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. `3.4 eV`
B. `6.8 eV`
C. 10.2 eV`
D. zero

Explanation

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

### Step 1: Understand the Ground State Energy
The ground state energy of the hydrogen atom is given as:
\[ E_1 = -13.6 \, \text{eV} \]

### Step 2: Determine the Energy of the First Excited State
The energy of an electron in a hydrogen atom at a certain energy level \( n \) is given by the formula:
\[ E_n = \frac{-13.6 \, \text{eV}}{n^2} \]

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 3: Calculate 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 found:
\[ \text{Excitation Energy} = (-3.4 \, \text{eV}) - (-13.6 \, \text{eV}) \]
\[ \text{Excitation Energy} = -3.4 \, \text{eV} + 13.6 \, \text{eV} \]
\[ \text{Excitation Energy} = 10.2 \, \text{eV} \]

### Final Answer
The excitation energy when the electron is in the first excited state is:
\[ \text{Excitation Energy} = 10.2 \, \text{eV} \]

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