Among the following transitions, which one results in the minimum...

The energy of a photon emitted during a transition from $n_2$ to $n_1$ is given by Rydberg's formula:

\Delta E = 13.6 Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) eV

The wavelength of the emitted photon is inversely proportional to the energy difference, since $\Delta E =$\dfrac{hc}{\lambda}. Thus, the minimum wavelength corresponds to the maximum energy difference.

Calculating the proportional energy difference for each given transition:

For $n = 5$ to $n = 4$: $\Delta E \propto$\dfrac{1}{4^2} - \dfrac{1}{5^2} = \dfrac{9}{400} =$0.0225$

For $n = 4$ to $n = 3$: $\Delta E \propto$\dfrac{1}{3^2} - \dfrac{1}{4^2} = \dfrac{7}{144} \approx$0.0486$

For $n = 3$ to $n = 2$: $\Delta E \propto$\dfrac{1}{2^2} - \dfrac{1}{3^2} = \dfrac{5}{36} \approx$0.1389$

For $n = 2$ to $n = 1$: $\Delta E \propto$\dfrac{1}{1^2} - \dfrac{1}{2^2} = \dfrac{3}{4} =$0.75$

The transition from $n = 2$ to $n = 1$ has the maximum energy difference and therefore results in the minimum wavelength.