• # question_answer In a decay series, $_{82}^{206}Pb$ is obtained at the end from $_{92}^{238}U.$ How many alpha and beta particles must have been emitted? A) $7\alpha ,10\beta$ B) $8\alpha ,6\beta$ C) $5\alpha ,6\beta$ D) $6\alpha ,8\beta$

 $\underset{92}{\mathop{238}}\,U\to \underset{82}{\mathop{206}}\,Pb$ Number of $\alpha$-particles $=\frac{chnages\,\,in\,\,mass\,\,number}{4}$ $=\frac{238-206}{4}$ $=\frac{38}{4}=8\alpha$ Number of $\beta$-particles $=2\times$ number of $\alpha$-particles - change in atomic number $=2\times 8-\left( 92-82 \right)$ $=16-10=6\beta$