Hydrogen emission spectrum — Lyman, Balmer, Paschen, Brackett, Pfund: Visual walkthrough

We assume you know only that atoms have a heavy centre (the nucleus) and a tiny electron near it. Everything else — energy levels, jumps, photons, wavelength — we draw as we go.

Step 1 — The electron lives on fixed steps

WHAT: Inside a hydrogen atom the electron cannot sit anywhere. It can only occupy certain allowed levels, labelled by a whole number $n = 1, 2, 3, \dots$

WHY this idea and not "anywhere it likes": if the electron could hold any energy, a heated gas would glow in every colour (a smooth rainbow). Experiment shows only specific colours. The only way to explain that is: energies come in fixed steps. This is quantization (see Quantization of Energy).

PICTURE: think of a staircase. $n = 1$ is the ground floor (lowest, most negative energy), $n = 2$ the next step up, and so on. The steps get closer together as you climb.

Figure — Hydrogen emission spectrum — Lyman, Balmer, Paschen, Brackett, Pfund

Step 2 — Emission is a fall

WHAT: Light comes out when the electron drops from a higher step $n_{2}$ to a lower step $n_{1}$ (so $n_{2} > n_{1}$ ).

WHY: an electron sitting high up carries extra energy. Nature prefers low energy, so it tumbles down. The energy it loses has to go somewhere — it leaves the atom as a packet of light.

PICTURE: a ball on step $n_{2} = 3$ falling to step $n_{1} = 1$ . The height of the fall is the energy released. Notice: the same fall always releases the same energy, because the steps never move.

Step 3 — Measure the drop (how much energy comes out)

WHAT: The energy released, $Δ E$ , is the height fallen: the higher starting level minus the lower ending level.

WHY (the sign, carefully): "how far did it fall?" = starting height − ending height = $E_{n_{2}} - E_{n_{1}}$ . Since $n_{2} > n_{1}$ , the start $E_{n_{2}}$ is less negative (higher up) than the end $E_{n_{1}}$ , so this difference is genuinely positive — energy leaves the atom. Substituting $E_{n} = - 13.6/ n^{2}$ :

$Δ E = E_{n_{2}} - E_{n_{1}} = - \frac{13.6}{n _{2}^{2}} - (- \frac{13.6}{n _{1}^{2}}) = 13.6 (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) eV$

$E_{n_{2}} - E_{n_{1}}$ ::: start minus end — the honest "height fallen."
the two minus signs from $E_{n}$ ::: cancel into a $+$ on the destination term, leaving $\frac{1}{n _{1}^{2}}$ first.
$\frac{1}{n _{1}^{2}}$ ::: the destination term; because $n_{1}$ is smaller, this term is the bigger one, so it dominates and keeps $Δ E > 0$ .
$\frac{1}{n _{2}^{2}}$ ::: the starting term, smaller, subtracted off.

So the "smaller $n$ first" ordering isn't a trick — it falls out of substituting the negative energies into start − end.

Step 4 — The lost energy becomes exactly one photon

WHAT: The packet of light is a photon. Its energy is tied to its wavelength $λ$ (how long one wiggle of the light wave is) by Planck's relation.

WHY this tool — Planck's relation and not something else: we need a bridge from energy (what the electron lost) to colour (what we see). The only law that connects them is:

$Δ E = \frac{h c}{λ}$

$h$ ::: Planck's constant — the fixed "exchange rate" between energy and frequency.
$c$ ::: speed of light.
$λ$ ::: wavelength. Sits in the bottom, so more energy ⇒ shorter wavelength. This inverse is the single most important fact on this page.

See Photon Energy and Planck Relation.

PICTURE: a big drop makes a short, tightly-packed (bluish/UV) wave; a small drop makes a long, stretched-out (reddish/IR) wave.

Step 5 — Set the two expressions for $Δ E$ equal and cancel

First, one shorthand so the algebra stays clean.

WHAT: The energy the electron lost (Step 3) equals the energy the photon carries (Step 4). Writing $13.6 eV = R_{H} h c$ on the left side of Step 3 and setting it equal to Step 4:

$\frac{h c}{λ} = R_{H} h c (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$

WHY: conservation of energy — nothing is created or lost, so the two sides must match. Now the same factor $h c$ sits on both sides, so it cancels:

$\frac{1}{λ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$

$\frac{1}{λ}$ ::: how many wave-wiggles fit in one metre; bigger energy ⇒ bigger value ⇒ shorter wave. (Chemists call this the wavenumber; we'll just use $1/ λ$ so no new symbol is needed.)
$R_{H} = 1.097 \times 1 0^{7} m^{- 1}$ ::: the Rydberg constant, just defined.
the bracket ::: the same "how far it fell" factor from Step 3, now controlling colour directly.

This is the parent's central result — derived from the staircase alone.

Step 6 — Group the falls into series (fix the landing step)

WHAT: Collect all falls that end on the same bottom step $n_{1}$ . Each such family is a series.

WHY: every line in one series shares the destination $\frac{1}{n _{1}^{2}}$ ; only the starting step $n_{2}$ changes. That makes the whole family live in one region of the spectrum.

PICTURE: colour-coded falls — all arrows landing on $n_{1} = 1$ (Lyman, UV), all landing on $n_{1} = 2$ (Balmer, visible), and higher landings (Paschen/Brackett/Pfund, IR).

Falls onto $n_{1} = 1$ start from the deepest part of the well ⇒ biggest $Δ E$ ⇒ shortest $λ$ (UV).
Falls onto higher $n_{1}$ are gentle drops ⇒ small $Δ E$ ⇒ long $λ$ (IR).

Step 7 — The two edge cases of every series

WHAT: Two limits bound each series.

First line ( $n_{2} = n_{1} + 1$ ): the smallest possible jump ⇒ smallest $Δ E$ ⇒ longest $λ$ in that series.

Series limit ( $n_{2} \to \infty$ ): the electron falls from infinitely far out. Then $\frac{1}{n _{2}^{2}} \to 0$ and the formula degenerates to

$\frac{1}{λ _{limit}} = \frac{R _{H}}{n _{1}^{2}} ⟹ λ_{limit} = \frac{n _{1}^{2}}{R _{H}}$

$\frac{1}{n _{2}^{2}} \to 0$ ::: an infinitely high electron has essentially $0$ energy, so nothing is subtracted.
this is the shortest $λ$ of the series — the lines pile up and converge here.

Degenerate check — no fall at all: if $n_{2} = n_{1}$ the bracket is $0$ , so $\frac{1}{λ} = 0$ : no photon, no line. Correct — nothing moved, nothing shines.

The one-picture summary

Everything above, on one canvas, with each region tagged by the step that built it: ① the staircase (Step 1) → ②–③ a labelled fall and its measured height (Steps 2–3) → ④ the photon it makes (Step 4) → ⑤ the boxed formula (Step 5) → ⑥–⑦ the colour-coded series and their limits (Steps 6–7).

Recall Feynman retelling — with the traps highlighted

Here's the walkthrough boiled to its decisions and its two easy-to-fall-into traps. The electron sits on a staircase whose steps are fixed, so a given fall always makes a given colour — that's the whole reason lines are sharp. Trap 1 (energy vs wavelength): a bigger fall makes a shorter wave, not a longer one, because $λ$ sits underneath $h c$ — energy and wavelength are opposites. Trap 2 (the sign): when you compute "start minus end," the two minus signs baked into $E_{n} = - 13.6/ n^{2}$ flip the destination term to the front, which is why the smaller $n$ ends up first in the bracket — it's not an arbitrary rule to memorise, it's forced by the algebra. Finally, don't forget the mirror image: run the arrow upward and the very same formula tells you which photon the atom will swallow (absorption). Emission and absorption are one formula read in two directions. Cancel the shared $h c$ , name the leftover $R_{H}$ , and you're done.

Connections

Bohr Model of the Atom — supplies the staircase $E_{n} = - 13.6/ n^{2}$ eV.
Quantization of Energy — why the steps exist at all.
Photon Energy and Planck Relation — the $Δ E = h c / λ$ bridge in Step 4.
Rydberg Constant and Spectra of Hydrogen-like Ions — same derivation with a $Z^{2}$ factor.
Ionization Energy of Hydrogen — the Lyman limit, $13.6$ eV.
Electromagnetic Spectrum — where each series lands.