The half-life of Tc^99 is 6.0 h. The delivery of a sample of Tc^99 from the reactor to the nuclear m — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Number of Half-Lives During Transit $$n = \frac{t}{t_{1/2}} = \frac{3.0\ \text{h}}{6.0\ \text{h}} = 0.5\ \text{half-lives}$$ Step 2 - Relate Initial to Final Mass Using the first-order decay equation: $$N = N_0 \left(\frac{1}{2}\right)^n \implies N_0 = N \cdot 2^n$$ Step 3 - Calculate Minimum Initial Mass $$N_0 = 10.0\ \text{mg} \times 2^{0.5} = 10.0 \times \sqrt{2} = 10.0 \times 1.4142$$ $$N_0 = \boxed{14.1\ \text{mg}}$$ Step 4 - Evaluate Options - **(A) 20.0 mg**: After transit: $20.0 / \sqrt{2} \approx 14.1$ mg received. Exceeds 10.0 mg — works, but not the *minimum*. Incorrect. - **(B) 15.0 mg**: After transit: $15.0 / \sqrt{2} \approx 10.6$ mg. Meets the target but exceeds it — not minimum. Incorrect. - **(C) 14.1 mg**: After transit: $14.1 / \sqrt{2} \approx 10.0$ mg. Exactly meets the requirement. **Correct.** - **(D) 12.5 mg**: After transit: $12.5 / \sqrt{2} \approx 8.84$ mg. Less than 10.0 mg required. Incorrect. $$\boxed{\text{Answer: C — 14.1 mg}}$$