Question

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

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

Answer 1

Answer: (B)

Let $x$ alpha particles and $y$ beta particles are emitted
$\therefore{ }_{88} A^{196} \rightarrow_{78} B^{164}+x_2 H e^4+y_{-1^{e^0}}$
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