A nucleus `._n X^m` emits one `alpha` and one `beta` particles. The resulting nucleus is.

Options

A. `._(n)X^(m-4)`
B. `._(n-2)Y^(m-4)`
C. `._(n-4)Z^(m-4)`
D. None of these

Answer

To solve the problem of determining the resulting nucleus after a nucleus \( _n X^m \) emits one alpha particle and one beta particle, we can follow these steps:

### Step 1: Understand the effect of alpha decay
When a nucleus emits an alpha particle, it loses 2 protons and 2 neutrons. This results in:
- **Mass number (A)** decreases by 4.
- **Atomic number (Z)** decreases by 2.

For the nucleus \( _n X^m \):
- New mass number after alpha decay: \( m - 4 \)
- New atomic number after alpha decay: \( n - 2 \)

### Step 2: Understand the effect of beta decay
When a nucleus emits a beta particle, a neutron is converted into a proton, which results in:
- **Mass number (A)** remains the same.
- **Atomic number (Z)** increases by 1.

Applying this to the nucleus after the alpha decay:
- Mass number remains: \( m - 4 \)
- Atomic number after beta decay: \( (n - 2) + 1 = n - 1 \)

### Step 3: Write the resulting nucleus
After both emissions, the resulting nucleus can be represented as:
\[ _{(n-1)} Y^{(m-4)} \]

### Conclusion
The resulting nucleus after the emission of one alpha particle and one beta particle from the original nucleus \( _n X^m \) is:
\[ _{(n-1)} Y^{(m-4)} \]