Total Alpha and Beta Particles Emitted in Uranium Decay

To find the total number of $\alpha$ and $\beta$ particles emitted in the given nuclear decay, let:

• $x$ be the number of $\alpha$-particles emitted ($^{4}_{2}\text{He}$)
• $y$ be the number of $\beta$-particles emitted ($^{0}_{-1}\text{e}$)

The nuclear reaction can be represented as:
$^{238}_{92}\text{U} \longrightarrow \, ^{214}_{82}\text{Pb} + x \, ^{4}_{2}\text{He} + y \, ^{0}_{-1}\text{e}$

Applying the principle of conservation of mass number:
$238 = 214 + 4x + 0y$
$4x = 238 - 214$
$4x = 24 \implies x = 6$

Thus, the number of $\alpha$-particles emitted is $6$.

Applying the principle of conservation of atomic number (charge):
$92 = 82 + 2x - y$

Substituting the value of $x = 6$:
$92 = 82 + 2(6) - y$
$92 = 94 - y \implies y = 2$

Thus, the number of $\beta$-particles emitted is $2$.

The total number of $\alpha$ and $\beta$ particles emitted is:
$\text{Total particles} = x + y = 6 + 2 = 8$