The groud state energy of hydrogen atom is `-13.6 eV`. When its electron is in first excited state, its exciation energy is

Options

A. `54.4` eV
B. `-27.2` eV
C. `10.2` eV
D. `-3.4` eV

Answer

To find the excitation energy of the hydrogen atom when its electron is excited from the ground state to the first excited state, we can follow these steps:

### Step 1: Understand the energy levels of the hydrogen atom
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

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

### Step 3: Calculate the first excited state energy
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: Determine the excitation energy
The excitation energy is the difference in energy between the first excited state and the ground state:
\[ \text{Excitation Energy} = E_2 - E_1 \]
Substituting the values we calculated:
\[ \text{Excitation Energy} = (-3.4 \, \text{eV}) - (-13.6 \, \text{eV}) = -3.4 \, \text{eV} + 13.6 \, \text{eV} = 10.2 \, \text{eV} \]

### Final Answer
The excitation energy required to move the electron from the ground state to the first excited state of the hydrogen atom is:
\[ \text{Excitation Energy} = 10.2 \, \text{eV} \]

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