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 Atomic Number (Z)**:
The atomic number of lithium (Li) is 3. Therefore, for `Li^(++)`, Z = 3.

2. **Determine the Ground State and First Excited State**:
- The ground state corresponds to n1 = 1.
- The first excited state corresponds to n2 = 2.

3. **Use the Formula for Excitation Energy**:
The energy difference (ΔE) between two energy levels in a hydrogen-like atom is given by the formula:
\[ \Delta E = 13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

4. **Substitute the Values into the Formula**:
- Here, Z = 3, n1 = 1, and n2 = 2.
- First, calculate \(Z^2\):
\[ Z^2 = 3^2 = 9 \]
- Now substitute into the formula:
\[ \Delta E = 13.6 \times 9 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

5. **Calculate the Terms Inside the Parentheses**:
- Calculate \( \frac{1}{1^2} - \frac{1}{2^2} \):
\[ \frac{1}{1} - \frac{1}{4} = 1 - 0.25 = 0.75 \]

6. **Final Calculation**:
- Now substitute back into the energy difference equation:
\[ \Delta E = 13.6 \times 9 \times 0.75 \]
- Calculate:
\[ = 13.6 \times 9 = 122.4 \]
\[ \Delta E = 122.4 \times 0.75 = 91.8 \, \text{eV} \]

### Conclusion:
The first excitation energy of the ground state of the `Li^(++)` atom is **91.8 eV**.

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