Step 1: Understanding the Concept:
Photons are emitted when an electron drops from a higher energy level ($n_{high}$) to a lower level ($n_{low}$). Frequency is directly proportional to the energy difference between these levels.
Step 2: Key Formula or Approach:
1. Energy of photon: $E = h\nu = 13.6 \left( \frac{1}{n_{low}^2} - \frac{1}{n_{high}^2} \right) \text{ eV}$
2. Higher energy difference $\implies$ Higher frequency.
Step 3: Detailed Explanation:
1. Transitions (A) and (C) go from lower to higher orbits, meaning they represent absorption of photons, not emission.
2. Transition (D): $n = 6$ to $n = 2$ (Balmer series). Energy $\approx 13.6 \times (1/4 - 1/36) \approx 3.02 \text{ eV}$.
3. Transition (B): $n = 2$ to $n = 1$ (Lyman series). Energy $\approx 13.6 \times (1 - 1/4) = 10.2 \text{ eV}$.
Since the energy jump from $n = 2$ to $n = 1$ is much larger than $n = 6$ to $n = 2$, it produces the photon with the highest frequency.
Step 4: Final Answer:
The transition $n = 2 \to 1$ emits the highest frequency photon.