If the binding energy of the electron in a hydrogen atom is `13.6 eV`, the energy required to remove the electron from the first excited state of `Li^(++)` is

Options

A. `30.6 eV`
B. `13.6 eV`
C. `3.4 eV`
D. `122.4 eV`

Explanation

To solve the problem of finding the energy required to remove the electron from the first excited state of \( \text{Li}^{++} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Binding Energy**: The binding energy of an electron in a hydrogen atom is given as \( 13.6 \, \text{eV} \). This value represents the energy required to remove the electron from the ground state of hydrogen.

2. **Formula for Binding Energy**: The energy required to remove an electron from an excited state in a hydrogen-like atom is given by the formula:
\[ E = \frac{13.6 \, \text{eV} \times Z^2}{n^2} \]
where \( Z \) is the atomic number of the element and \( n \) is the principal quantum number of the excited state.

3. **Identify Values for Lithium**: For the lithium ion \( \text{Li}^{++} \):
- The atomic number \( Z = 3 \) (since lithium has 3 protons).
- The first excited state corresponds to \( n = 2 \).

4. **Substituting Values into the Formula**:
\[ E = \frac{13.6 \, \text{eV} \times 3^2}{2^2} \]
\[ E = \frac{13.6 \, \text{eV} \times 9}{4} \]

5. **Calculating the Energy**:
\[ E = \frac{122.4 \, \text{eV}}{4} = 30.6 \, \text{eV} \]

6. **Final Result**: The energy required to remove the electron from the first excited state of \( \text{Li}^{++} \) is \( 30.6 \, \text{eV} \).

### Conclusion:
Thus, the energy required to remove the electron from the first excited state of \( \text{Li}^{++} \) is \( 30.6 \, \text{eV} \).

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