When a parent nucleus undergoes a specific radioactive emission, the resulting daughter nuclide is p — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

An $\alpha$ -decay decreases $A$ by 4 and $Z$ by 2. The new isotopic number $(A_{new} - 2Z_{new})$ becomes $(A-4) - 2(Z-2) = A - 4 - 2Z + 4 = A - 2Z$. The value remains perfectly constant.