Step 1: Understanding the Topic
This question involves a nuclear reaction, specifically a radioactive decay series. The key principle for solving this is the conservation of mass number (A, the superscript) and atomic number (Z, the subscript). We must ensure that the sum of the mass numbers and the sum of the atomic numbers are equal on both the reactant and product sides of the nuclear equation.
Step 2: Key Approach - Conservation Laws
We need to know the changes caused by alpha ($\alpha$) and beta-minus ($\beta^-$) decay:
An $\alpha$-particle is a helium nucleus, ${}^{4}_{2}He$. Its emission decreases the mass number by 4 and the atomic number by 2.
A $\beta^-$-particle is an electron, ${}^{0}_{-1}e$. Its emission does not change the mass number but increases the atomic number by 1.
We will set up two equations, one for the conservation of A and one for the conservation of Z.
Step 3: Detailed Calculation
A. Conservation of Mass Number (A):
Let's check if the mass numbers are balanced.
\[
\text{Initial A} = 238
\]
\[
\text{Final A} = \text{A of Pb} + 7 \times (\text{A of } \alpha) + x \times (\text{A of } \beta^-)
\]
\[
\text{Final A} = 210 + 7 \times (4) + x \times (0) = 210 + 28 = 238
\]
The mass numbers are already balanced, which serves as a good check.
B. Conservation of Atomic Number (Z):
Now we set up the equation for the atomic numbers to find $x$.
\[
\text{Initial Z} = 92
\]
\[
\text{Final Z} = \text{Z of Pb} + 7 \times (\text{Z of } \alpha) + x \times (\text{Z of } \beta^-)
\]
\[
92 = 82 + 7 \times (2) + x \times (-1)
\]
\[
92 = 82 + 14 - x
\]
\[
92 = 96 - x
\]
Now, solve for $x$:
\[
x = 96 - 92
\]
\[
x = 4
\]
Step 4: Final Answer
The number of $\beta^-$ particles emitted is 4.
\[
\boxed{x = 4}
\]