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 ionization energy and the energy of the electron in the atom. Here’s a step-by-step solution:

### Step 1: Understand the Ionization Energy
The ionization energy (IE) is the energy required to remove an electron from an atom or ion. For a hydrogen-like atom, the energy of the electron in the nth stationary state is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.
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