The ionisation energy of `He^(o+)` is `19.6 xx 10^(-18) J atom^(-1)` .The energy of the first stationary state of `Li^(2+)` will be

Options

A. `84.2 xx 10^(-18) J" atom" ^(-1)`
B. `84.10 xx 10^(-18) J "atom" ^(-1)`
C. `63.2 xx 10^(-18) J "atom" ^(-1)`
D. `21.2 xx 10^(-18 J) "atom" ^(-1)`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Ionization Energy of \( \text{He}^{+} \)**:
The ionization energy of \( \text{He}^{+} \) is given as:
\[
E_{\text{He}^{+}} = 19.6 \times 10^{-18} \, \text{J}
...