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)

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