In a process, `""_(92)U^(238)` is converted to `""_(90)Th^(234)`. Determine the particles emitted in the process and also write the equation for the given process.

Answer

To solve the problem of determining the particles emitted when Uranium-238 is converted to Thorium-234, we can follow these steps:

### Step 1: Identify the Initial and Final Nuclei
- The initial nucleus is Uranium-238, denoted as \( _{92}^{238}U \).
- The final nucleus is Thorium-234, denoted as \( _{90}^{234}Th \).

### Step 2: Analyze the Change in Mass Number and Atomic Number
- The mass number of Uranium-238 is 238, and the mass number of Thorium-234 is 234.
- The atomic number of Uranium is 92, and the atomic number of Thorium is 90.
- The change in mass number is \( 238 - 234 = 4 \).
- The change in atomic number is \( 92 - 90 = 2 \).

### Step 3: Determine the Type of Decay
- The decrease in mass number suggests that an alpha particle is emitted.
- An alpha particle consists of 2 protons and 2 neutrons, which has a mass number of 4 and an atomic number of 2.

### Step 4: Write the Decay Equation
- The emission of one alpha particle can be represented as:
\[ _{92}^{238}U \rightarrow _{90}^{234}Th + _{2}^{4}He \]
- Here, \( _{2}^{4}He \) represents the alpha particle emitted.

### Step 5: Verify Conservation of Mass Number and Atomic Number
- On the left side (reactants):
- Mass number: 238
- Atomic number: 92
- On the right side (products):
- Mass number: \( 234 + 4 = 238 \)
- Atomic number: \( 90 + 2 = 92 \)
- Both mass number and atomic number are conserved.

### Conclusion
- The particles emitted in the process are one alpha particle, represented as \( _{2}^{4}He \).
- The complete equation for the process is:
\[ _{92}^{238}U \rightarrow _{90}^{234}Th + _{2}^{4}He \]