The total energy of the electron in the first excited state of hydrogen is `-3.4 eV`. What is the kinetic energy of the electron in this state?

Options

A. `-3.4` eV
B. `-6.8` eV
C. `6.8` eV
D. `3.4` eV

Answer

To find the kinetic energy of the electron in the first excited state of hydrogen, we can use the relationship between total energy and kinetic energy in a hydrogen atom.

### Step-by-Step Solution:

1. **Understand the Total Energy of the Electron**:
The total energy (E) of an electron in a hydrogen atom is given as -3.4 eV for the first excited state (n=2).

2. **Use the Relationship Between Total Energy and Kinetic Energy**:
In a hydrogen atom, the total energy (E) is related to the kinetic energy (K) by the formula:
\[ K = -\frac{E}{2} \]
This relationship arises from the fact that the potential energy (U) is twice the negative of the kinetic energy in a bound system.

3. **Substitute the Total Energy**:
Given that the total energy \( E = -3.4 \, \text{eV} \), we can substitute this value into the equation:
\[ K = -\left(-3.4 \, \text{eV}\right) / 2 \]

4. **Calculate the Kinetic Energy**:
\[ K = \frac{3.4 \, \text{eV}}{2} = 1.7 \, \text{eV} \]

5. **Final Result**:
The kinetic energy of the electron in the first excited state of hydrogen is \( 1.7 \, \text{eV} \).

### Summary:
The kinetic energy of the electron in the first excited state of hydrogen is \( 1.7 \, \text{eV} \).