In the disintegration of a radioactive element, `alpha`- and `beta`-particles are evolved from the nucleus.
`._(0)n^(1) rarr ._(1)H^(1) + ._(-1)e^(0) +` Antineutrino + Energy
`4 ._(1)H^(1) rarr ._(2)He^(4) + 2 ._(+1)e^(0) +` Energy
Then, emission of these particles changes the nuclear configuration and results into a daughter nuclide. Emission of an `alpha`-particles results into a daughter element having atomic number lowered by 2 and mass number by 4, on the other hand, emission of a `beta`-particle yields an element having atomic number raised by 1.
A radioactive element belongs to `III B` group, it emits ona `alpha`- and `beta`-particle to form a daughter nuclide. The position of daughter nuclide will be in

Options

A. `II A`
B. `I A`
C. `II B`
D. `IVB`

Explanation

To solve the problem, we need to analyze the changes in atomic number and mass number that occur when a radioactive element from group III B emits one alpha particle and one beta particle.

### Step-by-Step Solution:

1. **Identify the Initial Element:**
Let's denote the radioactive element as \( X \) with an atomic number \( Z \) and mass number \( A \).

2. **Alpha Particle Emission:**
When \( X \) emits one alpha particle (\( \alpha \)), the changes are as follows:
- The mass number decreases by 4.
- The atomic number decreases by 2.

Therefore, after emitting an alpha particle, the new element can be represented as:
\[ X \rightarrow X' \quad \text{where} \quad X' = (A - 4) \quad \text{and} \quad Z' = (Z - 2) \]

3. **Beta Particle Emission:**
Next, when \( X' \) emits one beta particle (\( \beta \)), the changes are:
- The mass number remains the same.
- The atomic number increases by 1.

Thus, after emitting a beta particle, the new element becomes:
\[ X' \rightarrow Y \quad \text{where} \quad Y = (A - 4) \quad \text{and} \quad Z'' = (Z - 2 + 1) = (Z - 1) \]

4. **Determine the Group of the Daughter Nuclide:**
Since the original element \( X \) belongs to group III B (or group 3), and after the emissions, the new atomic number \( Z'' = Z - 1 \) indicates that the daughter nuclide \( Y \) will belong to group II A (or group 2).

5. **Conclusion:**
Therefore, the position of the daughter nuclide will be in group 2 A.

### Final Answer:
The daughter nuclide will be in **2nd A**.