A common example of `beta-"decay"` is
`""_(15)P^(32)to""_(16)P^(32)+x+y`
Then, x and y stand for

Options

A. electron and nectrino
B. positron and neutrino
C. electron and antineutrino
D. positron and antineutrino

Explanation

To solve the problem, we need to analyze the beta decay process described in the equation:

\[ _{15}^{32}P \rightarrow _{16}^{32}P + x + y \]

### Step-by-Step Solution:

1. **Identify the Initial and Final Nuclei**:
- The initial nucleus is Phosphorus-32 \((_{15}^{32}P)\), which has 15 protons and 17 neutrons (since 32 - 15 = 17).
- The final nucleus is Sulfur-32 \((_{16}^{32}P)\), which has 16 protons and 16 neutrons (since 32 - 16 = 16).

2. **Understand Beta Decay**:
- In beta decay, a neutron in the nucleus is converted into a proton. This process increases the atomic number by 1 (from 15 to 16) while keeping the mass number the same (32).
- The reaction can be summarized as:
\[ n \rightarrow p + e^- + \bar{\nu} \]
- Here, \(n\) is a neutron, \(p\) is a proton, \(e^-\) is the emitted beta particle (electron), and \(\bar{\nu}\) is the emitted antineutrino.

3. **Determine the Particles Emitted**:
- From the beta decay process, we can identify that:
- \(x\) corresponds to the emitted electron \((e^-)\).
- \(y\) corresponds to the emitted antineutrino \((\bar{\nu})\).

4. **Write the Final Answer**:
- Thus, we conclude that:
- \(x = e^-\) (electron)
- \(y = \bar{\nu}\) (antineutrino)

### Final Answer:
- \(x\) stands for the electron and \(y\) stands for the antineutrino.