Hydrogen emission spectrum — Lyman, Balmer, Paschen, Brackett, Pfund: Worked examples

This page is the drill hall for the parent topic. We take the one master tool you already met — the Rydberg formula — and throw every kind of question at it until nothing can surprise you.

Before symbols pile up, three plain-word anchors so nothing is used un-earned:

Wavelength $λ$ = the length of one full ripple of the emitted light. Small $λ$ = squished ripples = high energy (UV/blue). Large $λ$ = stretched ripples = low energy (red/IR).
Energy of the jump $Δ E$ = how much energy the electron loses falling. It becomes exactly one photon: $Δ E = h c / λ$ (from Photon Energy and Planck Relation). So more energy ⇒ shorter $λ$ — they are opposites.
Series = the family of lines that all end on the same lower level $n_{1}$ .

The scenario matrix

Every exam question about this topic is one of these cells. Each worked example below is tagged with the cell(s) it covers.

#	Case class	What makes it tricky	Covered by
A	First line of a series ( $n_{2} = n_{1} + 1$ )	smallest jump → longest $λ$ of the family	Ex 1
B	A deeper line ( $n_{2} = n_{1} + 2, 3, \dots$ )	plug bigger $n_{2}$ ; watch the shrinking bracket	Ex 2
C	Series limit ( $n_{2} \to \infty$ , degenerate)	one term vanishes — limiting behaviour	Ex 3
D	Reverse problem: given $λ$ , find $n_{2}$	solve backwards, must land on integer	Ex 4
E	Compare two lines (which is more energetic?)	no numbers needed, just the bracket	Ex 5
F	Cross-series clash / identify the line	same $λ$ region, different series	Ex 6
G	Real-world word problem (energy per photon in eV/J)	unit conversion m⁻¹ → J → eV	Ex 7
H	Exam twist: absorption vs emission, or an impossible line	sign of $Δ E$ , forbidden $n_{2} \leq n_{1}$	Ex 8
I	Zero / degenerate input ( $n_{2} = n_{1}$ )	the "no transition" case	Ex 9

Handy constants used below: $h = 6.626 \times 1 0^{- 34} J\cdots$ , $c = 3.00 \times 1 0^{8} m/s$ , $1 eV = 1.602 \times 1 0^{- 19} J$ .

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

Example 1 — First line of a series (Cell A)

Worked example Longest-wavelength line of the

Paschen series Paschen means $n_{1} = 3$ . Find the wavelength of its first (longest- $λ$ ) line.

Forecast: Paschen sits in the infrared. Guess: is $λ$ bigger or smaller than the 656 nm red Balmer line? (Bigger — IR is longer than red.)

Step 1 — Pick $n_{2}$ . First line ⇒ smallest allowed jump ⇒ $n_{2} = n_{1} + 1 = 4$ . Why this step? "First line" always means the gentlest fall, $n_{2} = n_{1} + 1$ ; that gives the smallest $Δ E$ and hence the largest $λ$ of the family.

Step 2 — Bracket. $\frac{1}{3 ^{2}} - \frac{1}{4 ^{2}} = \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} = 0.04861$ . Why this step? This bracket is the whole physics — it measures the size of the energy drop in units of $R_{H}$ .

Step 3 — Multiply and invert. $\frac{1}{λ} = 1.097 \times 1 0^{7} \times 0.04861 = 5.333 \times 1 0^{5} m^{- 1}$ $λ = \frac{1}{5.333 \times 1 0 ^{5}} = 1.875 \times 1 0^{- 6} m = 1875 nm$ Why this step? $λ = 1/ \overset{ν}{ˉ}$ ; the wavenumber is just the reciprocal of wavelength.

Verify: $1875 nm$ is deep infrared — much longer than the 656 nm red line, exactly as forecast. Units: $(m^{- 1})^{- 1} = m$ . ✓ (This is the famous Paschen- $α$ line.)

Example 2 — A deeper line (Cell B)

Worked example The

third Balmer line ( $n_{2} = 5 \to n_{1} = 2$ ), "H-γ" Find its wavelength and check it is still visible.

Forecast: As $n_{2}$ grows (3→4→5), lines pile toward the blue/violet end. Guess $λ$ smaller than 656 nm.

Step 1 — Identify levels. Balmer ⇒ $n_{1} = 2$ . "Third line" counts $n_{2} = 3, 4, 5$ , so $n_{2} = 5$ . Why this step? First line is $n_{2} = 3$ , second $n_{2} = 4$ , third $n_{2} = 5$ . Miscounting here is the #1 error.

Step 2 — Bracket. $\frac{1}{4} - \frac{1}{25} = \frac{25 - 4}{100} = \frac{21}{100} = 0.2100$ . Why this step? Bigger $n_{2}$ makes $1/ n_{2}^{2}$ tiny, so the bracket grows toward $1/4 = 0.25$ (the Balmer limit).

Step 3 — Wavelength. $\frac{1}{λ} = 1.097 \times 1 0^{7} \times 0.2100 = 2.304 \times 1 0^{6} m^{- 1}, λ = 4.34 \times 1 0^{- 7} m = 434 nm$ Why this step? Reciprocal as always.

Verify: 434 nm is violet — visible, and shorter than 656 nm red, as forecast. It is a real Balmer line (H-γ, violet). ✓

Example 3 — Series limit, a limiting/degenerate case (Cell C)

Worked example Series limit of

Balmer As $n_{2} \to \infty$ (electron falls from the very edge of the atom), find $λ$ .

Forecast: The limit is the shortest $λ$ of the whole series. Balmer's first line is 656 nm; the limit must be well below that.

Step 1 — Take the limit. As $n_{2} \to \infty$ , $\frac{1}{n _{2}^{2}} \to 0$ . Why this step? A huge denominator squared makes the term vanish — this is the limiting behaviour the matrix demands. No jump is bigger than "from infinity", so it caps the series.

Step 2 — Only one term survives. $\frac{1}{λ} = R_{H} \cdot \frac{1}{n _{1}^{2}} = \frac{1.097 \times 1 0 ^{7}}{2 ^{2}} = 2.7425 \times 1 0^{6} m^{- 1}$ Why this step? This is the general series-limit formula $λ_{limit} = n_{1}^{2} / R_{H}$ .

Step 3 — Invert. $λ = \frac{4}{1.097 \times 1 0 ^{7}} = 3.646 \times 1 0^{- 7} m = 365 nm$ .

Verify: 365 nm is near-UV, shorter than every visible Balmer line — the series "piles up" here. Matches the tabulated Balmer limit. ✓

Example 4 — Reverse problem: given $λ$ , find $n_{2}$ (Cell D)

Worked example A hydrogen lamp shows a line at

$λ = 486 nm$ in the Balmer series. Which transition is it?

Forecast: 486 nm is blue-green, between red (656) and violet (434). Guess $n_{2} = 4$ .

Step 1 — Compute $\overset{ν}{ˉ}$ . $\frac{1}{λ} = \frac{1}{486 \times 1 0 ^{- 9}} = 2.0576 \times 1 0^{6} m^{- 1}$ . Why this step? We must feed the formula a wavenumber to solve for $n_{2}$ .

Step 2 — Isolate the bracket. Balmer ⇒ $n_{1} = 2$ : $\frac{1}{4} - \frac{1}{n _{2}^{2}} = \frac{ν ˉ}{R _{H}} = \frac{2.0576 \times 1 0 ^{6}}{1.097 \times 1 0 ^{7}} = 0.18757$ Why this step? Dividing by $R_{H}$ strips it away, leaving pure integer physics.

Step 3 — Solve for $n_{2}$ . $\frac{1}{n _{2}^{2}} = 0.25 - 0.18757 = 0.06243 \Rightarrow n_{2}^{2} = 16.02 \Rightarrow n_{2} = 4.$ Why this step? $n_{2}$ must come out (near) an integer — that is the sanity gate for a real line.

Verify: $n_{2} = 4$ gives $\frac{1}{4} - \frac{1}{16} = 0.1875$ , so $λ = 486.3$ nm. Matches. This is H-β (blue-green). ✓

Example 5 — Compare two lines, no numbers needed (Cell E)

Example 6 — Cross-series identification (Cell F)

Worked example Two lines both sit near

$λ \approx 1875 nm$ (IR). One is Paschen- $α$ ( $4 \to 3$ ). Could a Brackett line coincide?

Forecast: Brackett ( $n_{1} = 4$ ) lines are even longer-wave IR, so probably no Brackett line is this short — unless it's a high- $n_{2}$ one? Let's test.

Step 1 — Brackett's shortest possible line is its series limit. $\frac{1}{λ} = R_{H} / 4^{2} = R_{H} /16$ . Why this step? The series limit is the smallest $λ$ Brackett can ever reach; if even that is longer than 1875 nm, no Brackett line can coincide.

Step 2 — Compute the Brackett limit. $\frac{1}{λ} = \frac{1.097 \times 1 0 ^{7}}{16} = 6.856 \times 1 0^{5} m^{- 1}, λ_{min} = 1459 nm .$ Why this step? This is the shortest Brackett wavelength.

Step 3 — Compare. Brackett lines run from $λ_{limit} = 1459$ nm upward to its first line ( $5 \to 4$ ) at ~4050 nm. So 1875 nm does fall inside Brackett's range! Why this step? We must check the full window, not just one line.

Verify — which Brackett $n_{2}$ ? Solve $\frac{1}{16} - \frac{1}{n _{2}^{2}} = \frac{1/ ( 1875 \times 1 0 ^{- 9} )}{R _{H}} = 0.04861$ . Then $\frac{1}{n _{2}^{2}} = 0.0625 - 0.04861 = 0.01389$ , $n_{2}^{2} = 72.0$ , $n_{2} = 8.49$ — not an integer. So no real Brackett line lands exactly at 1875 nm; only Paschen- $α$ genuinely lives there. Lesson: nearby wavelength ranges can overlap, but a real line demands integer $n_{2}$ . ✓

Example 7 — Real-world word problem: energy per photon (Cell G)

Worked example A UV sterilising lamp emits the

Lyman- $α$ line ( $2 \to 1$ ). What is each photon's energy in joules and eV?

Forecast: Lyman is high-energy UV; expect roughly 10 eV (ionization is 13.6 eV, and $2 \to 1$ is most of the way down).

Step 1 — Wavenumber. $\frac{1}{λ} = R_{H} (\frac{1}{1} - \frac{1}{4}) = 1.097 \times 1 0^{7} \times 0.75 = 8.2275 \times 1 0^{6} m^{- 1}$ . Why this step? Lyman ⇒ $n_{1} = 1$ ; first line ⇒ $n_{2} = 2$ .

Step 2 — Wavelength then energy. $λ = 1.2154 \times 1 0^{- 7} m = 121.5$ nm. $E = \frac{h c}{λ} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.00 \times 1 0 ^{8} )}{1.2154 \times 1 0 ^{- 7}} = 1.635 \times 1 0^{- 18} J .$ Why this step? $E = h c / λ$ converts the wavelength into the actual photon energy — this is the Photon Energy and Planck Relation in action.

Step 3 — Convert to eV. $\frac{1.635 \times 1 0 ^{- 18}}{1.602 \times 1 0 ^{- 19}} = 10.2 eV$ . Why this step? eV is the natural atomic unit; divide by the charge-of-electron conversion.

Verify: $10.2 eV = 13.6 \times 0.75$ eV exactly — because $Δ E = 13.6 \times (bracket)$ . Matches Ionization Energy of Hydrogen scaling; and 121.5 nm is real Lyman-α. ✓

Example 8 — Exam twist: absorption & an impossible line (Cell H)

Worked example (a) An electron

absorbs a photon and goes $1 \to 3$ . What wavelength does it absorb? (b) Is a "Balmer line with $n_{2} = 2$ " possible?

Forecast (a): Absorption is emission run backwards; same energy, same $λ$ as the $3 \to 1$ emission. Forecast: identical wavelength.

Step 1 (a) — Absorption uses the same $Δ E$ . The energy needed to climb $1 \to 3$ equals the energy released falling $3 \to 1$ : $\frac{1}{λ} = R_{H} (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) = 1.097 \times 1 0^{7} \times 0.8889 = 9.751 \times 1 0^{6} m^{- 1} .$ Why this step? Energy is conserved and level differences are fixed; the sign of $Δ E$ flips (electron gains energy) but its magnitude — and thus $λ$ — is unchanged. See Quantization of Energy.

Step 2 (a) — Wavelength. $λ = 1.026 \times 1 0^{- 7} m = 102.6 nm$ (UV).

Step 3 (b) — Check legality. A Balmer line has $n_{1} = 2$ . "Balmer with $n_{2} = 2$ " needs $n_{2} = n_{1}$ , violating $n_{2} > n_{1}$ . Why this step? If $n_{2} = n_{1}$ the bracket is $0$ ⇒ $Δ E = 0$ ⇒ no photon. The line does not exist.

Verify (a): 102.6 nm matches the tabulated $1 \to 3$ transition. (b): bracket $= 1/4 - 1/4 = 0$ , confirming no line. ✓

Example 9 — Zero / degenerate input (Cell I)

Recall

Recall First line vs series limit — which is longest / shortest

$λ$ ? First line ( $n_{2} = n_{1} + 1$ ) = longest $λ$ (smallest jump). Series limit ( $n_{2} \to \infty$ ) = shortest $λ$ (biggest jump).

Recall Reverse problem: how do you know your

$n_{2}$ is right? It must come out as a (near-)integer. A non-integer $n_{2}$ means no real line exists at that wavelength.

Recall Absorption vs emission for the same two levels?

Same $∣Δ E ∣$ , same $λ$ ; only the direction (electron climbs vs falls) differs.

Connections

Bohr Model of the Atom — source of $E_{n} = - 13.6/ n^{2}$ eV behind every bracket.
Quantization of Energy — why only integer $n$ , and why $n_{2} = n_{1}$ gives nothing.
Photon Energy and Planck Relation — $E = h c / λ$ used in Ex 7.
Rydberg Constant and Spectra of Hydrogen-like Ions — same drills with a $Z^{2}$ factor.
Ionization Energy of Hydrogen — the 13.6 eV that Ex 7's Lyman line is 75% of.
Electromagnetic Spectrum — UV/visible/IR labels used throughout.