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 a hydrogen atom, we can use the relationship between total energy, kinetic energy, and potential energy.

### Step-by-Step Solution:

1. **Understanding Total Energy**:
The total energy (E) of an electron in an atom is given by the sum of its kinetic energy (K) and potential energy (U):
\[ E = K + U \]

2. **Potential Energy in Hydrogen Atom**:
For a hydrogen atom, the potential energy (U) can be expressed as:
\[ U = -\frac{Ze^2}{4\pi \epsilon_0 r} \]
where \(Z\) is the atomic number (1 for hydrogen), \(e\) is the charge of the electron, \(\epsilon_0\) is the permittivity of free space, and \(r\) is the radius of the electron's orbit.

3. **Kinetic Energy Relation**:
The kinetic energy (K) of the electron can be related to the potential energy (U) by the formula:
\[ K = -\frac{1}{2} U \]

4. **Substituting Potential Energy**:
Using the expression for potential energy, we can substitute it into the kinetic energy formula:
\[ K = -\frac{1}{2} \left(-\frac{Ze^2}{4\pi \epsilon_0 r}\right) = \frac{Ze^2}{8\pi \epsilon_0 r} \]

5. **Total Energy in First Excited State**:
We know from the problem that the total energy in the first excited state of hydrogen is:
\[ E = -3.4 \, \text{eV} \]

6. **Relating Total Energy to Kinetic Energy**:
From the total energy equation, we can express kinetic energy as:
\[ K = E - U \]
Since \(U = -2E\) for hydrogen-like atoms, we can substitute:
\[ K = -3.4 \, \text{eV} - (-2 \times -3.4 \, \text{eV}) = -3.4 \, \text{eV} + 6.8 \, \text{eV} = 3.4 \, \text{eV} \]

7. **Conclusion**:
Therefore, 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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