In a hydrogen atom , If the energy of electron in the ground state is -x eV ., then that in the ` 2^(nd)` excited state of ` He^(+)` is

Options

A. `- x eV`
B. `-(4)/(9) x eV`
C. ` +2x eV `
D. ` -(9)/(4) x eV `

Answer

To solve the problem, we need to determine the energy of an electron in the second excited state of a helium ion (He⁺) given that the energy of an electron in the ground state of a hydrogen atom is -x eV.

### Step-by-Step Solution:

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

2. **Identify Parameters for Hydrogen**:
For hydrogen, \( Z = 1 \) and in the ground state, \( n = 1 \):
\[ E_1 = -\frac{13.6 \times 1^2}{1^2} = -13.6 \text{ eV} \]
We are given that this energy is -x eV, thus:
\[ x = 13.6 \text{ eV} \]

3. **Identify Parameters for Helium Ion (He⁺)**:
For the helium ion (He⁺), \( Z = 2 \). We need to find the energy in the second excited state, which corresponds to \( n = 3 \) (since the ground state is \( n = 1 \), the first excited state is \( n = 2 \), and the second excited state is \( n = 3 \)).

4. **Calculate Energy for Helium Ion**:
Using the energy formula for He⁺:
\[ E_n = -\frac{13.6 \times Z^2}{n^2} = -\frac{13.6 \times 2^2}{3^2} \]
Substituting the values:
\[ E_3 = -\frac{13.6 \times 4}{9} = -\frac{54.4}{9} \text{ eV} \]

5. **Relate to x**:
Since we found that \( x = 13.6 \text{ eV} \), we can express the energy in terms of \( x \):
\[ E_3 = -\frac{4}{9} x \text{ eV} \]

### Final Answer:
The energy of the electron in the second excited state of He⁺ is:
\[ E = -\frac{4}{9} x \text{ eV} \]