The total energy of eletcron in the ground state of hydrogen atom is `-13.6 eV`. The kinetic enegry of an electron in the first excited state is

Options

A. `3.4` eV
B. `6.8` eV
C. `13.6` eV
D. `1.7` eV

Answer

To solve the problem of finding the kinetic energy of an electron in the first excited state of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Total Energy Formula**:
The total energy \( E_n \) of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

2. **Identify the First Excited State**:
The first excited state corresponds to \( n = 2 \).

3. **Calculate Total Energy in the First Excited State**:
Substitute \( n = 2 \) into the total energy formula:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

4. **Relate Total Energy to Kinetic Energy**:
The kinetic energy \( K \) of the electron is related to the total energy by the equation:
\[ K = -E_n \]
Thus, substituting for \( E_2 \):
\[ K = -(-3.4 \, \text{eV}) = 3.4 \, \text{eV} \]

5. **Conclusion**:
The kinetic energy of the electron in the first excited state is:
\[ K = 3.4 \, \text{eV} \]

### Final Answer:
The kinetic energy of an electron in the first excited state is **3.4 eV**.

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