To analyze the transition of a hydrogen atom from the ground state to an excited state, we will consider the energy changes involved in this process.
### Step-by-Step Solution:
1. **Identify the Ground State and Excited State:**
- The ground state of a hydrogen atom corresponds to the principal quantum number \( n = 1 \).
- The first excited state corresponds to \( n = 2 \).
2. **Calculate the Total Energy for Each State:**
- The total energy \( E_n \) of a hydrogen atom in a state with quantum number \( n \) is given by the formula:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
- For the ground state (\( n = 1 \)):
\[
E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}
\]
- For the first excited state (\( n = 2 \)):
\[
E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV}
\]
3. **Determine the Kinetic Energy (KE) and Potential Energy (PE):**
- The kinetic energy \( KE \) is related to the total energy \( E \) and potential energy \( PE \) by the equation:
\[
KE = -\frac{1}{2} PE
\]
- For the ground state:
- Total energy \( E_1 = KE_1 + PE_1 \)
- Since \( E_1 = -13.6 \, \text{eV} \):
\[
PE_1 = 2 \times KE_1 \quad \text{and} \quad E_1 = KE_1 + PE_1
\]
Solving these gives:
\[
PE_1 = -27.2 \, \text{eV}, \quad KE_1 = 13.6 \, \text{eV}
\]
- For the first excited state:
- Total energy \( E_2 = KE_2 + PE_2 \)
- Since \( E_2 = -3.4 \, \text{eV} \):
\[
PE_2 = 2 \times KE_2 \quad \text{and} \quad E_2 = KE_2 + PE_2
\]
Solving these gives:
\[
PE_2 = -6.8 \, \text{eV}, \quad KE_2 = 3.4 \, \text{eV}
\]
4. **Analyze Changes in Kinetic and Potential Energy:**
- From ground state to first excited state:
- Kinetic Energy changes from \( 13.6 \, \text{eV} \) to \( 3.4 \, \text{eV} \) (decreases).
- Potential Energy changes from \( -27.2 \, \text{eV} \) to \( -6.8 \, \text{eV} \) (increases).
5. **Conclusion:**
- As the hydrogen atom transitions from the ground state to the excited state, the potential energy increases while the kinetic energy decreases. Therefore, the correct answer is:
- **Potential energy increases and kinetic energy decreases.**
