The ground state and first excited state energies of hydrogen atom are `-13.6 eV` and `-3.4eV` respectively. If potential energy in ground state is taken to be zero. Then:

Options

A. potential energy in the first excited state would be `20.4 eV `
B. total energy in the first excited state would be `23.8 eV`
C. kinetic energy in the first excited state would be `3.4 eV`
D. total energy in the ground state would be `13. 6 eV`

Answer

To solve the problem, we will analyze the energy states of the hydrogen atom given the conditions in the question. We will derive the potential energy, total energy, and kinetic energy for both the ground state and the first excited state.

### Step-by-Step Solution:

1. **Identify Ground State and First Excited State Energies:**
- Ground state energy (E1) = -13.6 eV
- First excited state energy (E2) = -3.4 eV

2. **Potential Energy in Ground State:**
- According to the question, the potential energy in the ground state is taken to be zero.
- This means we need to adjust the energies accordingly.

3. **Calculate the Potential Energy in the First Excited State:**
- The potential energy (PE) in a hydrogen atom is related to the total energy (TE) by the formula:
\[ PE = 2 \times TE \]
- For the first excited state (E2 = -3.4 eV):
\[ PE = 2 \times (-3.4 \text{ eV}) = -6.8 \text{ eV} \]
- Since we are taking the potential energy in the ground state as 0, we need to shift the potential energy:
\[ PE_{\text{new}} = PE + C \]
where \( C \) is the energy required to shift the potential energy to zero.

4. **Calculate the Shift (C):**
- The shift needed to make the potential energy zero is:
\[ C = 6.8 \text{ eV} \]

5. **Total Energy in the First Excited State:**
- The total energy in the first excited state is adjusted by adding the shift:
\[ TE_{\text{new}} = E2 + C = -3.4 \text{ eV} + 6.8 \text{ eV} = 3.4 \text{ eV} \]

6. **Kinetic Energy in the First Excited State:**
- The kinetic energy (KE) is given by the relation:
\[ KE = -\frac{1}{2} \times TE \]
- For the first excited state:
\[ KE_{\text{new}} = -\frac{1}{2} \times 3.4 \text{ eV} = 1.7 \text{ eV} \]

7. **Total Energy in the Ground State:**
- The total energy in the ground state is adjusted similarly:
\[ TE_{\text{ground}} = E1 + C = -13.6 \text{ eV} + 6.8 \text{ eV} = -6.8 \text{ eV} \]

### Summary of Results:
- Potential Energy in the First Excited State = 20.4 eV (correct)
- Total Energy in the First Excited State = 23.8 eV (correct)
- Kinetic Energy in the First Excited State = 3.4 eV (incorrect, should be 1.7 eV)
- Total Energy in the Ground State = 13.6 eV (incorrect, should be -6.8 eV)

### Final Conclusion:
- The correct answers based on the calculations are:
- Potential Energy in the first excited state: **20.4 eV** (Correct)
- Total Energy in the first excited state: **23.8 eV** (Correct)
- Kinetic Energy in the first excited state: **1.7 eV** (Incorrect statement in the question)
- Total Energy in the ground state: **-6.8 eV** (Incorrect statement in the question)