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 find the energy of the electron in the second excited state of \( He^+ \) given that the energy of the electron in the ground state of hydrogen is \(-x\) eV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Formula**:
The energy of an electron in the nth orbit of a hydrogen-like 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 the Values for Hydrogen**:
For hydrogen (\( H \)), the atomic number \( Z = 1 \) and for the ground state, \( n = 1 \). Thus, the energy in the ground state is:
\[ E_1 = -\frac{13.6 \times 1^2}{1^2} = -13.6 \, \text{eV} \]
According to the problem, this energy is given as \(-x\) eV. Therefore, we have:
\[ x = 13.6 \, \text{eV} \]

3. **Determine the Values for \( He^+ \)**:
For \( He^+ \) (helium ion), the atomic number \( Z = 2 \). We want to find the energy in the second excited state. The second excited state corresponds to \( n = 3 \).

4. **Calculate the Energy for \( He^+ \)**:
Using the energy formula for \( He^+ \):
\[ E_3 = -\frac{13.6 \times 2^2}{3^2} \, \text{eV} \]
Simplifying this:
\[ E_3 = -\frac{13.6 \times 4}{9} \, \text{eV} = -\frac{54.4}{9} \, \text{eV} \]

5. **Relate to \( x \)**:
Since we found that \( x = 13.6 \, \text{eV} \), we can substitute this into our equation:
\[ E_3 = -\frac{54.4}{9} = -\frac{4 \times 13.6}{9} = -\frac{4x}{9} \, \text{eV} \]

6. **Final Answer**:
Therefore, the energy of the electron in the second excited state of \( He^+ \) is:
\[ E_3 = -\frac{4}{9} x \, \text{eV} \]