The volume of blood in the patient is [(1.26)^3 = 2] — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Given Information (Passage Context) A solution containing $\ce{^{24}Na}$ with initial activity $A_0 = 1260\ \text{dps} = 1260 \times 60 = 75600\ \text{dpm}$ was injected. After 5 h, a blood sample has activity 15 dpm/mL. Half-life of $\ce{^{24}Na}$ = 15 h. Step 2 - Calculate Total Activity After 5 Hours Time elapsed = 5 h = $\frac{1}{3}$ of the half-life (since $t_{1/2} = 15$ h). $$A = A_0 \times \left(\frac{1}{2}\right)^{t/t_{1/2}} = 75600 \times \left(\frac{1}{2}\right)^{5/15} = \frac{75600}{2^{1/3}}$$ Given: $2^{1/3} = 1.26$ (or equivalently $(1.26)^3 = 2$): $$A = \frac{75600}{1.26} = 60000\ \text{dpm (total, in blood)}$$ Step 3 - Calculate Blood Volume The 60,000 dpm is uniformly distributed throughout the blood. The sample shows 15 dpm/mL: $$V = \frac{A_{\text{total}}}{A_{\text{per mL}}} = \frac{60000\ \text{dpm}}{15\ \text{dpm/mL}} = 4000\ \text{mL} = \boxed{4.0\ \text{L}}$$ Step 4 - Evaluate Options - **(A) 2.0 L**: Would require activity per mL to be 30 dpm. Incorrect. - **(B) 3.0 L**: Would require 20 dpm/mL. Incorrect. - **(C) 4.0 L**: Matches our calculation. **Correct.** - **(D) 5.0 L**: Would require 12 dpm/mL. Incorrect. $$\boxed{\text{Answer: C — Blood volume = 4.0 L}}$$