Question

A sample of ${}^{14}{CO}_2$ was to be mixed with ordinary ${CO}_2$ for a biological tracer experiment. In order that $10c{m}^3$ of the diluted gas at STP should have ${10}^4$ dis/min, how many $\mu Ci$ of radioactive carbon-14 are needed to prepare $0.06m^3$ of the diluted gas?

• A. $27.03\mu Ci$
• B. $22.03\mu Ci$
• C. $57.03\mu Ci$
• D. None of these

Answer 1

The correct option is A $27.03\mu Ci$

$10c{m}^3$ of the diluted gas has activity

$={10}^{4}dis/min=\frac{{10}^4}{60}$ dps

$\therefore 0.06m^3(=60,000c{m}^3)$ of the diluted gas has activity

$\frac{{10}^4}{60}\times \frac{60000}{10}$ dps

Hence, number of $\mu Ci$ needed $={10}^4\times \frac{60000}{600\times 3.7\times {10}^4}(1\mu Ci=3.7\times {10}^{4}dps)$

$=27.03\mu Ci$