Question

${}_{90}T{h}^{232}⟶{}_{82}P{b}^{208}$ 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)

${}_{90}T{h}^{232}⟶{}_{82}P{b}^{208}$
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
$=90-12=78$
$\left(\because \alpha ={}_{2}H{e}^4\right)$.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.