Co^57 decays to Fe^57 by β^+ emission. The resulting Fe^57 is in its excited state and comes to the — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - The Sequential Decay Chain $$\ce{^{57}_{27}Co ->[\beta^+][270\ \text{days}] ^{57}_{26}Fe^* ->[\gamma][10^{-8}\ \text{s}] ^{57}_{26}Fe}$$ Step 2 - Rate-Determining Step Analysis The $\gamma$-emission from excited $\ce{Fe^{57*}}$ has an extremely short half-life of $10^{-8}$ s. This means $\ce{Fe^{57*}}$ decays to the ground state almost instantaneously after it is formed. In such a fast-secondary/slow-primary consecutive decay, the rate of $\gamma$-emission is governed entirely by the **slow first step** (the $\beta^+$ decay of Co-57). For every Co-57 atom that undergoes $\beta^+$ decay, one $\ce{Fe^{57*}}$ is produced, which almost instantly emits one $\gamma$-ray. Step 3 - Conclusion Therefore, the rate of $\gamma$ emission $\approx$ rate of Co-57 decay, and both halve in exactly **one half-life of Co-57 = 270 days**. Step 4 - Evaluate Options - **(A) 135 days**: This would be half the Co-57 half-life — incorrect. - **(B) 270 days**: Equal to the half-life of Co-57. **Correct.** - **(C) $10^{-8}$ s**: This is the half-life of the intermediate $\ce{Fe^{57*}}$ — not relevant to the overall $\gamma$ rate drop. Incorrect. - **(D) $5 \times 10^{-9}$ s**: Half of $10^{-8}$ — incorrectly applies to Fe*. Incorrect. $$\boxed{\text{Answer: B — 270 days (one half-life of Co-57)}}$$