If the energy in the first excited state in hydrogen atom is `23.8 eV` then the potential energy of a hydrogen atom in the ground state can be assumed to be

Options

A. `10 eV`
B. `23.3 eV`
C. `-13.6 eV`
D. Zero

Answer

To find the potential energy of a hydrogen atom in the ground state, given that the energy in the first excited state (n=2) is 23.8 eV, we can follow these steps:

### Step 1: Understand the Energy Levels
According to the Bohr model of the hydrogen atom, the energy of an electron in the nth energy level is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

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

### Step 3: Calculate the Energy in the First Excited State (n=2)
For the first excited state (n=2):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 4: Relate the Given Energy to the Calculated Energy
The problem states that the energy in the first excited state is 23.8 eV. This is a reference energy, and we need to adjust our calculations based on this information.

### Step 5: Determine the Potential Energy
In the Bohr model, the potential energy (PE) is related to the total energy (E) by the equation:
\[ PE = 2E \]
Thus, for the ground state:
\[ PE_1 = 2 \times E_1 = 2 \times (-13.6 \, \text{eV}) = -27.2 \, \text{eV} \]

### Step 6: Adjust the Reference Point
Since the energy in the first excited state is given as 23.8 eV, we need to adjust our reference point. The potential energy in the ground state can be considered as:
\[ PE_{\text{ground state}} = 0 \, \text{eV} \]

### Conclusion
Thus, the potential energy of a hydrogen atom in the ground state can be assumed to be:
\[ \text{Potential Energy} = 0 \, \text{eV} \]

### Final Answer
The potential energy of a hydrogen atom in the ground state can be assumed to be **0 eV**.

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