Complete the decay reaction `._10Na^(23) to?+._-1e^0+?`
Also, find the maximum KE of electrons emitted during this decay. Given mass of `._10Na^(23)=22.994465 u`. mass of `._11Na^(23)=22.989768u`.

Explanation

To solve the decay reaction and find the maximum kinetic energy of the emitted electrons, we will follow these steps:

### Step 1: Write the decay reaction
The decay reaction can be represented as:
\[ _{10}^{23}\text{Na} \rightarrow \, ? + \, _{-1}^{0}e + \, ? \]
Here, we need to identify the products of the decay.

### Step 2: Determine the atomic number and mass number of the products
Let’s denote the product of the decay as \( _{Z}^{A}X \).

1. **Conservation of charge (atomic number)**:
The atomic number of sodium (Na) is 10. The emitted electron has an atomic number of -1. Therefore, we can set up the equation:
\[ 10 = Z - 1 \implies Z = 10 + 1 = 11 \]

2. **Conservation of mass number**:
The mass number of sodium is 23. Since the electron has a mass number of 0, we can set up the equation:
\[ 23 = A + 0 \implies A = 23 \]

Thus, the product of the decay is \( _{11}^{23}\text{Na} \).

### Step 3: Write the complete decay reaction
Now we can write the complete decay reaction:
\[ _{10}^{23}\text{Na} \rightarrow \, _{11}^{23}\text{Na} + \, _{-1}^{0}e + \, Q \]
Where \( Q \) represents the energy released during the decay.

### Step 4: Calculate the maximum kinetic energy of the emitted electrons
To find the maximum kinetic energy (KE) of the emitted electrons, we use the mass difference between the reactants and products. The kinetic energy can be calculated using the formula:
\[ KE = \Delta m \times 931 \, \text{MeV} \]
Where \( \Delta m \) is the mass difference.

1. **Calculate the mass difference**:
\[ \Delta m = m_{Na_{10}} - m_{Na_{11}} = 22.994465 \, u - 22.989768 \, u \]
\[ \Delta m = 0.004697 \, u \]

2. **Convert the mass difference to energy**:
\[ KE = 0.004697 \, u \times 931 \, \text{MeV/u} = 4.37 \, \text{MeV} \]

### Final Answer
The complete decay reaction is:
\[ _{10}^{23}\text{Na} \rightarrow \, _{11}^{23}\text{Na} + \, _{-1}^{0}e + \, Q \]
The maximum kinetic energy of the emitted electrons is:
\[ KE = 4.37 \, \text{MeV} \]