693}\log \frac{120\times {10}^3}{16\times V}$
Therefore, total volume of blood is $V=5.95\times {10}^{3}mL$
Let $\left(A\right)$ be the activity of blood after 10 hours, i.e., $t=10hrs$
$\because t=\frac{2.303}{\lambda }\log \frac{r_0}{r}$
$10=\frac{2.303\times 15}{0.693}\log \frac{120\times {10}^3}{A}$
$\therefore A=75.6\times {10}^3$ dpm for V mL
$A=\frac{75.6\times {10}^3}{5.95\times {10}^3}$ dpm for 1 mL
$A=12.71\text{dpm per mL}=\frac{12.71}{60}=0.2118\text{dps per mL}$