Question

${}_{92}{U}^{238}$ decays to a stable nucleus of ${}_{82}{Pb}^{206}$. In this process:

• A. $8\alpha$-particles and $6\beta$ -particles are emitted
• B. $6\alpha$ particles and $8\beta$ -particles are emitted
• C. $7\alpha$ -particles and $7\beta$ -particles are emitted
• D. $8\alpha$ -particles and $4\beta$ -particles are emitted

Answer 1

The correct option is A $8\alpha$-particles and $6\beta$ -particles are emitted

Difference in mass number is $238-206=32$

So, number of $\alpha$ particles emitted is $32/4=8$

Decrease in Atomic number due to $\alpha$-decay is $8\times 2=16$.

So, number of $\beta$-particles emitted is $16-(92-82)=16-10=6$

$\therefore 8\alpha$-particles and $6\beta$-particles.