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 is 23.8 eV, we can follow these steps:

### Step 1: Understand the Energy Levels of Hydrogen Atom
The energy levels of a hydrogen atom can be described by the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number (which is 1 for hydrogen) and \( n \) is the principal quantum number.

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

### Step 3: Calculate the Energy in the First Excited State
For the first excited state (\( n = 2 \)):
\[ E_2 = -\frac{1^2 \cdot 13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 4: Relate Total Energy, Kinetic Energy, and Potential Energy
In quantum mechanics, the total energy \( E \) is related to the kinetic energy \( T \) and potential energy \( V \) as follows:
\[ E = T + V \]
For a hydrogen atom, the kinetic energy is given by:
\[ T = -\frac{1}{2} V \]
Thus, we can express the total energy as:
\[ E = T + V = -\frac{1}{2} V + V = \frac{1}{2} V \]
From this, we can derive that:
\[ V = 2E \]

### Step 5: Calculate the Potential Energy in the Ground State
Using the total energy in the ground state:
\[ V_1 = 2 \cdot E_1 = 2 \cdot (-13.6 \, \text{eV}) = -27.2 \, \text{eV} \]

### Conclusion
The potential energy of a hydrogen atom in the ground state can be assumed to be:
\[ V_1 = -27.2 \, \text{eV} \]

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