The only stable nucleus of fluorine is F^19. What type of decay is expected from F^18? — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - n/p Ratio of Stable Fluorine-19 Stable $\ce{^{19}_{9}F}$: $p = 9$, $n = 10$ $$\left(\frac{n}{p}\right)_{\text{stable}} = \frac{10}{9} \approx 1.11$$ Step 2 - n/p Ratio of Fluorine-18 Unstable $\ce{^{18}_{9}F}$: $p = 9$, $n = 9$ $$\left(\frac{n}{p}\right)_{\text{actual}} = \frac{9}{9} = 1.00 < 1.11$$ F-18 is neutron-deficient (proton-rich) → must convert a proton to neutron. Step 3 - Decay Mode Selection For light nuclei ($Z < 20$), converting $p \to n$ occurs predominantly via positron ($\beta^+$) emission: $$\ce{^{18}_{9}F -> ^{18}_{8}O + ^0_{+1}e + \nu_e}$$ K-capture also converts $p \to n$, but positron emission is dominant for light elements. Step 4 - Option Analysis - (A) $\alpha$-decay: only in heavy nuclei ($Z > 83$) — incorrect - (B) $\beta$-decay: converts $n \to p$, increases $n/p$ → for neutron-rich nuclei — incorrect - (C) positron decay: converts $p \to n$, increases $n/p$ → correct for proton-rich F-18 ✓ - (D) K-capture: also converts $p \to n$, but secondary to $\beta^+$ for light nuclei — not dominant $$\boxed{C}$$