Ionisation energy of `He^+` is `19.6 xx 10^-18 J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^( 2 +)` is.

Options

A. `4.41 xx 10^-16 J "atom"^-1`
B. `- 4.41 xx 10^-17 J "atom"^-1`
C. `- 2.2 xx 10^-15 J "atom"^-1`
D. `8.82 xx 10^17 J "atom"^-1`

Explanation

To find the energy of the first stationary state (n = 1) of \( \text{Li}^{2+} \), we can use the relationship between the ionization energy of a hydrogen-like atom and its energy levels. Here’s a step-by-step solution:

### Step 1: Understand the relationship between ionization energy and energy levels
The energy of the nth stationary state of a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{Z^2 \cdot E_H}{n^2} \]
where:
- \( E_H \) is the ionization energy of hydrogen (approximately \( 2.18 \times 10^{-18} \) J),
- \( Z \) is the atomic number of the element,
- \( n \) is the principal quantum number.

### Step 2: Identify the given data
From the question, we know:
- The ionization energy of \( \text{He}^+ \) is \( 19.6 \times 10^{-18} \) J.
- For \( \text{Li}^{2+} \), \( Z = 3 \) (since lithium has an atomic number of 3).

### Step 3: Relate the ionization energy of \( \text{He}^+ \) to \( \text{Li}^{2+} \)
The ionization energy of a hydrogen-like atom can also be expressed as:
\[ E_{ionization} = -E_n \]
For \( \text{He}^+ \):
\[ E_{ionization}(\text{He}^+) = -E_1(\text{He}^+) = 19.6 \times 10^{-18} \text{ J} \]
Thus, we can say:
\[ E_1(\text{He}^+) = -19.6 \times 10^{-18} \text{ J} \]

### Step 4: Calculate the energy of \( \text{Li}^{2+} \)
Using the proportionality of the energies:
\[ \frac{E_{Li^{2+}}}{E_{He^+}} = \frac{Z_{Li^{2+}}^2}{Z_{He^+}^2} \]
Where \( Z_{He^+} = 2 \) and \( Z_{Li^{2+}} = 3 \):
\[ \frac{E_{Li^{2+}}}{-19.6 \times 10^{-18}} = \frac{3^2}{2^2} = \frac{9}{4} \]

### Step 5: Solve for \( E_{Li^{2+}} \)
Rearranging gives:
\[ E_{Li^{2+}} = -19.6 \times 10^{-18} \times \frac{9}{4} \]
Calculating this:
\[ E_{Li^{2+}} = -19.6 \times 10^{-18} \times 2.25 = -44.1 \times 10^{-18} \text{ J} \]

### Step 6: Final result
Thus, the energy of the first stationary state (\( n = 1 \)) of \( \text{Li}^{2+} \) is:
\[ E_{Li^{2+}} = -4.41 \times 10^{-17} \text{ J} \]