When a radioactive nucleus decays, the and values change, creating relationships with the parent nuc — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

A is true: $\alpha$ -decay decreases $A$ by 4 and $Z$ by 2, so $A - 2Z$ remains constant (isodiaphere). B is true: $\beta^-$ -decay increases $Z$ by 1 but leaves $A$ unchanged, forming an isobar. C is true: one $\alpha$ reduces $Z$ by 2, two $\beta$ s increase $Z$ back by 2; the final $Z$ equals the parent $Z$, forming an isotope. D is wrong: Positron emission leaves $A$ unchanged but decreases $Z$ by 1, making it an isobar, not an isotone.