Spectral series — Lyman, Balmer, Paschen: Worked examples

This page is a worked-example workshop for the parent topic. The parent gave you the master equation. Here we hit every kind of question an exam or a real spectrometer can throw at you — every series, both endpoints, the degenerate cases, a word problem, and an exam twist.

Everything rests on one equation, the Rydberg formula, which we restate so no symbol is unearned:

The scenario matrix

Before computing anything, let's map out every case class this topic contains. Each worked example below is tagged with the cell it fills.

Cell	Case class	What makes it different	Example
A	Lyman, first line ( $n_{1} = 1, n_{2} = 2$ )	deepest landing, shortest series → UV	Ex 1
B	Balmer, first line ( $n_{1} = 2, n_{2} = 3$ )	visible red H-alpha	Ex 2
C	Paschen, general line ( $n_{1} = 3, n_{2} = 5$ )	infrared, non-adjacent jump	Ex 3
D	Series limit ( $n_{2} \to \infty$ )	degenerate/limiting input	Ex 4
E	Reverse problem: given $λ$ , find the transition	inverse / identification	Ex 5
F	Absorption vs emission (same numbers, opposite direction)	sign/direction case	Ex 6
G	Hydrogen-like ion ( $Z > 1$ )	scaling case, $Z^{2}$ factor	Ex 7
H	Real-world word problem (spectrometer / astronomy)	applied context	Ex 8
I	Exam twist: which of two lines is shorter, no calculator	reasoning-only	Ex 9

Two facts control the whole matrix, so pin them now:

Figure — Spectral series — Lyman, Balmer, Paschen

Worked examples

Cell A — Lyman first line (UV)

Worked example Ex 1 — Longest wavelength of the Lyman series

Find the longest wavelength emitted in the Lyman series ( $n_{1} = 1$ ).

Forecast: Longest $λ$ means the smallest jump. What is the smallest jump that lands on level 1? Guess the pair $(n_{1}, n_{2})$ before reading on.

Pick the transition. Smallest jump onto level 1 is from level 2: $n_{1} = 1, n_{2} = 2$ . Why this step? Longest wavelength ⇔ lowest energy ⇔ nearest higher level, which is $n_{1} + 1 = 2$ .
Plug in. $\frac{1}{λ} = 1.097 \times 1 0^{7} (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = 1.097 \times 1 0^{7} \cdot \frac{3}{4} = 8.23 \times 1 0^{6} m^{- 1} .$ Why this step? The Rydberg formula turns the level pair directly into a wavenumber.
Invert to get $λ$ . $λ = \frac{1}{8.23 \times 1 0 ^{6}} = 1.215 \times 1 0^{- 7} m = 121.5 nm .$ Why this step? $λ$ is the reciprocal of the wavenumber.

Verify: 121.5 nm is well below the 400 nm edge of visible light ⇒ ultraviolet, exactly as the "deepest landing = shortest λ" rule predicts. ✓

Cell B — Balmer first line (visible)

Cell C — Paschen, non-adjacent jump (IR)

Cell D — The limiting case ( $n_{2} \to \infty$ )

Cell E — Reverse problem (identify the transition)

Worked example Ex 5 — Given the wavelength, name the jump

A hydrogen line is measured at $λ = 486 nm$ . Which transition produced it?

Forecast: 486 nm is blue-green — squarely visible — so which series must it belong to? Guess $n_{1}$ before computing.

Guess the series from colour: visible ⇒ Balmer ⇒ $n_{1} = 2$ . Why? Only Balmer lands in the visible band.
Compute the wavenumber: $\frac{1}{λ} = \frac{1}{486 \times 1 0 ^{- 9}} = 2.058 \times 1 0^{6} m^{- 1} .$
Solve for $n_{2}$ . Set $\frac{1}{λ} = R_{H} (\frac{1}{4} - \frac{1}{n _{2}^{2}})$ : $\frac{1}{n _{2}^{2}} = \frac{1}{4} - \frac{2.058 \times 1 0 ^{6}}{1.097 \times 1 0 ^{7}} = 0.25 - 0.1876 = 0.0624 \Rightarrow n_{2}^{2} = 16.02 \Rightarrow n_{2} = 4.$ Why this step? We rearrange to isolate the unknown level, then take the reciprocal-square-root.

Verify: $n_{2} = 4, n_{1} = 2$ is the Balmer $β$ line. Forward-check: $R_{H} (\frac{1}{4} - \frac{1}{16}) = 1.097 \times 1 0^{7} \cdot \frac{3}{16} = 2.057 \times 1 0^{6} \Rightarrow λ = 486 nm$ . ✓

Cell F — Absorption vs emission (direction/sign)

Worked example Ex 6 — Same numbers, opposite arrow

Cold hydrogen gas absorbs light and lifts an electron from $n = 1$ to $n = 3$ . What wavelength is absorbed, and how does it compare to the emission $3 \to 1$ ?

Forecast: Absorption goes up the staircase; emission goes down. Does the wavelength differ between "up 1→3" and "down 3→1"? Guess before computing.

Energy needed to go up equals energy released coming down — same gap: $\frac{1}{λ} = R_{H} (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) = 1.097 \times 1 0^{7} \cdot \frac{8}{9} = 9.75 \times 1 0^{6} m^{- 1} .$ Why lower level first even for absorption? The formula gives a positive wavenumber only when we write $1/ n_{1}^{2} - 1/ n_{2}^{2}$ ; the photon's energy is the same magnitude either direction.
Invert: $λ = 1.026 \times 1 0^{- 7} m = 102.6 nm$ (UV).

Verify: Absorption line and emission line sit at the identical wavelength 102.6 nm — the dark absorption line and the bright emission line coincide. This is why Emission vs Absorption Spectra look like negatives of each other. ✓ (This is the Lyman- $β$ line.)

Cell G — Hydrogen-like ion (the $Z^{2}$ scaling)

Cell H — Real-world word problem

Worked example Ex 8 — Astronomer's redshift-free ID

A star's spectrum shows a bright hydrogen line at $434 nm$ . The astronomer wants the transition to confirm it is hydrogen (not a Doppler-shifted impostor). Identify it.

Forecast: 434 nm is violet — visible — so Balmer again ( $n_{1} = 2$ ). Guess whether $n_{2}$ is larger or smaller than in Ex 5 (which was 486 nm).

Wavenumber: $\frac{1}{λ} = \frac{1}{434 \times 1 0 ^{- 9}} = 2.304 \times 1 0^{6} m^{- 1} .$
Solve for $n_{2}$ with $n_{1} = 2$ : $\frac{1}{n _{2}^{2}} = 0.25 - \frac{2.304 \times 1 0 ^{6}}{1.097 \times 1 0 ^{7}} = 0.25 - 0.2100 = 0.0400 \Rightarrow n_{2}^{2} = 25 \Rightarrow n_{2} = 5.$ Why? Same isolate-and-invert method as Ex 5; a shorter wavelength (bigger gap) means a higher starting level than the 486 nm line.

Verify: $n_{2} = 5 \to n_{1} = 2$ is the Balmer- $γ$ line. Forward check: $R_{H} (\frac{1}{4} - \frac{1}{25}) = 1.097 \times 1 0^{7} \cdot \frac{21}{100} = 2.304 \times 1 0^{6} \Rightarrow λ = 434 nm$ . ✓ It is hydrogen — genuine.

Cell I — Exam twist (no-calculator reasoning)

Worked example Ex 9 — Which is shorter, without computing?

Order these three lines by wavelength, shortest first, using reasoning only: Lyman-limit ( $\infty \to 1$ ), Balmer $3 \to 2$ , Paschen $4 \to 3$ .

Forecast: Write down your ordering guess now.

Rank by energy gap. Deeper landing = bigger $R_{H} / n_{1}^{2}$ scale. Lyman lands on 1, Balmer on 2, Paschen on 3. Plus the limit ( $n_{2} = \infty$ ) is the biggest jump within Lyman. Why this step? Shorter $λ$ ⇔ bigger energy gap; no numbers needed, just compare the dominant $1/ n_{1}^{2}$ term and the jump size.
Order: Lyman-limit (biggest gap, $1/ λ = R_{H}$ ) $>$ Balmer $3 \to 2$ $>$ Paschen $4 \to 3$ in energy, so in wavelength (reverse): Lyman-limit shortest, then Balmer, then Paschen longest.

Verify with numbers: Lyman-limit $= 91.2 nm$ , Balmer $3 \to 2 = 656 nm$ , Paschen $4 \to 3 = 1875 nm$ . Ordering shortest→longest: 91.2 < 656 < 1875 nm. ✓ Matches the reasoning.

Recall

Recall Fast checks across the matrix

First line of any series uses which start level ::: $n_{2} = n_{1} + 1$ (smallest gap, longest λ) Series limit uses which start level ::: $n_{2} \to \infty$ (biggest gap, shortest λ) Lyman first line wavelength ::: 121.5 nm (UV) Balmer H-alpha (3→2) wavelength ::: 656 nm (red) Paschen 5→3 wavelength ::: about 1282 nm (IR) Balmer series limit wavelength ::: 365 nm A 486 nm hydrogen line is which transition ::: Balmer 4→2 Absorption 1→3 wavelength ::: 102.6 nm, same as emission 3→1 He+ (Z=2) 3→2 wavelength ::: 164 nm, hydrogen value divided by Z squared Lyman series limit wavelength ::: 91.2 nm

Connections

Parent topic — the derivation these examples exercise.
Bohr Model of the Atom — where $E_{n} = - 13.6/ n^{2}$ comes from.
Photon Energy and Planck's Relation — the $λ = 1240/Δ E$ shortcut used throughout.
Hydrogen-like Ions and Z dependence — the $Z^{2}$ scaling of Ex 7.
Ionization Energy — the $n_{2} \to \infty$ series limit of Ex 4.
Emission vs Absorption Spectra — the same-wavelength coincidence of Ex 6.