A hydrogen atom in state `n = 6` makes two successive transition and reaches the ground state in the first transition a photon of `1.13eV` is emitted (a) Find the energy of the photon emitted in the second transition (b) what is the value of n in the intermediate state?

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as per the question.

### Step-by-Step Solution:

#### Part (a): Find the energy of the photon emitted in the second transition.

1. **Identify the initial and final states**:
- The hydrogen atom starts in the state \( n = 6 \) and transitions to the ground state \( n = 1 \) through an intermediate state \( n = n_i \).

2. **Calculate the energy of the initial state \( n = 6 \)**:
- The energy of an electron in a hydrogen atom at a given principal quantum number \( n \) is given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
- For \( n = 6 \):
\[ E_6 = -\frac{13.6 \, \text{eV}}{6^2} = -\frac{13.6 \, \text{eV}}{36} \approx -0.3778 \, \text{eV} \]

3. **Energy of the ground state \( n = 1 \)**:
- For \( n = 1 \):
\[ E_1 = -13.6 \, \text{eV} \]

4. **Energy of the first transition**:
- A photon of energy \( 1.13 \, \text{eV} \) is emitted during the first transition from \( n = 6 \) to \( n = n_i \).
- The energy difference can be expressed as:
\[ E_6 - E_{n_i} = 1.13 \, \text{eV} \]

5. **Calculate the energy of the intermediate state \( n_i \)**:
- Rearranging the equation gives:
\[ E_{n_i} = E_6 - 1.13 \, \text{eV} \]
- Substituting \( E_6 \):
\[ E_{n_i} = -0.3778 \, \text{eV} - 1.13 \, \text{eV} = -1.5078 \, \text{eV} \]

6. **Calculate the energy of the second transition**:
- The energy emitted in the second transition is:
\[ E_{2} = E_1 - E_{n_i} \]
- Substituting the values:
\[ E_{2} = -13.6 \, \text{eV} - (-1.5078 \, \text{eV}) = -13.6 + 1.5078 = -12.0922 \, \text{eV} \]
- The energy of the photon emitted in the second transition is:
\[ E_{2} \approx 12.09 \, \text{eV} \]

#### Part (b): Determine the value of \( n \) in the intermediate state.

1. **Use the energy of the intermediate state**:
- We already calculated the energy of the intermediate state \( E_{n_i} \approx -1.5078 \, \text{eV} \).

2. **Set up the equation for the intermediate state**:
- Using the formula for energy:
\[ E_{n_i} = -\frac{13.6 \, \text{eV}}{n_i^2} \]
- Setting this equal to the calculated energy:
\[ -\frac{13.6}{n_i^2} = -1.5078 \]

3. **Solve for \( n_i^2 \)**:
- Rearranging gives:
\[ n_i^2 = \frac{13.6}{1.5078} \approx 9.024 \]

4. **Calculate \( n_i \)**:
- Taking the square root:
\[ n_i \approx 3 \]

### Final Answers:
- (a) The energy of the photon emitted in the second transition is approximately **12.09 eV**.
- (b) The value of \( n \) in the intermediate state is **3**.