To solve the problem, we need to find the kinetic energy of an electron in the ground state of a hydrogen atom, given that the potential energy of the electron in the first excited state is taken to be zero.
### Step-by-Step Solution:
1. **Identify the Energy Levels**:
- The first excited state corresponds to \( n = 2 \).
- The ground state corresponds to \( n = 1 \).
2. **Calculate the Total Energy in the First Excited State**:
- The formula for the total energy of an electron in a hydrogen atom is given by:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
- For \( n = 2 \):
\[
E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV}
\]
3. **Relate Kinetic Energy and Potential Energy**:
- The potential energy (PE) is related to the kinetic energy (KE) by the equation:
\[
PE = -2 \times KE
\]
- The total energy can also be expressed as:
\[
E = PE + KE
\]
- Rearranging gives:
\[
E = -KE
\]
4. **Calculate Potential Energy in the First Excited State**:
- Given that the potential energy in the first excited state is taken to be zero, we need to adjust the energy levels accordingly.
- The potential energy in the first excited state is:
\[
PE = -2 \times KE
\]
- From the total energy calculated:
\[
E_2 = -3.4 \, \text{eV} \Rightarrow -KE = -3.4 \, \text{eV} \Rightarrow KE = 3.4 \, \text{eV}
\]
- Therefore, the potential energy in the first excited state is:
\[
PE = -2 \times 3.4 \, \text{eV} = -6.8 \, \text{eV}
\]
5. **Adjust for the Given Condition**:
- Since the problem states that the potential energy in the first excited state is zero, we need to add \( 6.8 \, \text{eV} \) to all energy levels to shift them.
- Thus, the new total energy in the ground state becomes:
\[
E_1 = -13.6 \, \text{eV} + 6.8 \, \text{eV} = -6.8 \, \text{eV}
\]
6. **Calculate the Kinetic Energy in the Ground State**:
- Using the relationship \( E = -KE \):
\[
-KE = -6.8 \, \text{eV} \Rightarrow KE = 6.8 \, \text{eV}
\]
- Therefore, the kinetic energy in the ground state is:
\[
KE = 6.8 \, \text{eV}
\]
7. **Final Adjustment**:
- Since we need to add \( 6.8 \, \text{eV} \) to the kinetic energy in the ground state as well:
\[
KE_{new} = 6.8 \, \text{eV} + 6.8 \, \text{eV} = 13.6 \, \text{eV}
\]
### Conclusion:
The kinetic energy of an electron in the ground state, after adjusting for the given condition, is \( 20.4 \, \text{eV} \).
