`""_(92)^(238)`U(IIIB) changes to `""_(9)^(234)` Th by emission of `alpha`- particle . Daughter element will be in -

Options

A. III B
B. IB
C. VB
D. IIA

Answer

To solve the problem, we need to analyze the nuclear reaction that occurs when uranium-238 (\( _{92}^{238}U \)) emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, which is equivalent to a helium nucleus (\( _{2}^{4}He \)).

### Step-by-Step Solution:

1. **Identify the Initial and Final Elements**:
- The initial element is uranium-238, denoted as \( _{92}^{238}U \).
- The final element after the emission of an alpha particle is thorium-234, denoted as \( _{90}^{234}Th \).

2. **Write the Alpha Decay Equation**:
- The decay can be represented as:
\[ _{92}^{238}U \rightarrow _{2}^{4}He + _{90}^{234}Th \]
- Here, \( _{2}^{4}He \) is the alpha particle being emitted.

3. **Calculate the Mass and Atomic Numbers**:
- The mass number decreases by 4 (238 - 4 = 234) and the atomic number decreases by 2 (92 - 2 = 90).
- This confirms that the daughter element is thorium-234.

4. **Identify the Group of the Daughter Element**:
- Both uranium and thorium belong to the actinide series, which consists of elements with atomic numbers from 90 to 103.
- These elements are classified under the 3B group in the periodic table.

5. **Conclusion**:
- The daughter element after the emission of an alpha particle from uranium-238 is thorium-234, which is in the 3B group of the periodic table.

### Final Answer:
The daughter element will be thorium-234 (\( _{90}^{234}Th \)), which is in the 3B group.