Radioactive emissions systematically transform the identity of a parent nuclide. According to the la — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

A: $\alpha$ -decay decreases $A$ by 4 and $Z$ by 2. The isotopic number $(A - 2Z)$ becomes $(A-4) - 2(Z-2) = A - 2Z$, keeping it constant (isodiaphere). B: $\beta^-$ -decay increases $Z$ by 1 but leaves $A$ unchanged (isobar). C: One $\alpha$ reduces $Z$ by 2, two $\beta^-$ raise $Z$ by 2, restoring the parent's $Z$ value (isotope). D: Positron emission reduces $Z$ by 1 and leaves $A$ unchanged, making it an isobar, not an isotone (neutrons increase by 1 while protons drop by 1).