What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?

Explanation

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

### Step 1: Understand the Energy Formula
The energy of an electron in the nth orbit of 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 (for hydrogen, \( Z = 1 \)) and \( n \) is the principal quantum number.

### Step 2: Calculate the Energy in the Ground State
For the ground state, \( n = 1 \):
\[ E_1 = -\frac{13.6 \times 1^2}{1^2} = -13.6 \, \text{eV} \]

### Step 3: Calculate the Potential Energy in the Ground State
The relationship between total energy (E), kinetic energy (K), and potential energy (U) is:
\[ E = K + U \]
Also, we know that:
\[ K = -\frac{1}{2} U \]
From the ground state energy, we can express the potential energy:
\[ E_1 = K + U \implies -13.6 = K + U \]
Substituting \( K = -\frac{1}{2} U \):
\[ -13.6 = -\frac{1}{2} U + U \implies -13.6 = \frac{1}{2} U \implies U = -27.2 \, \text{eV} \]

### Step 4: Adjust Potential Energy to be Zero in the Ground State
If we take the potential energy in the ground state to be zero, we need to add 27.2 eV to the potential energy:
\[ U_{\text{new}} = U + 27.2 = 0 \, \text{eV} \]

### Step 5: Calculate the Energy in the First Excited State
For the first excited state, \( n = 2 \):
\[ E_2 = -\frac{13.6 \times 1^2}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]

### Step 6: Adjust the Total Energy with New Potential Energy
Now, we need to adjust the total energy for the first excited state:
\[ E_{\text{total}} = E_2 + U_{\text{new}} = -3.4 + 27.2 = 23.8 \, \text{eV} \]

### Final Answer
The energy of the hydrogen atom in the first excited state, with the potential energy taken to be zero in the ground state, is:
\[ \boxed{23.8 \, \text{eV}} \]