Maximum frequency of emission is obtained for the transition:

Options

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

Answer

To determine the maximum frequency of emission for the given transitions, we need to analyze the energy levels of an electron in an atom, particularly in the context of the hydrogen atom. The frequency of the emitted photon during an electron transition is directly related to the energy difference between the two energy levels involved in the transition.

### Step-by-Step Solution:

1. **Understanding Energy Levels**:
The energy levels of an electron in a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Calculating Energy Differences**:
For each transition, we need to calculate the energy difference \( \Delta E \) between the two levels involved in the transition. The energy difference is given by:
\[ \Delta E = E_{n_i} - E_{n_f} = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2} \]
where \( n_i \) is the initial state and \( n_f \) is the final state.

3. **Evaluating Each Transition**:
- **Transition \( n = 2 \) to \( n = 1 \)**:
\[ \Delta E_{2 \to 1} = -\frac{13.6}{1^2} + \frac{13.6}{2^2} = -13.6 + 3.4 = -10.2 \, \text{eV} \]
- **Transition \( n = 6 \) to \( n = 2 \)**:
\[ \Delta E_{6 \to 2} = -\frac{13.6}{2^2} + \frac{13.6}{6^2} = -3.4 + 0.378 = -3.022 \, \text{eV} \]
- **Transition \( n = 1 \) to \( n = 2 \)**:
\[ \Delta E_{1 \to 2} = -\frac{13.6}{2^2} + \frac{13.6}{1^2} = -3.4 + 13.6 = 10.2 \, \text{eV} \]
- **Transition \( n = 2 \) to \( n = 6 \)**:
\[ \Delta E_{2 \to 6} = -\frac{13.6}{6^2} + \frac{13.6}{2^2} = -0.378 + 3.4 = 3.022 \, \text{eV} \]

4. **Finding the Maximum Energy Gap**:
From the calculated energy differences:
- \( \Delta E_{2 \to 1} = 10.2 \, \text{eV} \)
- \( \Delta E_{6 \to 2} = 3.022 \, \text{eV} \)
- \( \Delta E_{1 \to 2} = 10.2 \, \text{eV} \)
- \( \Delta E_{2 \to 6} = 3.022 \, \text{eV} \)

The maximum energy gap is \( 10.2 \, \text{eV} \) which occurs for the transitions \( n = 2 \to 1 \) and \( n = 1 \to 2 \).

5. **Determining Frequency**:
The frequency \( \nu \) of the emitted photon can be calculated using the relation:
\[ E = h \nu \implies \nu = \frac{E}{h} \]
Since we are looking for the maximum frequency, we can conclude that the transition with the maximum energy gap will yield the maximum frequency.

### Conclusion:
The maximum frequency of emission is obtained for the transition \( n = 2 \to 1 \).