The ionisation energy of `H` atom is `21.79xx10^(-19) J`. The the value of binding energy of second excited state of `Li^(2+)` ion

Options

A. `3^(2)xx21.7xx10^(-19)J`
B. `21.79xx10^(-19) J`
C. `1/3xx21.79xx10^(-19) J`
D. `1/3^(2)xx21.79xx10^(-19) J`

Explanation

To find the binding energy of the second excited state of the \( \text{Li}^{2+} \) ion, we can use the formula for the binding energy of an electron in a hydrogen-like atom:

\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]

Where:
- \( E_n \) is the binding energy of the electron in the nth energy level,
- \( Z \) is the atomic number of the ion,
- \( n \) is the principal quantum number.

### Step-by-step Solution:

1. **Identify the Atomic Number (Z)**:
- For \( \text{Li}^{2+} \), the atomic number \( Z \) is 3.

2. **Identify the Principal Quantum Number (n)**:
- The second excited state corresponds to \( n = 3 \).

3. **Substitute the Values into the Formula**:
- Using the formula for binding energy:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
- Substitute \( Z = 3 \) and \( n = 3 \):
\[ E_3 = -\frac{3^2 \cdot 13.6 \, \text{eV}}{3^2} \]

4. **Simplify the Expression**:
- The \( 3^2 \) in the numerator and denominator cancels out:
\[ E_3 = -13.6 \, \text{eV} \]

5. **Convert the Energy from eV to Joules**:
- To convert from eV to Joules, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E_3 = -13.6 \times 1.6 \times 10^{-19} \, \text{J} \]
\[ E_3 = -21.76 \times 10^{-19} \, \text{J} \]

### Final Answer:
The binding energy of the second excited state of the \( \text{Li}^{2+} \) ion is approximately \( -21.76 \times 10^{-19} \, \text{J} \).