Complete the series `.^6He to e^(-) + .^6Li+`

Options

A. neutrino
B. antineutrino
C. proton
D. neutron

Explanation

To complete the series for the reaction involving Helium-6, we need to analyze the given nuclear reaction step by step.

### Step-by-Step Solution:

1. **Identify the initial nucleus**:
The reaction starts with Helium-6, denoted as \(^6He\). This nucleus has:
- Atomic number (Z) = 2 (since Helium has 2 protons)
- Mass number (A) = 6 (total number of protons and neutrons)

2. **Identify the products**:
The products of the reaction are given as an electron (\(e^-\)) and Lithium-6 (\(^6Li^+\)).
- The electron has a charge of -1, and its mass number is 0.
- Lithium-6 has:
- Atomic number (Z) = 3 (since Lithium has 3 protons)
- Mass number (A) = 6

3. **Write the reaction**:
The reaction can be written as:
\[ ^6He \rightarrow e^- + ^6Li^+ \]

4. **Check conservation of charge**:
- Initial charge: Helium-6 has a charge of 0.
- Final charge: The electron has a charge of -1, and Lithium-6 has a charge of +1.
- Total final charge = -1 + 1 = 0, which matches the initial charge.

5. **Check conservation of mass number**:
- Initial mass number: 6 (from Helium-6).
- Final mass number: 0 (from the electron) + 6 (from Lithium-6) = 6.
- Total final mass number = 6, which matches the initial mass number.

6. **Include neutrino**:
In nuclear reactions, we often have to account for the emission of a neutrino. In this case, an anti-neutrino (\(\bar{\nu}\)) is also produced to conserve lepton number:
\[ ^6He \rightarrow e^- + ^6Li^+ + \bar{\nu} \]

7. **Final complete reaction**:
The complete reaction can be expressed as:
\[ ^6He \rightarrow e^- + ^6Li^+ + \bar{\nu} \]

### Final Answer:
The complete series is:
\[ ^6He \rightarrow e^- + ^6Li^+ + \bar{\nu} \]