The ionisation energy of `He^+" is "19.6 xx 10^(-12)` J/atom. Calculate the energy of the first stationary state of `Li^(2+)`.

Explanation

To solve the problem of calculating the energy of the first stationary state of \( \text{Li}^{2+} \), we will use the relationship between the energy levels of hydrogen-like ions. The energy of the first stationary state (ground state) can be derived from the ionization energy of \( \text{He}^{+} \).

### Step-by-Step Solution:

1. **Identify the given data:**
- Ionization energy of \( \text{He}^{+} = 19.6 \times 10^{-12} \, \text{J/atom} \)
- Atomic number of \( \text{He} (Z_{He}) = 2 \)
- Atomic number of \( \text{Li} (Z_{Li}) = 3 \)

2. **Understand the relationship between energy levels:**
The energy of the first stationary state (ground state) for hydrogen-like ions can be expressed as:
\[ E_n = -\frac{Z^2 \cdot k}{n^2} \]
where \( Z \) is the atomic number, \( n \) is the principal quantum number, and \( k \) is a constant. For our purposes, we can relate the energies of different ions using the ionization energy.

3. **Set up the ratio of energies:**
The relationship between the energies of the first stationary states of \( \text{He}^{+} \) and \( \text{Li}^{2+} \) can be written as:
\[ \frac{E_{1, \text{He}^{+}}}{E_{1, \text{Li}^{2+}}} = \frac{Z_{He}^{2}}{Z_{Li}^{2}} \]
Since we are looking for \( E_{1, \text{Li}^{2+}} \), we can rearrange this to:
\[ E_{1, \text{Li}^{2+}} = E_{1, \text{He}^{+}} \cdot \frac{Z_{Li}^{2}}{Z_{He}^{2}} \]

4. **Substitute the known values:**
- \( Z_{He} = 2 \) and \( Z_{Li} = 3 \)
\[ E_{1, \text{Li}^{2+}} = -19.6 \times 10^{-12} \cdot \frac{3^2}{2^2} \]
\[ E_{1, \text{Li}^{2+}} = -19.6 \times 10^{-12} \cdot \frac{9}{4} \]

5. **Calculate the energy:**
\[ E_{1, \text{Li}^{2+}} = -19.6 \times 10^{-12} \cdot 2.25 \]
\[ E_{1, \text{Li}^{2+}} = -44.1 \times 10^{-12} \, \text{J/atom} \]

6. **Final answer:**
\[ E_{1, \text{Li}^{2+}} = -4.41 \times 10^{-11} \, \text{J/atom} \]