Question

A small quantity of solution containing $24Na$ radionuclide (half-life 15 h) of activity 1.0 microcurie is injected into the blood of a person. A sample of the blood of volume $1c{m}^3$ taken after 5 h shows an activity of 296 disintegrations $mi{n}^{-1}$. Determine the total volume of blood in the body of the person. Assume that the radioactivity solution mixes uniformly in the blood of the person. (1 curie $=3.7\times {10}^{10}disintegrations{s}^{-1}$)

Answer 1

Initial activity of ${}^{24}Na$,
$A=\frac{dN}{dt}=1.0\mu C$
$=1.0\times {10}^{-6}\times 3.7\times {10}^{10}$
$=3.7\times {10}^{4}disintegrations{s}^{-1}$

Half-life, $T=15h=15\times 3600s$

Initial activity,
$A=\frac{dN}{dt}=\lambda N_0$

$3.7\times {10}^4=\frac{0.693}{15\times 3600}{N}_{)}$

$\therefore N_0=\frac{3.7\times {10}^4\times 15\times 3600}{0.693}$$=2.883\times {10}^9$

Let the number of radioactive nuclei present after 5 h be N' in $1c{m}^3$ of sample of blood. Then,
$\frac{dN}{dt}=\lambda N^{\prime }\Rightarrow \frac{dN}{dt}\frac{296}{60}=\frac{0.693}{15\times 3600}{N}^{\prime }$

$\therefore N^{\prime }=\frac{296\times 15\times 3600}{60\times 0.693}=3.844\times {10}^5$

If $N_0^{\prime }$ is initial number of radioactive nuclei is $1c{m}^3$ of sample, then
$\frac{N^{\prime }}{N_0^{\prime }}={\left(\frac{1}{2}\right)}^{t/T}$

$N_0^{\prime }=(2{)}^{t/T}{N}^{\prime }$$=(2{)}^{5/15}{N}^{\prime }=(2{)}^{1/3}\times 3.844\times {10}^5$

Let $y=(2{)}^{1/3}$
$\therefore logy=\frac{1}{3}log2=\frac{1}{3}\times 0.3010=0.1003$

$\Rightarrow y=Antilog\left(0.1003\right)=1.2598$

$\therefore N_0^{\prime }=1.2598\times 3.844\times {10}^5$ $=4.843\times {10}^5$

Volume of blood $=\frac{N_0}{N_0^{\prime }}=\frac{2.883\times {10}^9}{4.833\times {10}^5}=0.595\times {10}^{4}c{m}^3$$=5.95L$