Na^23 is the more stable isotope of Na. Find out the process by which Na^24 can undergo radioactive — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Compare n/p Ratios of Na Isotopes Stable $\ce{^{23}_{11}Na}$: $p=11$, $n=12$, ratio $= 12/11 \approx 1.09$ Radioactive $\ce{^{24}_{11}Na}$: $p=11$, $n=13$, ratio $= 13/11 \approx 1.18$ Since $1.18 > 1.09$: $\ce{^{24}Na}$ is neutron-rich → must decrease $n/p$ ratio. Step 2 - Decay Mode for Neutron-Rich Nuclei To decrease $n/p$: convert $n \to p$ via $\beta^-$ emission: $$\ce{^1_0n -> ^1_1p + ^0_{-1}e + \bar{\nu}_e}$$ Nuclear equation: $$\ce{^{24}_{11}Na -> ^{24}_{12}Mg + ^0_{-1}e + \bar{\nu}_e}$$ Result: $\ce{^{24}_{12}Mg}$ with $n/p = 12/12 = 1.0$ — stable ✓ Step 3 - Option Analysis - (A) $\beta$-emission: converts $n \to p$, decreases $n/p$ → correct for neutron-rich $\ce{^{24}Na}$ ✓ - (B) $\alpha$-emission: only heavy nuclei ($Z > 83$) — incorrect - (C) $\beta^+$ emission: converts $p \to n$, increases $n/p$ → for proton-rich nuclei — incorrect - (D) electron capture: also $p \to n$ (proton-rich) — incorrect $$\boxed{A}$$