Find the Q-value of the given `beta^(+)` decay.
`""_(11)N^(22) rarr ""_(10)Ne^(22) + ""_(+1) e^(0)` Given that :
Atomic mass of `""_(11)N^(22)` is 21.994435 u, Atomic mass of `""_(11) Ne^(22)` is 21.991384 u, Mass of `""_(+1)e^(0)` is 0.0005486`

Options

A. 2.82 MeV
B. 1.82 MeV
C. 0.82 MeV
D. 3.82 MeV

Explanation

To find the Q-value of the given beta-plus decay reaction, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: The beta-plus decay reaction is given as:
\[ _{11}N^{22} \rightarrow _{10}Ne^{22} + _{+1}e^{0} \]
Here, nitrogen-22 decays into neon-22 and a positron (beta-plus particle).

2. **Write the Q-value Formula**: The Q-value for a nuclear reaction can be calculated using the formula:
\[ Q = (M_x - M_y - M_e) c^2 \]
where:
- \( M_x \) is the mass of the parent nucleus (nitrogen-22),
- \( M_y \) is the mass of the daughter nucleus (neon-22),
- \( M_e \) is the mass of the emitted positron,
- \( c \) is the speed of light.

3. **Substitute the Mass Values**:
- Mass of nitrogen-22, \( M_x = 21.994435 \, u \)
- Mass of neon-22, \( M_y = 21.991384 \, u \)
- Mass of positron, \( M_e = 0.0005486 \, u \)

Now substituting these values into the Q-value formula:
\[ Q = (21.994435 \, u - 21.991384 \, u - 0.0005486 \, u) c^2 \]

4. **Calculate the Mass Difference**:
\[ M_x - M_y - M_e = 21.994435 - 21.991384 - 0.0005486 = 0.0025024 \, u \]

5. **Convert Mass Difference to Energy**: To convert the mass difference to energy, we use the conversion factor \( 1 \, u \approx 931.5 \, MeV/c^2 \):
\[ Q = 0.0025024 \, u \times 931.5 \, MeV/u \]
\[ Q \approx 2.331 \, MeV \]

6. **Final Result**: The Q-value of the beta-plus decay reaction is approximately:
\[ Q \approx 2.331 \, MeV \]