The potential energy of an electron in the hydrogen atom is -6.8 eV. Indicate in which excited state, the electron is present?

Options

A. first
B. second
C. third
D. fourth

Answer

To determine the excited state of an electron in a hydrogen atom given its potential energy, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The potential energy (PE) of the electron in the hydrogen atom is given as -6.8 eV.

2. **Relate Potential Energy to Total Energy**:
- The relationship between potential energy (PE) and total energy (TE) in a hydrogen atom is given by the formula:
\[ PE = 2 \times TE \]
- Rearranging this gives us:
\[ TE = \frac{PE}{2} \]

3. **Calculate Total Energy**:
- Substitute the given potential energy into the equation:
\[ TE = \frac{-6.8 \, \text{eV}}{2} = -3.4 \, \text{eV} \]

4. **Use the Formula for Total Energy in Hydrogen Atom**:
- The total energy for an electron in a hydrogen atom is given by:
\[ TE = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
- For hydrogen, \( Z = 1 \), so:
\[ TE = -\frac{13.6 \, \text{eV}}{n^2} \]

5. **Set the Total Energy Equal to the Calculated Value**:
- We can set the total energy we calculated equal to the formula:
\[ -3.4 \, \text{eV} = -\frac{13.6 \, \text{eV}}{n^2} \]

6. **Solve for \( n^2 \)**:
- Rearranging gives:
\[ n^2 = \frac{13.6 \, \text{eV}}{3.4 \, \text{eV}} = 4 \]

7. **Calculate \( n \)**:
- Taking the square root of both sides:
\[ n = \sqrt{4} = 2 \]

8. **Determine the Excited State**:
- The principal quantum number \( n = 1 \) corresponds to the ground state, and \( n = 2 \) corresponds to the first excited state. Therefore, the electron is in the **first excited state**.

### Final Answer:
The electron is present in the **first excited state** (n = 2).

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