Question

The number of $\alpha$ and ${\beta }^{-}$ particles emitted in nuclear reaction, ${}_{90}T{h}^{228}⟶{}_{83}B{i}^{212}$ are respectively :

• A. $4,1$
• B. $3,7$
• C. $8,1$
• D. $4,7$

Answer 1

The correct option is A $4,1$

Number of $\alpha$ particles $=\frac{\text{difference of mass number}}{4}$

$=\frac{228-212}{4}=\frac{16}{4}=4$

Number of ${\beta }^{-}$ particles $=2\times \alpha$ (difference in atomic numbers)

$=2\times 4-(90-83)$

$=2\times 4-7=1$

${}_{90}T{h}^{228}{⟶}_{83}B{i}^{212}+4{}_2{\alpha }^4{+}_{-1}{\beta }^0$

Hence, option $\text{A}$ is correct.