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

Options

A. `._(n)^(m-6)Z`
B. `._(n)^(m-4)X`
C. `._(n-4)^(m-4)Y`
D. `._(n-4)^(m-6)Z`

Answer

To solve the problem, we need to determine the resulting nucleus after a nucleus represented as `._n^m X` emits one alpha particle and two beta particles.

### Step-by-Step Solution:

1. **Identify the initial properties of the nucleus**:
- The nucleus is represented as `._n^m X`, where:
- `m` is the mass number (total number of protons and neutrons).
- `n` is the atomic number (number of protons).

2. **Understand the effects of alpha decay**:
- An alpha particle consists of 2 protons and 2 neutrons.
- When an alpha particle is emitted:
- The mass number decreases by 4 (because 2 protons and 2 neutrons are lost).
- The atomic number decreases by 2 (because 2 protons are lost).

Therefore, after emitting one alpha particle, the new mass number and atomic number will be:
- New mass number = `m - 4`
- New atomic number = `n - 2`

3. **Understand the effects of beta decay**:
- A beta particle is an electron emitted from the nucleus when a neutron converts into a proton.
- When a beta particle is emitted:
- The mass number remains the same (the total number of nucleons does not change).
- The atomic number increases by 1 (because one neutron is converted into one proton).

Since we are emitting two beta particles, the changes will be:
- New mass number remains the same = `m - 4` (from the alpha decay).
- New atomic number after two beta emissions = `n - 2 + 2 = n`.

4. **Combine the results**:
- After one alpha particle and two beta particles are emitted, the resulting nucleus will have:
- Mass number = `m - 4`
- Atomic number = `n`.

5. **Write the resulting nucleus**:
- The resulting nucleus can be represented as `._n^(m-4) Y`, where `Y` is the new element with atomic number `n`.

### Final Result:
The resulting nucleus after the emission of one alpha particle and two beta particles is `._(n)^(m-4) Y`.