The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV.
What is the potential energy of the electron in this state ?

Options

A. `-6.8 eV`
B. `-8.6 eV`
C. `6.8 eV`
D. `8.6 eV`

Answer

To find the potential energy of an electron in the first excited state of the hydrogen atom, we can follow these steps:

### Step 1: Understand the relationship between total energy, kinetic energy, and potential energy
In a hydrogen atom, the total energy (E) of the electron is related to its kinetic energy (K) and potential energy (U) by the equation:
\[ E = K + U \]

### Step 2: Use the known values
We know from the problem that the total energy of the electron in the first excited state is:
\[ E = -3.4 \, \text{eV} \]

### Step 3: Relate kinetic energy and potential energy
For an electron in a hydrogen atom, the potential energy (U) is related to the kinetic energy (K) by the following relationships:
- The kinetic energy is given by:
\[ K = -\frac{1}{2} U \]
- Therefore, we can express the total energy in terms of potential energy:
\[ E = K + U = -\frac{1}{2} U + U = \frac{1}{2} U \]

### Step 4: Solve for potential energy
From the equation \( E = \frac{1}{2} U \), we can rearrange it to find the potential energy:
\[ U = 2E \]
Substituting the value of total energy:
\[ U = 2 \times (-3.4 \, \text{eV}) = -6.8 \, \text{eV} \]

### Conclusion
The potential energy of the electron in the first excited state of the hydrogen atom is:
\[ U = -6.8 \, \text{eV} \]

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