Assuming Bohr's model for `Li^(++)` atom, the first excitation energy of ground state of `Li^(++)` atom is

Options

A. `10.2 eV`
B. `91.8 eV`
C. `13.6 eV`
D. `3.4 eV`

Answer

To find the first excitation energy of the ground state of the `Li^(++)` atom using Bohr's model, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ground State**:
- The ground state corresponds to the principal quantum number \( n = 1 \).

2. **Determine the Excited State**:
- The first excitation means the electron moves from \( n = 1 \) to \( n = 2 \).

3. **Use the Energy Formula for Hydrogen-like Atoms**:
- The energy of an electron in a hydrogen-like atom is given by:
\[ E_n = -\frac{Z^2}{n^2} E_0 \]
where \( Z \) is the atomic number, \( n \) is the principal quantum number, and \( E_0 = 13.6 \, \text{eV} \).

4. **Calculate the Energy for \( n = 1 \) and \( n = 2 \)**:
- For Lithium ion \( Li^{++} \), the atomic number \( Z = 3 \).
- Calculate \( E_1 \) (for \( n = 1 \)):
\[ E_1 = -\frac{3^2}{1^2} \times 13.6 \, \text{eV} = -9 \times 13.6 \, \text{eV} = -122.4 \, \text{eV} \]
- Calculate \( E_2 \) (for \( n = 2 \)):
\[ E_2 = -\frac{3^2}{2^2} \times 13.6 \, \text{eV} = -\frac{9}{4} \times 13.6 \, \text{eV} = -30.6 \, \text{eV} \]

5. **Find the Excitation Energy**:
- The excitation energy is the difference between the energy of the excited state and the ground state:
\[ \text{Excitation Energy} = E_2 - E_1 \]
\[ \text{Excitation Energy} = (-30.6 \, \text{eV}) - (-122.4 \, \text{eV}) = 91.8 \, \text{eV} \]

6. **Conclusion**:
- The first excitation energy of the ground state of the `Li^(++)` atom is \( 91.8 \, \text{eV} \).