The ionisation energy of the ionised sodium atom `Na^(+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`

Answer

To find the ionization energy of the ionized sodium atom \( Na^{+10} \), we can use the formula for ionization energy:

\[ IE = \frac{13.6 \, Z^2}{n^2} \]

### Step 1: Identify the values of \( Z \) and \( n \)

1. **Determine \( Z \)**: The atomic number of sodium (Na) is 11. Since we are considering \( Na^{+10} \), it means that 10 electrons have been removed from the neutral sodium atom. Therefore, the effective nuclear charge \( Z \) remains 11.

2. **Determine \( n \)**: The ionized sodium atom \( Na^{+10} \) has lost 10 electrons. The remaining electrons are in the first shell (1s orbital), which corresponds to \( n = 1 \).

### Step 2: Substitute the values into the formula

Now we can substitute the values of \( Z \) and \( n \) into the ionization energy formula:

\[ IE = \frac{13.6 \times (11)^2}{(1)^2} \]

### Step 3: Calculate \( Z^2 \)

Calculate \( Z^2 \):

\[ Z^2 = 11^2 = 121 \]

### Step 4: Calculate the ionization energy

Now substitute \( Z^2 \) back into the formula:

\[ IE = \frac{13.6 \times 121}{1} = 1643.6 \, \text{eV} \]

### Conclusion

The ionization energy of the ionized sodium atom \( Na^{+10} \) is:

\[ IE = 1643.6 \, \text{eV} \]