Question

${{}_{90}Th}^{232}⟶{{}_{82}Pb}^{208}$. The number of $\alpha$ and $\beta$- particles emitted during the above reaction is:

• A. $8\alpha$ and $4\beta$
• B. $8\alpha$ and $16\beta$
• C. $4\alpha$ and $2\beta$
• D. $6\alpha$ and $4\beta$

Answer 1

The correct option is D $6\alpha$ and $4\beta$

no. of $\alpha$ particles $=\frac{232-208}{4}=\frac{24}{4}=6\alpha$

No. of $\beta$ particles $=2\times 6-(90-82)$

$=12-8$

$=4\beta$

Hence, $6\alpha$ and $4\beta$ particles are emitted.