A man weighing 80 kg was injected with 5 ml of water containing tritium giving counts per minute. Af — Nuclear Chemistry and Radioactivity Chemistry Question
Question
A man weighing 80 kg was injected with 5 ml of water containing tritium giving $9 \times 10^9$ counts per minute. After some time, the tritiated water had equilibrium with the body water of the patient. A 1.0 ml sample of plasma water then showed an activity of $2.25 \times 10^5$ counts per minute. The mass per cent of water in the body is
💡 Solution & Explanation
The total initial activity introduced is $9 \times 10^9 \text{ cpm}$. At equilibrium, the activity is distributed uniformly at $2.25 \times 10^5 \text{ cpm/mL}$. The total volume of water in the body is $V = \frac{9 \times 10^9}{2.25 \times 10^5} = 4 \times 10^4 \text{ mL} = 40 \text{ L} \approx 40 \text{ kg}$. The mass percent of water in the 80 kg man is $\frac{40}{80} \times 100\% = 50\%$. Therefore, correct answer is 0050.