In the nuclear reaction `._92 U^238 rarr ._z Th^A + ._2He^4`, the values of `A` and `Z` are.

Options

A. `A = 234, Z = 94`
B. `A = 234, Z = 90`
C. `A = 238, Z = 94`
D. `A = 238, Z = 90`

Explanation

To solve the nuclear reaction \( _{92}^{238}U \rightarrow _{Z}^{A}Th + _{2}^{4}He \), we need to determine the values of \( A \) and \( Z \).

### Step-by-Step Solution:

1. **Identify the Components of the Reaction:**
- The reactant is Uranium-238, denoted as \( _{92}^{238}U \).
- The products are Thorium (denoted as \( _{Z}^{A}Th \)) and Helium-4 (denoted as \( _{2}^{4}He \)).

2. **Apply Conservation of Mass Number:**
- The mass number (A) must be conserved in nuclear reactions. This means that the total mass number on the left side must equal the total mass number on the right side.
- For the left side: The mass number of Uranium is 238.
- For the right side: The mass number of Thorium is \( A \) and the mass number of Helium is 4.
- Therefore, we can write the equation:
\[ 238 = A + 4 \]

3. **Solve for A:**
- Rearranging the equation gives:
\[ A = 238 - 4 = 234 \]

4. **Apply Conservation of Atomic Number:**
- The atomic number (Z) must also be conserved. The atomic number of Uranium is 92.
- For the right side: The atomic number of Thorium is \( Z \) and the atomic number of Helium is 2.
- Therefore, we can write the equation:
\[ 92 = Z + 2 \]

5. **Solve for Z:**
- Rearranging the equation gives:
\[ Z = 92 - 2 = 90 \]

6. **Final Values:**
- From our calculations, we find:
- \( A = 234 \)
- \( Z = 90 \)

### Conclusion:
The values of \( A \) and \( Z \) are:
- \( A = 234 \)
- \( Z = 90 \)