In the complete natural disintegration of Uranium-238 () to the stable isotope Lead-206 (), calculat — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Change in mass $\Delta A = 238 - 206 = 32$. Since only $\alpha$ -particles change mass, $n_\alpha = 32/4 = 8$. Emitting $8\alpha$ reduces $Z$ by $16$ ($92 - 16 = 76$). The final $Z$ is $82$. The difference ($82 - 76 = 6$) represents the number of $\beta^-$ particles emitted.