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.
How many `alpha`- and `beta`-particle should be emitted from a radioactive nuclide so that an isobar is formed?

Options

A. `1 alpha, 1 beta`
B. `1 alpha, 2 beta`
C. `2 alpha, 2 beta`
D. `n beta`

Explanation

To determine how many alpha (α) and beta (β) particles should be emitted from a radioactive nuclide to form an isobar, we need to understand the changes in atomic and mass numbers that occur during these emissions.

### Step-by-Step Solution:

1. **Understanding Isobars**:
Isobars are nuclides that have the same mass number (A) but different atomic numbers (Z).

2. **Effects of Alpha Emission**:
- When an alpha particle (which consists of 2 protons and 2 neutrons) is emitted, the atomic number (Z) decreases by 2 and the mass number (A) decreases by 4.
- Therefore, if we emit \( n \) alpha particles, the changes can be summarized as:
- New Atomic Number: \( Z - 2n \)
- New Mass Number: \( A - 4n \)

3. **Effects of Beta Emission**:
- When a beta particle (an electron) is emitted, the atomic number (Z) increases by 1, while the mass number (A) remains unchanged.
- Therefore, if we emit \( m \) beta particles, the changes can be summarized as:
- New Atomic Number: \( Z + m \)
- New Mass Number: \( A \) (remains the same)

4. **Setting Up the Equations**:
To form an isobar, we need the mass number to remain the same:
\[ A - 4n = A \]
This simplifies to:
\[ 4n = 0 \quad \Rightarrow \quad n = 0 \]
Thus, no alpha particles can be emitted if we want to keep the mass number the same.

5. **Finding the Atomic Number**:
Now, we need to adjust the atomic number using beta particles:
\[ Z - 2n + m = Z' \]
where \( Z' \) is the new atomic number after emissions. Since \( n = 0 \):
\[ Z + m = Z' \]
To form an isobar, \( Z' \) must be different from \( Z \) but still yield the same mass number.

6. **Conclusion**:
Since we established that \( n = 0 \) (no alpha particles emitted), we can emit any number of beta particles \( m \) to achieve the desired atomic number while keeping the mass number the same.

### Final Answer:
- **Number of Alpha Particles (n)**: 0
- **Number of Beta Particles (m)**: Any positive integer (1 or more)