Question

${ }_{90} Th ^{232} \longrightarrow { }_{82} Pb ^{208}$ 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

Answer: (D)

${ }_{90} Th ^{232} \longrightarrow { }_{82} Pb ^{208}$
In the above disintegration.
Change in atomic weight $=232-208=24$.
We know that for the emission of one $\alpha$-particle the atomic weight is decreased by $4 .$
$\therefore $ number of $\alpha$-particles emitted
$=\frac{24}{4}=6$
After the emission of $6 \alpha$-particles, the change in atomic number
$=90-12=78$
$\left(\because \alpha={ }_{2} He ^{4}\right)$.The increase in atomic number from $78$ to $82=4$
$\therefore $ number of $\beta$-particles emitted $=4$
$\left(\because \beta=-e^{\circ}\right)$
Hence, $6 \alpha$-particles and $4 \beta$-particles are emitted.