A freshly prepared radioactive source of half-life 2 h, emits radiations of intensity which is 64 ti — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - First-Order Decay and Activity Radioactive decay is first-order. After $n$ half-lives, the intensity drops to: $$A = A_0 \left(\frac{1}{2}\right)^n$$ Step 2 - Set Up the Safety Condition Given: $A_0 = 64 \cdot A_\text{safe}$, $t_{1/2} = 2$ h. We need $A = A_\text{safe}$. $$A_\text{safe} = 64 \cdot A_\text{safe} \cdot \left(\frac{1}{2}\right)^n$$ $$\left(\frac{1}{2}\right)^n = \frac{1}{64} = \left(\frac{1}{2}\right)^6$$ $$n = 6 \text{ half-lives}$$ Step 3 - Calculate Minimum Time $$t = n \times t_{1/2} = 6 \times 2\ \text{h} = \boxed{12\ \text{h}}$$ Step 4 - Evaluate Options - **(A) 6 h**: Only 3 half-lives → $A = A_0/8 = 8 \cdot A_\text{safe}$. Still 8× the safe limit. Incorrect. - **(B) 12 h**: 6 half-lives → $A = A_0/64 = A_\text{safe}$. Exactly at the safe threshold. **Correct.** - **(C) 24 h**: 12 half-lives. Safe, but not the *minimum* time required. Incorrect. - **(D) 128 h**: Far longer than necessary. Incorrect. $$\boxed{\text{Answer: B — 12 h}}$$