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?

Options

A. 10.2 eV
B. 13.6 eV
C. 23.8 eV
D. 27.2 eV

Explanation

To find the energy of a hydrogen atom in the first excited state when the potential energy is taken to be zero in the ground state, we can follow these steps:

### Step 1: Understand the Energy Formula
The energy of an electron in a hydrogen atom at the nth energy level is given by the formula:
\[ E_n = -\frac{Z^2}{n^2} \times 13.6 \text{ eV} \]
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 (where \( n = 1 \)):
\[ E_1 = -\frac{1^2}{1^2} \times 13.6 \text{ eV} = -13.6 \text{ eV} \]

### Step 3: Calculate the Potential Energy in the Ground State
The potential energy (PE) in the ground state can be calculated using the relation:
\[ PE = 2 \times E \]
Thus, for the ground state:
\[ PE_1 = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV} \]

### Step 4: Adjust the Potential Energy to Zero
Since the problem states that the potential energy is taken to be zero in the ground state, we need to add 27.2 eV to the energy values to shift the potential energy to zero:
\[ PE_1 = 0 \text{ eV} \]
This means we need to add 27.2 eV to the total energy to make the potential energy zero.

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

### Step 6: Adjust the Energy for the First Excited State
Now, we need to add the same adjustment of 27.2 eV to the energy of the first excited state:
\[ E_2 = -3.4 \text{ eV} + 27.2 \text{ eV} = 23.8 \text{ eV} \]

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