Question

The number of $\alpha$-particles and $\beta$-particles respectively emitted in the reaction ${_{88}A^{196} -> _{78}B^{164}}$

• A. $8$ and $8$
• B. $8$ and $6$
• C. $6$ and $8$
• D. $8$ and $4$

Answer 1

Answer: (B)

$8$ and $6$

Answer 2

Let $x$ alpha particles and $y$ beta particles are emitted $\therefore\, _{88}A^{196} \to_{78}B^{164}+x_{2} He^{4}+y_{-1^{e^0}}$ According to conservation of charge number, we get $88 = 78 + 2x - y \ldots\left(i\right)$ According to conservation of mass number, weget $196 = 164 + 4x + 0 \ldots\left(ii\right)$ Solving $\left(i\right)$ and $\left(ii\right)$, we get $x = 8, y = 6$ Hence, $8\alpha$-particles and $6\beta$-particles are emitted