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

Answer

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 (with unknown mass number \( A \) and atomic number \( Z \)) and Helium-4, denoted as \( _{2}^{4}He \).

2. **Conservation of Atomic Number**:
- The atomic number (number of protons) must be conserved in the reaction.
- For Uranium, the atomic number is 92. For Helium, the atomic number is 2.
- Therefore, we can set up the equation:
\[ 92 = Z + 2 \]
- Rearranging gives:
\[ Z = 92 - 2 = 90 \]

3. **Conservation of Mass Number**:
- The mass number (total number of protons and neutrons) must also be conserved.
- For Uranium, the mass number is 238. For Helium, the mass number is 4.
- We can set up the equation:
\[ 238 = A + 4 \]
- Rearranging gives:
\[ A = 238 - 4 = 234 \]

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

5. **Conclusion**:
- The values of \( A \) and \( Z \) for Thorium are \( A = 234 \) and \( Z = 90 \).

### Answer:
- \( A = 234 \)
- \( Z = 90 \)