Spectral series — Lyman, Balmer, Paschen: Exercises

Figure 1 — the wavelength map. The picture below is a single horizontal axis measured in nanometres (nm), short wavelengths on the left, long on the right. Three coloured blocks mark the three main series: the violet block on the far left is Lyman (ultraviolet, ~91–122 nm), the magenta block in the middle is Balmer (visible, ~365–656 nm), and the orange block on the right is Paschen (infrared, ~820–1875 nm). The small navy tick-marks below the axis label the key wavelengths you compute in these exercises. Notice the blocks do not touch — a fact you will prove in Exercise 3.1. Keep this map beside you.

Figure — Spectral series — Lyman, Balmer, Paschen

Level 1 — Recognition

Exercise 1.1

A hydrogen line is observed at $434 nm$ (blue-green, visible). Name the series it belongs to and state its landing level $n_{1}$ .

Recall Solution

What we ask: which family owns a visible line. Why: each series is defined by its landing floor $n_{1}$ , and only one series lands in the visible band.

$121.5 nm$ region = ultraviolet = Lyman ( $n_{1} = 1$ ).
$\approx 400$ – $700 nm$ = visible = Balmer ( $n_{1} = 2$ ).
$> 800 nm$ = infrared = Paschen ( $n_{1} = 3$ ). Since $434 nm$ is visible, this is the Balmer series, landing level $n_{1} = 2$ . (It is in fact the $5 \to 2$ line, called H- $γ$ .)

Exercise 1.2

Order these three transitions from shortest to longest emitted wavelength, using reasoning only (no arithmetic): (a) $2 \to 1$ , (b) $3 \to 2$ , (c) $4 \to 3$ .

Recall Solution

Idea: shorter wavelength ⇔ larger energy gap ⇔ deeper landing floor. The floors get closer together as $n$ grows (spacing shrinks like $1/ n^{2}$ ), so a jump landing on floor 1 is the biggest, landing on floor 3 the smallest.

$2 \to 1$ : lands on floor 1 → biggest gap → shortest $λ$ .
$3 \to 2$ : lands on floor 2 → medium.
$4 \to 3$ : lands on floor 3 → smallest gap → longest $λ$ . Order (short → long): $2 \to 1, 3 \to 2, 4 \to 3$ .

Level 2 — Application

Exercise 2.1

Compute the wavelength of the H- $γ$ line, the Balmer transition $5 \to 2$ . Give the answer in nm.

Recall Solution

What: plug $n_{1} = 2$ , $n_{2} = 5$ into Rydberg. Why these numbers: Balmer ⇒ landing floor $n_{1} = 2$ ; the "5" is the starting floor. $\frac{1}{λ} = R_{H} (\frac{1}{2 ^{2}} - \frac{1}{5 ^{2}}) = 1.097 \times 1 0^{7} (\frac{1}{4} - \frac{1}{25}) .$ $\frac{1}{4} - \frac{1}{25} = \frac{25 - 4}{100} = \frac{21}{100} = 0.21.$ $\frac{1}{λ} = 1.097 \times 1 0^{7} \times 0.21 = 2.304 \times 1 0^{6} m^{- 1} .$ $λ = \frac{1}{2.304 \times 1 0 ^{6}} = 4.34 \times 1 0^{- 7} m = 434 nm .$ Matches the observed blue-green line of Exercise 1.1. ✓

Exercise 2.2

Find the energy (in eV) of the photon released in the Lyman $3 \to 1$ transition, then convert to a wavelength in nm.

Recall Solution

What: use the energy-difference form. Why: it lets us answer in eV directly, then bridge to nm. $Δ E = 13.6 (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) = 13.6 (1 - \frac{1}{9}) = 13.6 \times \frac{8}{9} = 12.09 eV .$ Bridge to wavelength: $λ = \frac{1240}{12.09} = 102.5 nm (ultraviolet) .$ Deep landing (floor 1) + high start (floor 3) ⇒ big gap ⇒ short UV wavelength, exactly as expected.

Level 3 — Analysis

Exercise 3.1

Show that every Lyman-series wavelength is shorter than every Balmer-series wavelength. (Prove the two bands never overlap.)

Recall Solution

What we prove: the longest Lyman wavelength is still shorter than the shortest Balmer wavelength. If the worst-case Lyman line beats the best-case Balmer line, the bands can't touch. Why this pairing: within a series, longest $λ$ ↔ smallest wavenumber (first line, $n_{2} = n_{1} + 1$ ); shortest $λ$ ↔ largest wavenumber (series limit, $n_{2} \to \infty$ ). We compare the smallest Lyman wavenumber to the largest Balmer wavenumber — if even the smallest Lyman exceeds the largest Balmer, done.

Smallest Lyman wavenumber ( $2 \to 1$ , its first line): $\frac{1}{λ} = R_{H} (1 - \frac{1}{4}) = \frac{3}{4} R_{H} = 0.75 R_{H} .$
Largest Balmer wavenumber (series limit, $n_{2} \to \infty$ , so $1/ n_{2}^{2} \to 0$ ): $\frac{1}{λ} = R_{H} (\frac{1}{4} - 0) = \frac{1}{4} R_{H} = 0.25 R_{H} .$ Since $0.75 R_{H} > 0.25 R_{H}$ , the smallest Lyman wavenumber already exceeds the largest Balmer one. Bigger wavenumber ⇔ shorter wavelength, so the whole Lyman band lies at shorter wavelengths than the whole Balmer band. They never overlap. ∎

Figure 2 — the no-overlap proof, drawn in wavenumbers. Here the horizontal axis is $1/ λ$ measured in units of $R_{H}$ (so it runs from 0 to 1), with bigger values = shorter wavelength on the right. The magenta block on the left is the entire Balmer band, stretching from $0.14 R_{H}$ (its longest line, $3 \to 2$ ) to $0.25 R_{H}$ (its series limit). The violet block on the right is the entire Lyman band, from $0.75 R_{H}$ (its longest line, $2 \to 1$ ) to $1.0 R_{H}$ (its limit). The orange double-headed arrow marks the empty gap between $0.25 R_{H}$ and $0.75 R_{H}$ — visible proof that the two bands never overlap.

Exercise 3.2

For the Paschen series ( $n_{1} = 3$ ), find the wavelength range: the longest line and the series limit.

Recall Solution

What: find the two edges of the Paschen band — its longest and shortest wavelength. Why we pick $n_{2} = 4$ for the longest line: the longest wavelength means the lowest photon energy, which comes from the smallest energy gap; the smallest gap for a series landing on $n_{1} = 3$ is a fall from the very next floor, $n_{2} = n_{1} + 1 = 4$ .

Longest line = smallest jump = $n_{2} = 4$ : $\frac{1}{λ} = R_{H} (\frac{1}{9} - \frac{1}{16}) = 1.097 \times 1 0^{7} \times \frac{16 - 9}{144} = 1.097 \times 1 0^{7} \times \frac{7}{144} .$ $= 1.097 \times 1 0^{7} \times 0.04861 = 5.333 \times 1 0^{5} m^{- 1} \Rightarrow λ = 1875 nm .$

What: now the other edge. Why $n_{2} \to \infty$ : the shortest wavelength = largest energy gap = electron arriving from the very edge of being free, i.e. $n_{2} \to \infty$ so $1/ n_{2}^{2} \to 0$ . $\frac{1}{λ} = R_{H} \times \frac{1}{9} = 1.219 \times 1 0^{6} m^{- 1} \Rightarrow λ = 820 nm .$ So Paschen lines span roughly 820 nm (limit, shortest) up to 1875 nm (first line, longest) — all infrared.

Level 4 — Synthesis

Exercise 4.1

A photon of energy $2.55 eV$ is emitted by hydrogen. Which transition produced it? Identify $n_{1}$ and $n_{2}$ .

Recall Solution

Strategy: $2.55 eV$ is small enough to be visible, so guess Balmer ( $n_{1} = 2$ ) and solve for $n_{2}$ . $2.55 = 13.6 (\frac{1}{4} - \frac{1}{n _{2}^{2}}) \Rightarrow \frac{2.55}{13.6} = 0.25 - \frac{1}{n _{2}^{2}} .$ $0.1875 = 0.25 - \frac{1}{n _{2}^{2}} \Rightarrow \frac{1}{n _{2}^{2}} = 0.0625 = \frac{1}{16} \Rightarrow n_{2} = 4.$ Transition: $4 \to 2$ (the Balmer H- $β$ line). Check: $λ = 1240/2.55 = 486 nm$ , blue — correct.

Exercise 4.2

The Lyman first line ( $2 \to 1$ ) has $λ = 121.5 nm$ . A doubly-ionized lithium ion $Li^{2 +}$ is hydrogen-like with $Z = 3$ . Predict the wavelength of its $2 \to 1$ line.

Recall Solution

Tool: for a hydrogen-like ion of nuclear charge $Z$ , every wavenumber scales by $Z^{2}$ (see Hydrogen-like Ions and Z dependence): $\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) .$ Why $Z^{2}$ : the electron feels $Z$ times the nuclear pull, and the energy levels deepen as $Z^{2}$ , so gaps — hence wavenumbers — grow by $Z^{2}$ . Same $n$ 's ⇒ same bracket, so the wavenumber is $Z^{2} = 9$ times larger, meaning the wavelength is $9$ times smaller: $λ_{Li^{2 +}} = \frac{121.5 nm}{9} = 13.5 nm (extreme UV / soft X-ray) .$

Level 5 — Mastery

Exercise 5.1

Hydrogen gas is bombarded so that electrons are excited to the $n = 4$ level. When they cascade back down, how many distinct spectral lines appear, and which of them are visible (Balmer)?

Recall Solution

Counting the lines: every downward pair of levels among ${1, 2, 3, 4}$ gives one line. The number of ways to pick 2 floors out of 4 is $(2 4) = \frac{4 \times 3}{2} = 6 lines .$ Listing them: $4 \to 3, 4 \to 2, 4 \to 1, 3 \to 2, 3 \to 1, 2 \to 1$ . Which are Balmer (visible)? Those landing on $n_{1} = 2$ : $3 \to 2$ and $4 \to 2$ . So 2 of the 6 lines are visible; the rest are $2 \to 1, 3 \to 1, 4 \to 1$ (Lyman, UV) and $4 \to 3$ (Paschen, IR).

Figure 3 — the cascade from $n = 4$ . This is an energy-level diagram: four horizontal navy lines mark the floors $n = 1, 2, 3, 4$ , drawn at their true energies $E_{n} = - 13.6/ n^{2}$ eV (so floor 1 sits deep at $- 13.6$ eV and the floors bunch up near the top). Each downward arrow is one emitted line, colour-coded by which series it belongs to: violet arrows land on floor 1 (Lyman, UV), magenta arrows land on floor 2 (Balmer, visible), and the single orange arrow lands on floor 3 (Paschen, IR). Count the arrows and you see all six lines; count the magenta ones and you see the two visible Balmer lines.

Exercise 5.2

Find the wavelength of the $4 \to 2$ Balmer line and confirm it lies in the visible band, then verify it is one of the two visible lines counted in Exercise 5.1.

Recall Solution

$\frac{1}{λ} = R_{H} (\frac{1}{4} - \frac{1}{16}) = 1.097 \times 1 0^{7} \times \frac{4 - 1}{16} = 1.097 \times 1 0^{7} \times \frac{3}{16} .$ $= 1.097 \times 1 0^{7} \times 0.1875 = 2.057 \times 1 0^{6} m^{- 1} \Rightarrow λ = 486 nm (blue, visible) .$ Yes — this is exactly the H- $β$ line, one of the two Balmer lines from the $n = 4$ cascade (the other being $3 \to 2$ at $656 nm$ , H- $α$ ).

Exercise 5.3

The ionization energy of hydrogen from the ground state is $13.6 eV$ (see Ionization Energy). Using the Lyman series limit, show that the shortest Lyman wavelength corresponds exactly to a photon of energy $13.6 eV$ .

Recall Solution

What: Lyman series limit is $n_{1} = 1$ , $n_{2} \to \infty$ . $\frac{1}{λ} = R_{H} (\frac{1}{1} - 0) = R_{H} = 1.097 \times 1 0^{7} m^{- 1} \Rightarrow λ = 91.2 nm .$ Convert to energy: $E = \frac{1240 eV\cdotnm}{91.2 nm} = 13.6 eV .$ This equals the energy to lift the electron from the ground state ( $n = 1$ ) out to freedom ( $n = \infty$ ) — i.e. the ionization energy. The series limit is the ionization threshold seen from the light side. ∎

Recall One-line summary of every answer

1.1 ::: Balmer, $n_{1} = 2$
1.2 ::: order $2 \to 1, 3 \to 2, 4 \to 3$
2.1 ::: $434 nm$
2.2 ::: $12.09 eV$ , $102.5 nm$
3.1 ::: no overlap ( $0.75 R_{H}$ vs $0.25 R_{H}$ )
3.2 ::: $820$ – $1875 nm$
4.1 ::: $4 \to 2$
4.2 ::: $13.5 nm$
5.1 ::: six lines, 2 visible
5.2 ::: $486 nm$
5.3 ::: $91.2 nm = 13.6 eV$

Connections

Bohr Model of the Atom — where $E_{n} = - 13.6/ n^{2}$ comes from.
Photon Energy and Planck's Relation — the $E = h c / λ$ bridge used all over this page.
Hydrogen-like Ions and Z dependence — the $Z^{2}$ scaling of Exercise 4.2.
Ionization Energy — the series-limit link of Exercise 5.3.
Emission vs Absorption Spectra — why these appear as bright emission lines.
Quantization of Angular Momentum — the deeper reason the floors are discrete.