The ionisation energy of `He^(o+)` is `19.6 xx 10^(-18)` J per atom .The energy of the first stationary state of `Li^(2+)` will be

Options

A. `21.2 xx10^(-18)"J/atom"`
B. `44.10xx10^(-18)"J/atom"`
C. `63.2xx10^(-18)"J/atom"`
D. `84.2xx10^(-18)"J/atom"`

Answer

To find the energy of the first stationary state of the \( \text{Li}^{2+} \) ion, we can use the relationship between ionization energy and atomic number. The ionization energy is proportional to the square of the atomic number (\( Z \)).

### Step-by-Step Solution:

1. **Identify the given data:**
- Ionization energy of \( \text{He}^{+} \) = \( 19.6 \times 10^{-18} \) J/atom.
- Atomic number of \( \text{He} \) (\( Z_{\text{He}} \)) = 2.
- Atomic number of \( \text{Li} \) (\( Z_{\text{Li}} \)) = 3.

2. **Set up the proportionality relation:**
The ionization energy (\( IE \)) is proportional to \( Z^2 \):
\[ \frac{IE_{\text{He}^{+}}}{IE_{\text{Li}^{2+}}} = \frac{Z_{\text{He}}^2}{Z_{\text{Li}}^2} \]

3. **Substitute the values:**
\[ \frac{19.6 \times 10^{-18}}{IE_{\text{Li}^{2+}}} = \frac{2^2}{3^2} \]
\[ \frac{19.6 \times 10^{-18}}{IE_{\text{Li}^{2+}}} = \frac{4}{9} \]

4. **Cross-multiply to solve for \( IE_{\text{Li}^{2+}} \):**
\[ 19.6 \times 10^{-18} \times 9 = 4 \times IE_{\text{Li}^{2+}} \]
\[ IE_{\text{Li}^{2+}} = \frac{19.6 \times 10^{-18} \times 9}{4} \]

5. **Calculate \( IE_{\text{Li}^{2+}} \):**
\[ IE_{\text{Li}^{2+}} = \frac{176.4 \times 10^{-18}}{4} = 44.1 \times 10^{-18} \text{ J/atom} \]
\[ IE_{\text{Li}^{2+}} = 4.41 \times 10^{-18} \text{ J/atom} \]

6. **Conclusion:**
The energy of the first stationary state of \( \text{Li}^{2+} \) is \( 4.41 \times 10^{-18} \) J per atom.