The energy of an emitted photon is determined by the difference in energy between the initial and final orbits: $\Delta E = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.

The energy separation between adjacent energy levels decreases as the value of $n$ increases (i.e., energy levels get closer together at higher $n$).

Comparing the given transitions: $n=6 \to n=1$: Ends in ground state (Lyman series), corresponds to the highest energy (UV region). $n=5 \to n=3$: Ends in $n=3$ (Paschen series). $n=5 \to n=4$: Transition between adjacent higher levels. $n=6 \to n=5$: Transition between the highest adjacent levels given.

Calculating the factor $\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$ for the closest contenders:

• For $5 \to 4$: $\frac{1}{16} - \frac{1}{25} \approx 0.0625 - 0.04 = 0.0225$
• For $6 \to 5$: $\frac{1}{25} - \frac{1}{36} \approx 0.04 - 0.0278 = 0.0122$

The transition $n=6 \to n=5$ yields the smallest value, implying the emission of the least energetic photon (longest wavelength).