Determine the energy of the photon (in eV) emitted when the electron in a hydrogen atom undergoes a transition from the n = 8 level to then = 6 level.

Answer

To determine the energy of the photon emitted when the electron in a hydrogen atom transitions from the n = 8 level to the n = 6 level, we can use the formula for the energy change associated with the transition between energy levels in a hydrogen atom.

### Step-by-Step Solution:

1. **Identify the initial and final energy levels:**
- Initial level (n_initial) = 8
- Final level (n_final) = 6

2. **Use the formula for energy change:**
The energy change (ΔE) for a transition in a hydrogen atom can be calculated using the formula:
\[ \Delta E = -13.6 \, \text{eV} \times Z^2 \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \]
where \( Z \) is the atomic number. For hydrogen, \( Z = 1 \).

3. **Substitute the values into the formula:**
\[ \Delta E = -13.6 \, \text{eV} \times 1^2 \left( \frac{1}{6^2} - \frac{1}{8^2} \right) \]
\[ = -13.6 \, \text{eV} \left( \frac{1}{36} - \frac{1}{64} \right) \]

4. **Calculate the fractions:**
- Calculate \( \frac{1}{36} \) and \( \frac{1}{64} \):
\[ \frac{1}{36} \approx 0.02778 \]
\[ \frac{1}{64} = 0.015625 \]

5. **Find the difference:**
\[ \frac{1}{36} - \frac{1}{64} = 0.02778 - 0.015625 \approx 0.012155 \]

6. **Calculate the energy change:**
\[ \Delta E = -13.6 \, \text{eV} \times 0.012155 \approx -0.165 \, \text{eV} \]

7. **Interpret the result:**
Since the energy change is negative, it indicates that energy is emitted when the electron transitions from a higher energy level (n=8) to a lower energy level (n=6). Thus, the energy of the emitted photon is:
\[ E_{\text{photon}} \approx 0.165 \, \text{eV} \]

### Final Answer:
The energy of the photon emitted during the transition from n = 8 to n = 6 in a hydrogen atom is approximately **0.165 eV**.