Only the gap leaves as light

$-13.6$ sits below $-0.85$. So $n = 1$ is the ground state at the bottom, and the levels stack upward toward $0$ eV at ionization.

Ranking by magnitude puts the ground state at the top of the diagram and reverses the direction of every arrow that follows. Dropping the minus signs entirely and computing transitions between positive numbers does the same damage more quietly.

What the minus sign records is that the electron is bound. A deeper number means harder to remove, not more energy available to give away. That distinction is worth stating explicitly, because "more energy" is exactly what a large magnitude sounds like.

$$E_{\text{photon}} = \left| E_{\text{final}} - E_{\text{initial}} \right|.$$

Grabbing the level the atom lands on, or the one it started from, and using that magnitude directly is the error. For hydrogen falling from $n = 3$ to $n = 2$ it yields $1.51$ eV or $3.40$ eV in place of the correct value, and the wavelength computed from it is wrong by a factor no unit check will catch.

Subtract before anything else:

$$E_3 - E_2 = (-1.51) - (-3.40) = 1.89\ \text{eV}, \qquad \lambda = \frac{1240}{1.89} = 656\ \text{nm},$$

which is the red hydrogen line, visible through a diffraction grating in any school laboratory.

The one-sentence version: the level energies belong to the atom; only the gap between two of them leaves as light.

Follow the energy and the direction is never in doubt.

1. Absorption takes energy in, so the atom moves up a level. It must swallow a photon of exactly $E_{\text{final}} - E_{\text{initial}}$.
2. Emission gives energy out, so the atom drops down, releasing that same amount as a photon.

Draw absorption arrows pointing up and emission arrows pointing down, labelled with the identical difference. An element therefore shows the same set of wavelengths either way, which is what makes the two easy to trade:

1. Emission spectrum: bright lines on a dark field, from a hot gas radiating.
2. Absorption spectrum: dark gaps cut from a continuum, where a cooler gas removed those wavelengths on the way to the detector.

So dark lines in a stellar spectrum are not light the gas produced. They are light the gas took out, and their positions identify the elements in the star's outer layers.

Ionization moves the electron to the free state at $E = 0$, not to the highest level someone happened to draw. From the ground state of hydrogen it costs

$$0 - (-13.6) = 13.6\ \text{eV},$$

and nothing smaller will do it. Using the last line on a printed diagram, or substituting a bound-to-bound transition such as $n = 1$ to $n = 2$ at $10.2$ eV, understates it.

From $n = 2$ it costs only $3.4$ eV, since the electron is already less tightly bound. The lowest level always demands the most energy to ionize, which is what binding energy means.

One structural difference from a bound-to-bound transition is worth having. Those require an exact match, since the atom has nowhere to put the surplus. Ionization does not: a photon carrying more than the ionization energy is still absorbed, with the surplus becoming kinetic energy of the freed electron. That accounting is precisely the one the photoelectric effect uses two topics from now.