Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²): Worked examples

The scenario matrix

Every question this formula can throw is one (or a blend) of these cells. Each is covered by at least one worked example below.

Cell	What changes	Danger / degenerate case	Example
A. Standard emission	electron falls $n_{2} \to n_{1}$ , both finite	none — the warm-up	Ex 1
B. Absorption	electron climbs $n_{1} \to n_{2}$	same $λ$ , opposite energy flow	Ex 2
C. Series limit	$n_{2} \to \infty$ (degenerate upper)	$1/ n_{2}^{2} \to 0$ ; shortest $λ$	Ex 3
D. Ionisation	start at finite $n_{1}$ , end at $n_{2} = \infty$	energy = ionisation energy	Ex 4
E. Reverse problem	given $λ$ , find $n_{2}$ (or the series)	must solve for an integer	Ex 5
F. Hydrogen-like ion	nuclear charge $Z \neq = 1$	multiply by $Z^{2}$	Ex 6
G. Sign trap	numbers given in wrong order	keep $n_{2} > n_{1}$ or $λ < 0$	Ex 7
H. Word problem / real world	dressed-up spectroscopy	translate words → $n_{1}, n_{2}$	Ex 8
I. Comparison / limiting	which line is bluest? gap $\to 0$ ?	large- $n$ crowding	Ex 9

The diagram below is the map for this whole page. Each coloured arrow is one cell of the matrix drawn on the real hydrogen energy ladder. Notice the rungs are not evenly spaced — they crowd together near the top ( $E = 0$ , the red dashed ionisation line). A downward arrow (blue) is emission (Cells A/G); the upward orange arrow is absorption (Cell B); the green arrow starts at the very top ( $n = \infty$ ) and is a series limit (Cell C); the red arrow climbs all the way to $E = 0$ and is ionisation (Cell D). Every example below is just "measure the length of one of these arrows."

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

Intuition How to read the map above (referenced throughout)

Alt text / caption: A hydrogen energy-level diagram. Horizontal grey lines mark levels $n = 1$ to $7$ at heights $E_{n} = - 13.6/ n^{2}$ eV; a red dashed line at $E = 0$ is the ionisation edge ( $n = \infty$ ). Four coloured arrows show the scenario cells: a blue downward arrow $4 \to 2$ (emission, Ex 1), an orange upward arrow $1 \to 3$ (absorption, Ex 2), a green downward arrow from the top $\infty \to 2$ (series limit, Ex 3), and a red upward arrow $1 \to \infty$ (ionisation, Ex 4). The level lines bunch together as they approach $E = 0$ , illustrating the large- $n$ crowding of Ex 9. Read the vertical length of each arrow as the photon energy: a long arrow = big gap = big $\overset{ν}{ˉ}$ = small $λ$ (bluer/UV); a short arrow = small gap = large $λ$ (redder/IR). Ex 1–4 each literally measure one of these four arrows; Ex 9 uses the visible squishing near the top.

Worked Examples

Ex 1 — Cell A: Standard emission ( $4 \to 2$ )

Worked example

A hydrogen electron falls from $n = 4$ to $n = 2$ . Find the emitted wavelength and its colour. (This is the blue arrow in the figure above.)

Forecast: Bigger drop than the red $3 \to 2$ line — guess: shorter wavelength, so bluer than 656 nm. Try to name a colour.

Step 1 — Assign $n_{1}, n_{2}$ . Lower level = final = $n_{1} = 2$ ; upper = start = $n_{2} = 4$ . Why this step? $n_{1}$ is always the smaller number so the bracket stays positive (Cell G shows what breaks if you swap them).

Step 2 — Evaluate the bracket. $\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}} = \frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16} = 0.1875$ Why this step? The bracket is a pure number carrying the whole $n$ -dependence; compute it alone so the constant $R_{H}$ and units stay separate and error-free.

Step 3 — Multiply by $R_{H}$ . $\frac{1}{λ} = 1.097 \times 1 0^{7} \times 0.1875 = 2.057 \times 1 0^{6} m^{- 1}$ Why this step? Multiplying the pure bracket by $R_{H}$ (units $m^{- 1}$ ) converts it into an actual wavenumber we can invert.

Step 4 — Invert. $λ = \frac{1}{2.057 \times 1 0 ^{6}} = 4.86 \times 1 0^{- 7} m = 486 nm (blue-green)$ Why this step? The formula gives $1/ λ$ ; we want $λ$ itself, so take the reciprocal.

Verify: $486 < 656$ nm ✓ (shorter than H-alpha, as forecast). Units: $(m^{- 1})^{- 1} = m$ ✓. This is the H-beta line — a real Balmer line.

Ex 2 — Cell B: Absorption ( $1 \to 3$ )

Worked example

A ground-state hydrogen atom absorbs a photon and the electron climbs from $n = 1$ to $n = 3$ . What wavelength was absorbed? (This is the orange upward arrow in the figure above.)

Forecast: Absorption is emission run backwards. Does the wavelength depend on the direction? Guess yes/no.

Step 1 — Assign $n_{1}, n_{2}$ . Even for absorption, $n_{1}$ = lower level $= 1$ , $n_{2}$ = higher $= 3$ . Why this step? The formula only knows the two levels, not which way energy flowed. Absorption and emission of the same pair give the identical $λ$ .

Step 2 — Bracket. $\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}} = 1 - \frac{1}{9} = \frac{8}{9} = 0.8889$ Why this step? Isolate the pure number first so we can multiply cleanly by $R_{H}$ next.

Step 3 — Wavenumber. $\frac{1}{λ} = 1.097 \times 1 0^{7} \times 0.8889 = 9.751 \times 1 0^{6} m^{- 1}$ Why this step? Attach the constant $R_{H}$ to turn the bracket into a physical wavenumber in $m^{- 1}$ .

Step 4 — Wavelength. $λ = 1.026 \times 1 0^{- 7} m = 102.6 nm (UV, Lyman β)$ Why this step? Invert the wavenumber to get the wavelength the atom actually swallowed.

Verify: Landing on/leaving $n = 1$ ⇒ Lyman ⇒ UV ✓. Answer forecast: direction does not change $λ$ ✓.

Ex 3 — Cell C: Series limit of Balmer ( $\infty \to 2$ )

Worked example

Find the shortest wavelength possible in the Balmer series. (This is the green arrow from the top in the figure above.)

Forecast: "Shortest wavelength" = "biggest energy gap." Which upper level gives the biggest gap? Guess a value near the ionisation edge.

Step 1 — Recognise the limit. The largest bracket for fixed $n_{1} = 2$ comes when $n_{2}$ is as large as possible: $n_{2} \to \infty$ . Why this step? $\frac{1}{n _{2}^{2}}$ shrinks toward $0$ as $n_{2}$ grows, so the bracket grows toward its ceiling $\frac{1}{n _{1}^{2}}$ .

Step 2 — Take the limit. $\frac{1}{λ _{m i n}} = R_{H} (\frac{1}{2 ^{2}} - 0) = \frac{R _{H}}{4} = 2.7425 \times 1 0^{6} m^{- 1}$ Why this step? Setting $1/ n_{2}^{2} = 0$ gives the maximum wavenumber, which corresponds to the minimum wavelength — the "edge" of the series.

Step 3 — Invert. $λ_{m i n} = 3.646 \times 1 0^{- 7} m = 364.6 nm$ Why this step? Convert the maximum wavenumber back to a wavelength by taking the reciprocal.

Verify: This is just past the violet edge of visible (≈380 nm), into near-UV — the well-known Balmer edge ✓. It is longer than the Lyman limit (91.1 nm) because $n_{1} = 2$ gives a smaller ceiling than $n_{1} = 1$ ✓.

Ex 4 — Cell D: Ionisation energy from the ground state

Worked example

Using the Rydberg formula, find the wavelength of the photon that just ionises a hydrogen atom from $n = 1$ , and confirm the ionisation energy is $13.6 eV$ . (This is the red upward arrow to $E = 0$ in the figure above.)

Forecast: Ionisation means "kick the electron to $n = \infty$ ." What is $n_{2}$ ? What series limit is this?

Step 1 — Set $n_{1} = 1$ , $n_{2} = \infty$ . Why this step? Ionisation from the ground state is exactly the Lyman series limit — the electron is lifted from $n = 1$ to just barely free. See Ionisation Energy.

Step 2 — Wavenumber, then invert algebraically. $\frac{1}{λ} = R_{H} (1 - 0) = 1.097 \times 1 0^{7} m^{- 1}$ $λ = \frac{1}{1.097 \times 1 0 ^{7} m ^{- 1}} = 9.116 \times 1 0^{- 8} m = 91.16 nm$ Why this step? With $n_{2} = \infty$ the bracket is simply $1$ , so the wavenumber equals $R_{H}$ itself; taking the reciprocal of $1.097 \times 1 0^{7}$ (a $1 0^{7}$ becomes $1 0^{- 7}$ , and $1/1.097 = 0.9116$ ) gives $λ$ directly.

Step 3 — Energy of that photon. Using $E = \frac{h c}{λ}$ with $h c = 1.986 \times 1 0^{- 25} J\cdotm$ (defined at the top of this page): $E = \frac{1.986 \times 1 0 ^{- 25} J\cdotm}{9.116 \times 1 0 ^{- 8} m} = 2.179 \times 1 0^{- 18} J$ Why this step? The ionising photon's energy equals the binding energy of the $n = 1$ electron, so converting $λ$ to energy should reproduce $13.6$ eV. The metres cancel, leaving joules.

Step 4 — Convert to eV (divide by $1.602 \times 1 0^{- 19} J/eV$ ): $E = \frac{2.179 \times 1 0 ^{- 18}}{1.602 \times 1 0 ^{- 19}} = 13.6 eV ✓$ Why this step? Joules are unwieldy for atoms; dividing by the electron charge converts to electron-volts, the natural atomic unit.

Verify: Matches $E_{1} = - 13.6 eV$ from the Bohr model (binding energy = $+ 13.6$ eV) ✓. Full circle: the spectral constant $R_{H}$ predicts the ionisation energy.

Ex 5 — Cell E: Reverse problem (find $n_{2}$ from $λ$ )

Worked example

A hydrogen line has $λ = 434 nm$ and belongs to the Balmer series. Which upper level $n_{2}$ produced it?

Forecast: Balmer means $n_{1} = 2$ . The visible Balmer lines are $3, 4, 5, \dots \to 2$ . Guess whether $n_{2}$ is 5 or 6 for a violet-ish line.

Step 1 — Write the equation with the unknown. $\frac{1}{λ} = R_{H} (\frac{1}{4} - \frac{1}{n _{2}^{2}})$ Why this step? We now solve for $n_{2}$ instead of $λ$ — the only new algebra in a reverse problem.

Step 2 — Compute the left side. $\frac{1}{λ} = \frac{1}{434 \times 1 0 ^{- 9}} = 2.304 \times 1 0^{6} m^{- 1}$ Why this step? Turn the given wavelength into a wavenumber so both sides speak the same language ( $m^{- 1}$ ).

Step 3 — Isolate $1/ n_{2}^{2}$ . $\frac{1}{n _{2}^{2}} = \frac{1}{4} - \frac{1}{λ R _{H}} = 0.25 - \frac{2.304 \times 1 0 ^{6}}{1.097 \times 1 0 ^{7}} = 0.25 - 0.2100 = 0.0400$ Why this step? Divide the wavenumber by $R_{H}$ to strip the constant, then subtract from $\frac{1}{4}$ to leave only the unknown term.

Step 4 — Solve. $n_{2}^{2} = \frac{1}{0.0400} = 25.0 \Rightarrow n_{2} = 5$ Why this step? Reciprocate to free $n_{2}^{2}$ , then take the positive square root — quantum numbers are positive integers.

Verify: $n_{2} = 5$ is a whole number ✓ (if it weren't, we'd have mis-assigned the series). This is the H-gamma line — indeed 434 nm, violet ✓.

Ex 6 — Cell F: Hydrogen-like ion (He⁺, $3 \to 2$ )

Worked example

Singly-ionised helium (He⁺, one electron, nuclear charge $Z = 2$ ) undergoes $n_{2} = 3 \to n_{1} = 2$ . Find the wavelength and compare to hydrogen's red H-alpha (656 nm).

Forecast: He⁺ has a doubly-charged nucleus. Its energies scale as $Z^{2}$ . Predict whether the line moves to shorter or longer $λ$ , and roughly by what factor.

Step 1 — Use the $Z$ -scaled formula. From Hydrogen-like Ions and Z-scaling: $\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ Why this step? A charge- $Z$ nucleus pulls $Z$ times harder; energies (and hence wavenumbers) scale by $Z^{2}$ .

Step 2 — Same bracket as H-alpha. $\frac{1}{4} - \frac{1}{9} = \frac{5}{36} = 0.13889$ Why this step? The level pair $3 \to 2$ is identical to H-alpha, so the bracket is unchanged; only the $Z^{2}$ prefactor differs.

Step 3 — Multiply by $R_{H} Z^{2} = R_{H} \cdot 4$ . $\frac{1}{λ} = 1.097 \times 1 0^{7} \times 4 \times 0.13889 = 6.094 \times 1 0^{6} m^{- 1}$ Why this step? Insert $Z^{2} = 4$ to scale the hydrogen wavenumber up fourfold.

Step 4 — Invert. $λ = 1.641 \times 1 0^{- 7} m = 164.1 nm (UV)$ Why this step? Reciprocate the wavenumber to get the He⁺ wavelength.

Verify: Exactly $\frac{656}{4} = 164 nm$ : wavenumber $\times 4$ , so wavelength $\div 4$ ✓. Forecast confirmed — shifted to shorter $λ$ by factor $Z^{2} = 4$ ✓.

Ex 7 — Cell G: The sign trap (swapped levels)

Worked example

A student computes the $2 \to 5$ emission but writes $n_{1} = 5, n_{2} = 2$ by habit. Show what goes wrong, then fix it.

Forecast: What sign will $1/ λ$ have if the levels are swapped? What does a negative $λ$ physically mean?

Step 1 — Plug in the wrong order. $\frac{1}{λ} = R_{H} (\frac{1}{5 ^{2}} - \frac{1}{2 ^{2}}) = R_{H} (0.04 - 0.25) = R_{H} (- 0.21) = - 2.30 \times 1 0^{6} m^{- 1}$ Why this step? We deliberately break the rule $n_{2} > n_{1}$ to expose the failure.

Step 2 — Read the disaster. A negative wavenumber ⇒ negative $λ$ , which is physically impossible. Why this step? The sign is the diagnostic: it screams "you swapped $n_{1}$ and $n_{2}$ ."

Step 3 — Fix: put the smaller level in $n_{1}$ . Correct order $n_{1} = 2, n_{2} = 5$ : $\frac{1}{λ} = R_{H} (\frac{1}{4} - \frac{1}{25}) = R_{H} (0.25 - 0.04) = R_{H} (0.21) = 2.30 \times 1 0^{6} m^{- 1}$ Why this step? Restoring $n_{2} > n_{1}$ flips the bracket positive, giving a physical wavenumber.

Step 4 — Wavelength. $λ = 4.341 \times 1 0^{- 7} m = 434.1 nm$ Why this step? Invert the (now positive) wavenumber to read off the real wavelength.

Verify: Same magnitude as Step 1 but positive ✓, and it matches Ex 5's H-gamma line (434 nm) ✓. Rule: $\frac{1}{λ}$ is right either way, but only $n_{2} > n_{1}$ gives the physical sign.

Ex 8 — Cell H: Word problem (a star's spectrum)

Worked example

Astronomers see a bright emission line at $λ = 1875 nm$ (infrared) from hot hydrogen gas. Identify the transition (which series, which levels).

Forecast: Infrared ⇒ small energy gap ⇒ small $\overset{ν}{ˉ}$ . Which series lives in the IR — Lyman, Balmer, or Paschen? Guess $n_{1}$ .

Step 1 — Compute the wavenumber. $\frac{1}{λ} = \frac{1}{1875 \times 1 0 ^{- 9}} = 5.333 \times 1 0^{5} m^{- 1}$ Why this step? Turn the observed colour into a number we can match to $R_{H} \times (bracket)$ .

Step 2 — Divide by $R_{H}$ to get the bracket. $\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}} = \frac{5.333 \times 1 0 ^{5}}{1.097 \times 1 0 ^{7}} = 0.04862$ Why this step? Stripping $R_{H}$ leaves the pure $n$ -bracket, which we can test against integer level pairs.

Step 3 — Try Paschen ( $n_{1} = 3$ ). Then $\frac{1}{9} - \frac{1}{n _{2}^{2}} = 0.04862$ : $\frac{1}{n _{2}^{2}} = 0.11111 - 0.04862 = 0.06249 \Rightarrow n_{2}^{2} = 16.0 \Rightarrow n_{2} = 4$ Why this step? IR points to Paschen ( $n_{1} = 3$ , IR region); test it and see if $n_{2}$ lands on an integer.

Step 4 — Confirm the identification. $n_{2} = 4$ is a clean integer, so the line is Paschen- $α$ : $4 \to 3$ . Why this step? A whole-number $n_{2}$ confirms our guessed series was right; a fraction would have meant trying another $n_{1}$ .

Verify: Recompute $λ$ for $3 \to 4$ : $\frac{1}{λ} = R_{H} (1/9 - 1/16) = R_{H} \cdot 0.04861 = 5.333 \times 1 0^{5} m^{- 1} \Rightarrow λ = 1875 nm$ ✓. Paschen is indeed IR ✓. (See Hydrogen Spectrum and Spectral Series.)

Ex 9 — Cell I: Comparison / limiting (which line is bluest, and why lines crowd)

Worked example

In the Lyman series ( $n_{1} = 1$ ), (a) decide which transition among $2 \to 1$ , $3 \to 1$ , $10 \to 1$ gives the bluest (shortest- $λ$ ) light, and (b) compare the gap in wavenumber between the first two lines ( $2 \to 1$ and $3 \to 1$ ) with the gap between two high lines ( $10 \to 1$ and $11 \to 1$ ) to show lines crowd together at large $n_{2}$ . (This uses the visible squishing of the rungs near $E = 0$ in the figure above.)

Forecast: (a) Bigger $n_{2}$ = bigger gap? Guess which is bluest. (b) As $n_{2}$ grows, do consecutive lines spread apart or pile up near the series limit $R_{H}$ ? Guess.

Step 1 — Bluest line (part a). Wavenumber grows with the bracket, and the bracket $1 - \frac{1}{n _{2}^{2}}$ is largest for the largest $n_{2}$ : $\overset{ν}{ˉ}_{2 \to 1} = 0.75 R_{H}, \overset{ν}{ˉ}_{3 \to 1} = 0.8889 R_{H}, \overset{ν}{ˉ}_{10 \to 1} = 0.9900 R_{H} .$ So $10 \to 1$ has the largest $\overset{ν}{ˉ}$ ⇒ smallest $λ$ ⇒ bluest (deepest UV). Why this step? "Bluest" means shortest $λ$ means largest $\overset{ν}{ˉ}$ ; ranking the brackets ranks the colours without ever converting to nm.

Step 2 — First-pair gap (part b). Adjacent-line spacing is the difference of brackets (in units of $R_{H}$ ): $Δ_{low} = \overset{ν}{ˉ}_{3 \to 1} - \overset{ν}{ˉ}_{2 \to 1} = (0.8889 - 0.75) R_{H} = 0.1389 R_{H} .$ Why this step? The gap between two lines is just the difference of their brackets — a pure multiple of $R_{H}$ , so we can compare spacings directly.

Step 3 — High-pair gap. $\overset{ν}{ˉ}_{10 \to 1} = R_{H} (1 - \frac{1}{100}) = 0.9900 R_{H}, \overset{ν}{ˉ}_{11 \to 1} = R_{H} (1 - \frac{1}{121}) = 0.99174 R_{H}$ $Δ_{high} = (0.99174 - 0.99000) R_{H} = 0.001736 R_{H} .$ Why this step? Repeat the same difference far up the ladder to see how the spacing has changed.

Step 4 — Compare and explain the limit. $\frac{Δ _{low}}{Δ _{high}} = \frac{0.1389}{0.001736} \approx 80.$ The low- $n$ gap is about 80 times wider than the high- $n$ gap. As $n_{2} \to \infty$ , every $\overset{ν}{ˉ} \to R_{H}$ , so infinitely many lines pile up against the series limit $R_{H}$ . Why this step? The ratio makes the crowding quantitative; the $1/ n^{2}$ ceiling forces convergence to $R_{H}$ — exactly the squished rungs near $E = 0$ in the figure.

Verify: Both gaps positive and $Δ_{high} ≪ Δ_{low}$ ✓; ratio $\approx 80$ ✓; bluest line is $10 \to 1$ (largest $n_{2}$ ) ✓. This is the "squished staircase" of the intuition box, made numerical.

Active Recall

Recall Which cell is "find

$n_{2}$ from a given $λ$ ", and what's the extra step? Question ::: Cell E — you divide the observed wavenumber by $R_{H}$ , subtract from $1/ n_{1}^{2}$ , then take the reciprocal-and-square-root to get an integer $n_{2}$ . If it isn't an integer, you guessed the wrong series.

Recall Why does He⁺'s

$3 \to 2$ line sit at exactly one-quarter of hydrogen's? Question ::: Because wavenumber scales as $Z^{2}$ ; for He⁺, $Z = 2$ gives $\times 4$ wavenumber, so $λ$ is $\div 4$ (656 nm → 164 nm).

Recall What is the tell-tale sign you swapped

$n_{1}$ and $n_{2}$ ? Question ::: A negative wavenumber (and hence a negative wavelength). A physical $λ$ is always positive, so a minus sign means you must put the smaller quantum number in $n_{1}$ and the larger in $n_{2}$ .

Recall Within one series, which line is bluest?

Question ::: The one with the largest $n_{2}$ (closest to the series limit): its bracket $1 - 1/ n_{2}^{2}$ is biggest, so $\overset{ν}{ˉ}$ is biggest and $λ$ shortest.

Parent: Rydberg formula (topic note) · Prereqs: Bohr Model of the Atom, Energy of a Photon E = hc over lambda, Hydrogen Spectrum and Spectral Series, Ionisation Energy, Hydrogen-like Ions and Z-scaling, Reduced Mass in Two-Body Systems, Quantisation of Angular Momentum.