In beta decay, `""_(10)Ne^(23)` is converted into `""_(11)Na^(23)`. Calculate the maximum kinetic energy of the electrons emitted, given atomic masses of `""_(10)Ne^(23)` and `""_(11)Na^(23)` arc 22.994466 u and 22.989770 u, respectively.

Answer

To solve the problem of calculating the maximum kinetic energy of the electrons emitted during the beta decay of \( _{10}^{23}\text{Ne} \) to \( _{11}^{23}\text{Na} \), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Atomic mass of \( _{10}^{23}\text{Ne} \) = 22.994466 u
- Atomic mass of \( _{11}^{23}\text{Na} \) = 22.989770 u

2. **Calculate the Mass Defect (\( \Delta m \)):**
\[ \Delta m = \text{mass of Ne} - \text{mass of Na} = 22.994466 \, \text{u} - 22.989770 \, \text{u} \]
\[ \Delta m = 0.004696 \, \text{u} \]

3. **Convert the Mass Defect to Energy:**
The energy equivalent of the mass defect can be calculated using the formula:
\[ E = \Delta m \cdot c^2 \]
where \( 1 \, \text{u} \) is equivalent to \( 931.5 \, \text{MeV} \).
\[ E = 0.004696 \, \text{u} \times 931.5 \, \text{MeV/u} \]

4. **Perform the Calculation:**
\[ E = 0.004696 \times 931.5 \approx 4.3743 \, \text{MeV} \]

5. **Conclusion:**
The maximum kinetic energy of the electrons emitted during the beta decay of \( _{10}^{23}\text{Ne} \) to \( _{11}^{23}\text{Na} \) is approximately \( 4.3743 \, \text{MeV} \).