USAAAO 2026, I кръг — Задача 25

• D

Solution: Without prior knowledge of these lines, one can determine that the Lyman-$\alpha$ line has a shorter wavelength than the Balmer-$\alpha$ line using the Rydberg equation:

Since $\frac{1}{1^2} - \frac{1}{2^2}$ is greater than $\frac{1}{2^2} - \frac{1}{3^2}$, the transition from $n = 2$ to $n = 1$ will have a smaller wavelength than $n = 3$ to $n = 2$.

The H-I line comes from an energy difference between two hyperfine levels in the same orbital, while the Lyman-$\alpha$ and Balmer-$\alpha$ lines come from transitions between two orbitals, so the H-I line will have less energy, and thus a longer wavelength, than both of the other lines.

Of course, one can also determine the answer directly if they already know which parts of the electromagnetic spectrum each line is in (Lyman-$\alpha$ is ultraviolet, Balmer-$\alpha$ is visible, and H-I is in the microwave range).

Answer: D