The total energy of an electron in the first excited state of hydrogen is about `-3.4 eV`. Its kinetic energy in this state is:

Options

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

Answer

To find the kinetic energy of an electron in the first excited state of hydrogen, we can use the relationship between total energy and kinetic energy derived from Bohr's model of the atom.

### Step-by-Step Solution:

1. **Understand the Total Energy**:
The total energy (E) of an electron in the first excited state of hydrogen is given as:
\[ E = -3.4 \, \text{eV} \]

2. **Use the Relationship Between Total Energy and Kinetic Energy**:
According to Bohr's model, the total energy (E) of an electron in a hydrogen atom is related to its kinetic energy (K) by the equation:
\[ E = K + U \]
where U is the potential energy. For hydrogen, the kinetic energy is half the magnitude of the total energy:
\[ K = -\frac{E}{2} \]
However, we can also directly state that:
\[ K = -E \]
because the potential energy (U) is twice the kinetic energy (K) in magnitude but negative.

3. **Calculate the Kinetic Energy**:
Using the relationship \( K = -E \):
\[ K = -(-3.4 \, \text{eV}) = 3.4 \, \text{eV} \]

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

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

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