A radioactive nucleus X converts into nucleus Y by emitting `B^+`. If atomic masses of X and Y are `M_X` and `M_Y` then Q value of reaction will be :-

Options

A. `Q=M_x C^2`
B. `Q = (M_X - M_Y - M_e) C^2`
C. `Q=(M_X - M_Y)C^2`
D. `Q = (M_X-M_Y-2m_e)C^2`

Answer

To find the Q value of the reaction where a radioactive nucleus X converts into nucleus Y by emitting a positron (B+), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The nucleus X emits a positron (B+) and transforms into nucleus Y.
- During this process, a proton is lost from nucleus X, which means that the atomic number decreases by 1.

2. **Defining the Q Value**:
- The Q value of a nuclear reaction is defined as the energy released during the reaction. It can be calculated using the formula:
\[ Q = \Delta m \cdot c^2 \]
- Here, \(\Delta m\) is the change in mass during the reaction, and \(c\) is the speed of light.

3. **Calculating the Change in Mass (\(\Delta m\))**:
- The change in mass due to the reaction can be expressed as:
\[ \Delta m = M_X - M_Y - \text{mass of positron} \]
- Since a positron is emitted, we need to account for the mass of the positron. However, the mass of the positron is equivalent to the mass of an electron (\(m_e\)).
- Additionally, since the emission of a positron results in the loss of a proton, we also need to account for the loss of two electrons (one to neutralize the charge of the positron emitted).
- Therefore, the total change in mass can be expressed as:
\[ \Delta m = M_X - M_Y - 2m_e \]

4. **Substituting into the Q Value Formula**:
- Now substituting \(\Delta m\) into the Q value formula gives:
\[ Q = (M_X - M_Y - 2m_e) \cdot c^2 \]

5. **Final Expression for Q Value**:
- Thus, the Q value of the reaction can be summarized as:
\[ Q = (M_X - M_Y - 2m_e) \cdot c^2 \]

### Conclusion:
The Q value of the reaction where nucleus X converts to nucleus Y by emitting a positron is given by:
\[ Q = (M_X - M_Y - 2m_e) \cdot c^2 \]