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

Options

A. `.^(m-6)._(n)Z`
B. `.^(m-4)._(n)Z`
C. `.l_(n-2)^(m-4)Y`
D. `._(n-4)^(m-6)Z`

Explanation

To determine the resulting nucleus after the emission of one alpha particle and two beta particles from a nucleus represented as \( _n^m X \), we can follow these steps:

### Step 1: Understand the emission of an alpha particle.
An alpha particle is a helium nucleus, which consists of 2 protons and 2 neutrons. Therefore, when an alpha particle is emitted from the nucleus \( _n^m X \):
- 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).

So, after emitting one alpha particle, the new nucleus will be:
\[ _n^m X \rightarrow _{n-2}^{m-4} Y \]

### Step 2: Understand the emission of beta particles.
A beta particle is essentially an electron, which is emitted when a neutron in the nucleus is converted into a proton. Each beta particle emission results in:
- The mass number remaining the same (since a neutron is converted to a proton).
- The atomic number increases by 1 (since a proton is added).

Since two beta particles are emitted, the changes will be:
- The mass number remains \( m-4 \).
- The atomic number increases by 2 (1 for each beta particle).

So, after emitting two beta particles from the nucleus \( _{n-2}^{m-4} Y \):
\[ _{n-2}^{m-4} Y \rightarrow _{(n-2)+2}^{m-4} Z \]
This simplifies to:
\[ _{n}^{m-4} Z \]

### Step 3: Write the resulting nucleus.
After both emissions, the resulting nucleus is:
\[ _{n}^{m-4} Z \]

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