Four vessels 1, 2, 3 and 4 contain respectively, 10 g-atom (t_1/2 = 10 h), 1 g-atom (t_1/2 = 5 h), 5 — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Four vessels 1, 2, 3 and 4 contain respectively, 10 g-atom (t_1/2 = 10 h), 1 g-atom (t_1/2 = 5 h), 5 g-atom (t_1/2 = 2 h) and 2 g-atom (t_1/2 = 1 h) of different radioactive nuclides. In the beginning, the maximum radioactivity would be exhibited by the vessel
💡 Solution & Explanation
Step 1 - Activity Formula Initial activity $A_0 = \lambda N_0 = \frac{0.693}{t_{1/2}} \times N_0$ Since $0.693$ is the same constant for all, compare $\frac{N_0}{t_{1/2}}$ for each vessel. Here $N_0$ (g-atom) is proportional to moles of radioactive nuclei. Step 2 - Calculate $N_0/t_{1/2}$ for Each Vessel | Vessel | $N_0$ (g-atom) | $t_{1/2}$ (h) | $N_0/t_{1/2}$ | |--------|----------------|----------------|----------------| | 1 | 10 | 10 | **1.0** | | 2 | 1 | 5 | **0.2** | | 3 | 5 | 2 | **2.5** ← maximum | | 4 | 2 | 1 | **2.0** | Step 3 - Conclusion $$2.5 > 2.0 > 1.0 > 0.2$$ Vessel 3 has the highest initial activity. - (A) Vessel 4: ratio 2.0 — second highest, not maximum — incorrect - (B) Vessel 3: ratio 2.5 — highest ✓ - (C) Vessel 2: ratio 0.2 — lowest — incorrect - (D) Vessel 1: ratio 1.0 — third — incorrect $$\boxed{B}$$