`M_(x) and M_(y)` denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q - value for a `beta-` decay is `Q_(1)` and that for a `beta^(+)` decay is `Q_(2)`. If `m_(e)` denotes the mass of an electrons, then which of the following statements is correct?

Options

A. `Q_(1) =(M_(x)-M_(y)) c^(2) " and " Q_(2) =[M_(x)-M_(y) -2m_(e)]c^(2)`
B. `Q_(1) =(M_(x)-M_(y))c^(2) " and " Q_(2) =(M_(x) -M_(y))c^(2)`
C. `Q_(1)=(M_(x) -M_(y)) c^(2) " and " Q_(2) = (M_(x) -M_(y) +2c_(e))c^(2)`
D. `Q_(1) =(M_(x)-M_(y) +2m_(e)) c^(2) " and " Q_(2) =(M_(x)-m_(y) +2m_(e))c^(2)`

Explanation

To solve the problem regarding the Q-values for beta decay processes, we will analyze both beta minus (β-) decay and beta plus (β+) decay step by step.

### Step 1: Understanding Beta Minus Decay
In beta minus decay, a parent nucleus \( X \) with atomic mass \( M_X \) and atomic number \( Z \) decays into a daughter nucleus \( Y \) with atomic mass \( M_Y \) and atomic number \( Z + 1 \). During this process, a beta minus particle (an electron) and an antineutrino are emitted.

The reaction can be represented as:
\[ X \rightarrow Y + e^- + \bar{\nu} \]

### Step 2: Mass Defect in Beta Minus Decay
The mass defect (\( \Delta m \)) for beta minus decay can be calculated as follows:

1. The mass of the reactants (parent nucleus and emitted particles):
\[ \text{Mass of reactants} = M_X - Z \cdot m_e \]
where \( m_e \) is the mass of the electron.

2. The mass of the products (daughter nucleus and emitted particles):
\[ \text{Mass of products} = M_Y - (Z + 1) \cdot m_e + m_e = M_Y - Z \cdot m_e \]

3. Therefore, the mass defect is:
\[ \Delta m_1 = (M_X - Z \cdot m_e) - (M_Y - Z \cdot m_e) = M_X - M_Y \]

### Step 3: Q-Value for Beta Minus Decay
The Q-value for beta minus decay (\( Q_1 \)) is given by:
\[ Q_1 = \Delta m_1 \cdot c^2 = (M_X - M_Y) \cdot c^2 \]

### Step 4: Understanding Beta Plus Decay
In beta plus decay, the parent nucleus \( X \) decays into a daughter nucleus \( Y \) with atomic mass \( M_Y \) and atomic number \( Z - 1 \). During this process, a beta plus particle (a positron) and a neutrino are emitted.

The reaction can be represented as:
\[ X \rightarrow Y + e^+ + \nu \]

### Step 5: Mass Defect in Beta Plus Decay
The mass defect (\( \Delta m \)) for beta plus decay can be calculated similarly:

1. The mass of the reactants:
\[ \text{Mass of reactants} = M_X - Z \cdot m_e \]

2. The mass of the products:
\[ \text{Mass of products} = M_Y - (Z - 1) \cdot m_e + m_e = M_Y - (Z - 1) \cdot m_e + m_e = M_Y - (Z - 1) \cdot m_e + m_e \]

3. Therefore, the mass defect is:
\[ \Delta m_2 = (M_X - Z \cdot m_e) - (M_Y - (Z - 1) \cdot m_e + m_e) = M_X - M_Y - 2m_e \]

### Step 6: Q-Value for Beta Plus Decay
The Q-value for beta plus decay (\( Q_2 \)) is given by:
\[ Q_2 = \Delta m_2 \cdot c^2 = (M_X - M_Y - 2m_e) \cdot c^2 \]

### Step 7: Conclusion
From the calculations, we have:
- \( Q_1 = (M_X - M_Y) \cdot c^2 \)
- \( Q_2 = (M_X - M_Y - 2m_e) \cdot c^2 \)

Thus, the correct statement comparing \( Q_1 \) and \( Q_2 \) is:
\[ Q_2 = Q_1 - 2m_e \cdot c^2 \]
This indicates that \( Q_2 \) is less than \( Q_1 \) by an amount equal to the energy equivalent of the mass of two electrons.