Of the following isotopes, which one is likely to be the most stable? — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Of the following isotopes, which one is likely to be the most stable?
💡 Solution & Explanation
Step 1 - The Even-Odd Pairing Rule Nuclear stability is strongly influenced by the pairing of protons and neutrons. Nucleons pair with antiparallel spins, releasing "pairing energy" that increases binding energy. In decreasing order of stability: 1. **Even $Z$, Even $N$** (even-even): most stable; all nucleons can pair completely. 2. **Even $Z$, Odd $N$** or **Odd $Z$, Even $N$**: moderately stable. 3. **Odd $Z$, Odd $N$** (odd-odd): least stable; both protons and neutrons have unpaired particles. Step 2 - Neutron Numbers for Each Zinc Isotope ($Z = 30$, even) | Option | Isotope | $A$ | $N = A - Z$ | Z parity | N parity | Type | |--------|---------|-----|------------|----------|----------|------| | A | $\ce{^{63}Zn}$ | 63 | 33 | even | **odd** | even-odd | | B | $\ce{^{67}Zn}$ | 67 | 37 | even | **odd** | even-odd | | C | $\ce{^{71}Zn}$ | 71 | 41 | even | **odd** | even-odd | | D | $\ce{^{64}Zn}$ | 64 | **34** | even | **even** | **even-even** ✓ | Step 3 - Conclusion $\ce{^{64}Zn}$ is the only even-even isotope; it has complete proton and neutron pairing, giving it the maximum binding energy per nucleon among the options. It is also the most naturally abundant isotope of Zn (48.6%). $$\boxed{\text{Answer: D} \quad \ce{^{64}_{30}Zn}}$$ Step 4 - Evaluate Options - **(A) $\ce{Zn_{63}}$** — even-odd; radioactive ($\beta^+$ decay, $t_{1/2} = 38.5$ min). Incorrect. - **(B) $\ce{Zn_{67}}$** — even-odd; stable but less abundant. Incorrect. - **(C) $\ce{Zn_{71}}$** — even-odd; radioactive ($\beta^-$ decay, $t_{1/2} = 2.4$ min). Incorrect. - **(D) $\ce{Zn_{64}}$** — even-even; most stable. Correct.