Question

A sample of ${}^{14}{CO}_2$ was to be mixed with ordinary ${CO}_2$ for a biological tracer experiment. In order that ${10}^3{cm}^3$ of the diluted gas at NTP should have ${10}^4$ $dis/min$, how many $\mu Ci$ of radiocarbon$-14$ are needed to prepare $60L$ of the diluted gas?

Answer 1

$10{cm}^3$ of the diluted gas at NTP $={10}^{4}dis/min$
$=\frac{{10}^4}{60}dpsl$
$\therefore 60L(60,000{cm}^3)$ of the dilute gas at NTP $=\frac{{10}^4\times 60,000}{60\times 10}dps$
Thus, number of $\mu Ci$ of ${}^{14}{CO}_2$ needed $=\frac{{10}^4\times 60,000}{60\times 10\times 3.7\times {10}^4}(1\mu Ci=3.7\times {10}^{4}dps)$
$=27.03\mu Ci$