Which of the following transitions gives photon of maximum energy

Options

A. n=1 to n=2
B. n=2 to n=1
C. n=2 to n=6
D. n=6 to n=2

Answer

To determine which transition gives a photon of maximum energy, we need to analyze the energy transitions between different energy levels in an atom. The energy of a photon emitted during a transition can be calculated using the formula for the energy levels of a hydrogen atom:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

where \( E_n \) is the energy of the level corresponding to the principal quantum number \( n \).

### Step-by-step Solution:

1. **Identify the transitions**: The given transitions are:
- Transition 1: \( n = 1 \) to \( n = 2 \)
- Transition 2: \( n = 2 \) to \( n = 1 \)
- Transition 3: \( n = 2 \) to \( n = 6 \)
- Transition 4: \( n = 6 \) to \( n = 1 \)

2. **Determine the energy change for each transition**:
- For the transition \( n = 1 \) to \( n = 2 \):
\[ E_{1 \to 2} = E_2 - E_1 = \left(-\frac{13.6}{2^2}\right) - \left(-\frac{13.6}{1^2}\right) = -\frac{13.6}{4} + 13.6 = 13.6 \left(1 - \frac{1}{4}\right) = 13.6 \times \frac{3}{4} = 10.2 \, \text{eV} \]

- For the transition \( n = 2 \) to \( n = 1 \):
\[ E_{2 \to 1} = E_1 - E_2 = \left(-\frac{13.6}{1^2}\right) - \left(-\frac{13.6}{2^2}\right) = -13.6 + \frac{13.6}{4} = 13.6 \left(-1 + \frac{1}{4}\right) = 13.6 \times -\frac{3}{4} = -10.2 \, \text{eV} \]
(Note: This transition emits a photon, so we take the absolute value for energy.)

- For the transition \( n = 2 \) to \( n = 6 \):
\[ E_{2 \to 6} = E_6 - E_2 = \left(-\frac{13.6}{6^2}\right) - \left(-\frac{13.6}{2^2}\right) = -\frac{13.6}{36} + \frac{13.6}{4} = 13.6 \left(\frac{1}{4} - \frac{1}{36}\right) \]
\[ = 13.6 \left(\frac{9}{36} - \frac{1}{36}\right) = 13.6 \times \frac{8}{36} = 3.02 \, \text{eV} \]

- For the transition \( n = 6 \) to \( n = 1 \):
\[ E_{6 \to 1} = E_1 - E_6 = \left(-\frac{13.6}{1^2}\right) - \left(-\frac{13.6}{6^2}\right) = -13.6 + \frac{13.6}{36} = 13.6 \left(-1 + \frac{1}{36}\right) \]
\[ = 13.6 \left(-\frac{36}{36} + \frac{1}{36}\right) = 13.6 \times -\frac{35}{36} \approx -13.2 \, \text{eV} \quad (\text{taking absolute value}) \]

3. **Compare the energies**:
- \( E_{1 \to 2} = 10.2 \, \text{eV} \)
- \( E_{2 \to 1} = 10.2 \, \text{eV} \)
- \( E_{2 \to 6} = 3.02 \, \text{eV} \)
- \( E_{6 \to 1} \approx 13.2 \, \text{eV} \)

4. **Conclusion**: The transition that gives the photon of maximum energy is \( n = 6 \) to \( n = 1 \) with an energy of approximately \( 13.2 \, \text{eV} \).

### Final Answer:
The transition that gives a photon of maximum energy is from \( n = 6 \) to \( n = 1 \).