The ionization of the ionized sodium atom `N a^(10)` is:

Options

A. `13.6 eV`
B. `13.6 xx 11 eV`
C. `(13.6//11) eV`
D. `13.6 xx (11^(2)) eV`

Explanation

To find the ionization energy of the ionized sodium atom \( Na^{10} \), we can use the formula for the energy of hydrogen-like atoms. Here’s the step-by-step solution:

### Step 1: Identify the parameters
- The atomic number \( Z \) for sodium (Na) is 11.
- The principal quantum number \( n \) for the ground state is 1.
- The ground state energy \( E_0 \) for hydrogen is \( 13.6 \, \text{eV} \).

### Step 2: Use the ionization energy formula
The ionization energy \( E \) for hydrogen-like atoms can be calculated using the formula:
\[ E = -\frac{Z^2}{n^2} \cdot E_0 \]
For ionization, we consider the transition from \( n = 1 \) to \( n = \infty \). The formula for the change in energy is:
\[ \Delta E = E(n = \infty) - E(n = 1) = 0 - \left(-\frac{Z^2}{n^2} \cdot E_0\right) \]
Thus,
\[ \Delta E = \frac{Z^2}{n^2} \cdot E_0 \]

### Step 3: Substitute the values
Substituting \( Z = 11 \), \( n = 1 \), and \( E_0 = 13.6 \, \text{eV} \):
\[ \Delta E = \frac{11^2}{1^2} \cdot 13.6 \, \text{eV} \]
Calculating \( 11^2 \):
\[ 11^2 = 121 \]
Now substituting this back into the equation:
\[ \Delta E = 121 \cdot 13.6 \, \text{eV} \]

### Step 4: Calculate the final value
Now, we perform the multiplication:
\[ \Delta E = 121 \cdot 13.6 = 1645.6 \, \text{eV} \]

### Conclusion
The ionization energy of the ionized sodium atom \( Na^{10} \) is:
\[ \Delta E = 1645.6 \, \text{eV} \]

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