Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²): Visual walkthrough

This is the visual companion to the parent Rydberg note. Read that for the summary; read this to see it grow from nothing.

Step 0 — The words before the symbols

Before we start, let's agree on plain-language meaning for four words. If you already met these in Bohr Model of the Atom, this is just a refresher; if not, this is where you learn them.

Nothing here needs maths yet. Keep the staircase image in your head — the whole page is about that staircase.

Step 1 — Draw the lopsided staircase (the energy levels)

WHAT. The Bohr model says the electron can only sit on tracks whose energy is $E_{n} = - \frac{13.6 eV}{n ^{2}}$ Here $E_{n}$ is the energy of track number $n$ , the minus sign means "trapped", and $13.6 eV$ is just the depth of the deepest well ( $n = 1$ ). Dividing by $n^{2}$ makes higher tracks shallower.

WHY this shape and not evenly-spaced? Look at the numbers the $1/ n^{2}$ produces: $- 13.6, - 3.4, - 1.51, - 0.85, \dots$ The gaps between neighbours shrink fast. Physically this comes from Coulomb attraction fighting the quantised angular momentum $L = n ℏ$ from Quantisation of Angular Momentum — but you don't need that machinery here; you just need to see that the ladder crowds together at the top.

PICTURE. Each rung's height is $- 13.6/ n^{2}$ . Notice how the top rungs pile up toward $E = 0$ (the "escaped" line).

Figure — Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²)

Step 2 — Fix the mass we plug in (reduced mass)

WHAT. That $13.6 eV$ secretly hides a mass. The full Bohr expression is $E_{n} = - \frac{μ e ^{4}}{8 ε _{0}^{2} h ^{2}} \cdot \frac{1}{n ^{2}}$ and the mass inside is not the bare electron mass $m_{e}$ but the reduced mass $μ = \frac{m _{e} m _{p}}{m _{e} + m _{p}}$ .

WHY use $μ$ and not $m_{e}$ ? The proton is heavy but not infinitely heavy, so it wobbles too — both particles orbit a shared centre of mass. The reduced mass is the honest single mass that makes a two-body wobble behave like one body going round a fixed centre (see Reduced Mass in Two-Body Systems). Since $m_{p} \approx 1836 m_{e}$ , we get $μ \approx 0.99946 m_{e}$ — a tiny correction, but it's the difference between theory and the measured spectrum.

PICTURE. The electron doesn't circle a nailed-down proton; they both circle the balance point, the proton making a tiny loop, the electron a big one.

Step 3 — Drop the electron: energy released

WHAT. Let the electron fall from a higher rung $n_{2}$ to a lower rung $n_{1}$ (so $n_{2} > n_{1}$ ). The energy it sheds is the height of the fall: $E_{released} = E_{n_{2}} - E_{n_{1}} = \frac{μ e ^{4}}{8 ε _{0}^{2} h ^{2}} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$

WHY is this bracket positive? Since $n_{1} < n_{2}$ , we have $n_{1}^{2} < n_{2}^{2}$ , so $\frac{1}{n _{1}^{2}} > \frac{1}{n _{2}^{2}}$ and the bracket is $> 0$ . A falling electron releasing positive energy — exactly what we want. If you ever get a negative answer, you swapped $n_{1}$ and $n_{2}$ .

PICTURE. The vertical red drop between two rungs is the released energy. A big drop (down to $n_{1} = 1$ ) is tall; a small drop (down to $n_{1} = 3$ ) is short.

Step 4 — The released energy leaves as ONE photon

WHAT. All that released energy becomes a single photon. From Energy of a Photon E = hc over lambda: $E_{photon} = \frac{h c}{λ}$ where $h$ is Planck's constant, $c$ the speed of light, and $λ$ the wavelength (the length of one wave). Setting released energy equal to photon energy: $\frac{h c}{λ} = \frac{μ e ^{4}}{8 ε _{0}^{2} h ^{2}} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$

WHY equate them? Energy conservation: the atom loses exactly what the photon carries away. Nothing is created or destroyed — the drop becomes light.

WHY write $\frac{h c}{λ}$ and not $h ν$ ? Both are the same photon energy ( $ν = c / λ$ ), but we want a formula in wavelength because that's what a spectroscope measures on a screen. So we pick the $λ$ form deliberately.

PICTURE. The tall red arrow (the fall) turns into a wavy photon flying off. A taller fall ⇒ more energy ⇒ shorter, bluer wave; a shorter fall ⇒ longer, redder wave.

Step 5 — Solve for $1/ λ$ and name the constant

WHAT. Divide both sides of the Step 4 equation by $h c$ . The whole clump of constants in front becomes a single number we christen the Rydberg constant: $\frac{1}{λ} = R_{H} \frac{μ e ^{4}}{8 ε _{0}^{2} h ^{3} c} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$

WHY solve for $\frac{1}{λ}$ instead of $λ$ ? Because $1/ λ$ (the wavenumber) is directly proportional to the bracket, which is directly proportional to energy. Keeping it as $1/ λ$ makes the physics linear and clean; you invert at the very end only if you want $λ$ in nanometres.

PICTURE. Everything except the two step-numbers is now frozen into one constant $R_{H}$ . The formula is a straight-line machine: feed in $(\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ , read off $1/ λ$ .

Step 6 — The edge case: $n_{2} \to \infty$ (the series limit)

WHAT. Push the starting rung all the way to the top, $n_{2} \to \infty$ . Then $\frac{1}{n _{2}^{2}} \to 0$ and the formula collapses to $\frac{1}{λ _{m i n}} = R_{H} (\frac{1}{n _{1}^{2}} - 0) = \frac{R _{H}}{n _{1}^{2}}$

WHY does this matter? This is the shortest wavelength (highest energy) line the series can ever make — because the bracket hit its biggest possible value. Beyond it, the electron isn't falling from a rung at all; it's falling from freedom (it was ionised). This ties directly to Ionisation Energy: for $n_{1} = 1$ , $λ_{m i n} = 1/ R_{H} \approx 91.1 nm$ .

PICTURE. As $n_{2}$ climbs, the rungs crowd toward $E = 0$ , so the drops toward $n_{1}$ approach one fixed maximum height — the series piles up against a wall (the series limit), then stops.

Step 7 — The degenerate check: $n_{2} = n_{1}$

WHAT. What if start and end rungs are the same? Then $\frac{1}{λ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{1}^{2}}) = 0$

WHY include this? A zero wavenumber means $λ \to \infty$ and no photon at all — which is exactly right: an electron that doesn't move between rungs emits nothing. This is the sanity anchor of the whole formula. It also flags the sign rule: if you accidentally put the bigger $n$ as $n_{1}$ , the bracket flips negative, giving an impossible negative $1/ λ$ — nature's way of telling you the arrow points the wrong way.

PICTURE. No drop, no light. A flat "stay put" and a dark screen.

Worked check — the red $H_{α}$ line ( $3 \to 2$ )

The one-picture summary

Everything on this page — staircase, drop, photon, formula, edge cases — compressed into a single diagram. Read it left to right: pick two rungs, take their height difference, that difference is the photon, and $1/ λ$ is just that difference wearing a Rydberg-constant costume.

Recall Feynman retelling of the whole walkthrough

Picture a staircase inside the atom whose steps squeeze closer together the higher you climb — that squeezing is the $1/ n^{2}$ rule (Step 1). The steps are built from a mass, and the honest mass is a blend of electron and proton called the reduced mass, because the proton wobbles too (Step 2). Now let the electron slide down from a high step to a low step; the height it loses is real energy (Step 3), and that energy flies away as one packet of light — a big fall makes blue light, a small fall makes red (Step 4). Divide that energy by $h c$ and all the constants clump into one number $R_{H}$ ; what's left is the Rydberg formula (Step 5). Climb the starting step all the way to the top and the falls stop growing — they pile against the series limit, which is really the ionisation edge (Step 6). And if the electron doesn't move at all, no light comes out — the formula gives zero, exactly as it should (Step 7). That's the whole machine: a lopsided staircase turning falls into colours.

Active Recall

Recall Why is the bracket written

$\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}$ and not the other way? Question ::: Because $n_{1} < n_{2}$ makes $\frac{1}{n _{1}^{2}}$ the larger term; subtracting the smaller keeps the wavenumber positive, matching a real (positive-energy) photon.

Recall What physically happens at the series limit

$n_{2} \to \infty$ ? Question ::: The $1/ n_{2}^{2}$ term vanishes, giving the shortest wavelength / highest energy of the series; the electron falls from a free (ionised) state onto rung $n_{1}$ .