Activity of blood sample (in dpm per ml) drawn after a further time of 5 h is — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

**Step 1: Find the activity in blood at $t = 5$ h.** Na-24 has $t_{1/2} = 15$ h. Activity of Na-24 in blood at $t = 5$ h: $$A(5\text{h}) = 1260 \times \left(\frac{1}{2}\right)^{5/15} = 1260 \times 2^{-1/3} = 1260 \times 0.7937 = 1000 \text{ dps} = 60{,}000 \text{ dpm}$$ **Step 2: Determine the blood volume.** At $t = 5$ h, the blood sample shows 15 dpm/mL. Since Na-24 distributes uniformly in blood: $$V_{\text{blood}} = \frac{A(5\text{h})}{15 \text{ dpm/mL}} = \frac{60{,}000}{15} = 4000 \text{ mL} = 4 \text{ L}$$ **Step 3: Find activity per mL after a further 10 h (at $t = 15$ h total).** From $t = 5$ h, an additional 10 h elapses (i.e., $\frac{10}{15} = \frac{2}{3}$ of a half-life): $$A(15\text{h per mL}) = 15 \times \left(\frac{1}{2}\right)^{10/15} = 15 \times 2^{-2/3} = 15 \times 0.6300 = 9.45 \text{ dpm/mL}$$ **Answer: A — The activity of the blood sample after a further 10 hours is 9.45 dpm/mL.**