The isotope decays through a succession of steps to form the stable isotope (the Actinium series). D — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Change in mass number $\Delta A = 235 - 207 = 28$. Since only $\alpha$ -particles change the mass, the number of $\alpha$ -particles is $n_\alpha = \frac{28}{4} = 7$. Emitting $7\alpha$ -particles drops the atomic number by $7 \times 2 = 14$. The expected atomic number without $\beta$ emissions would be $92 - 14 = 78$. The actual final atomic number is $82$. The difference is $82 - 78 = 4$, which corresponds exactly to the number of $\beta^-$ -particles emitted.