The energy of an electron in first Bohr's orbit of H atom is `-13.6eV`. The energy value of electron in the first excited state of `Li^(2+)` is :

Options

A. 27.2 eV
B. `-30.6eV`
C. 30.6 eV
D. `-27.2eV`

Answer

To solve the problem of finding the energy value of an electron in the first excited state of \( \text{Li}^{2+} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the energy of the electron in the first Bohr orbit of hydrogen (H atom)**:
The energy of an electron in the first Bohr orbit (ground state) of hydrogen is given as:
\[ E_0 = -13.6 \, \text{eV} \]

2. **Use the formula for energy in the nth shell**:
The energy of an electron in the nth shell of a hydrogen-like atom is given by the formula:
\[ E_n = E_0 \cdot \frac{Z^2}{n^2} \]
where:
- \( E_n \) is the energy in the nth shell,
- \( E_0 \) is the energy of the electron in the first shell of hydrogen,
- \( Z \) is the atomic number of the element,
- \( n \) is the principal quantum number.

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

4. **Substitute the values into the formula**:
Now we can substitute \( Z \) and \( n \) into the energy formula:
\[ E_2 = E_0 \cdot \frac{Z^2}{n^2} = -13.6 \, \text{eV} \cdot \frac{3^2}{2^2} \]

5. **Calculate \( E_2 \)**:
First, calculate \( \frac{3^2}{2^2} \):
\[ \frac{3^2}{2^2} = \frac{9}{4} \]
Now substitute this back into the equation:
\[ E_2 = -13.6 \, \text{eV} \cdot \frac{9}{4} = -13.6 \, \text{eV} \cdot 2.25 \]
Now calculate:
\[ E_2 = -30.6 \, \text{eV} \]

6. **Final Answer**:
The energy value of the electron in the first excited state of \( \text{Li}^{2+} \) is:
\[ E_2 = -30.6 \, \text{eV} \]