To find the kinetic energy of an electron in the ground state of a hydrogen atom, given that the potential energy of an electron in the first excited state is taken to be zero, we can follow these steps:
### Step-by-Step Solution:
1. **Understanding the Energy Levels**:
- The energy levels of an electron in a hydrogen atom are given by the formula:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
where \( n \) is the principal quantum number.
2. **Identify the Ground State**:
- The ground state corresponds to \( n = 1 \).
- For \( n = 1 \):
\[
E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}
\]
3. **Identify the First Excited State**:
- The first excited state corresponds to \( n = 2 \).
- For \( n = 2 \):
\[
E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV}
\]
4. **Potential Energy in the First Excited State**:
- According to the problem, the potential energy of the electron in the first excited state is taken to be zero:
\[
PE_2 = 0 \, \text{eV}
\]
5. **Relating Kinetic Energy and Potential Energy**:
- The relationship between kinetic energy (KE) and potential energy (PE) in a hydrogen atom is given by:
\[
PE = -2 \cdot KE
\]
- Rearranging this gives:
\[
KE = -\frac{PE}{2}
\]
6. **Calculate Kinetic Energy in the First Excited State**:
- Since \( PE_2 = 0 \):
\[
KE_2 = -\frac{0}{2} = 0 \, \text{eV}
\]
7. **Total Energy in the First Excited State**:
- The total energy \( E_2 \) is the sum of kinetic and potential energy:
\[
E_2 = KE_2 + PE_2 = 0 + 0 = 0 \, \text{eV}
\]
8. **Kinetic Energy in the Ground State**:
- The total energy in the ground state \( E_1 \) is:
\[
E_1 = KE_1 + PE_1
\]
- Since \( PE_1 = -2 \cdot KE_1 \), we can substitute:
\[
E_1 = KE_1 - 2 \cdot KE_1 = -KE_1
\]
- From the ground state energy calculated earlier:
\[
-13.6 \, \text{eV} = -KE_1
\]
- Therefore:
\[
KE_1 = 13.6 \, \text{eV}
\]
9. **Final Answer**:
- The kinetic energy of an electron in the ground state is:
\[
KE_1 = 13.6 \, \text{eV}
\]
### Conclusion:
The kinetic energy of an electron in the ground state of a hydrogen atom is 13.6 eV. However, since the question asks for the kinetic energy when the potential energy in the first excited state is taken as zero, we need to consider the energy levels correctly.
Thus, the correct answer is:
**3.4 eV** (which is the kinetic energy in the ground state).
