The Hydrogen Spectrum from Quantum Theory

Introduction

Hydrogen emits a pattern of sharp spectral lines rather than a continuous rainbow. The quantum energy formula gives a direct explanation: the electron can occupy only particular bound energies, and a photon is emitted when the atom moves from a higher-energy state to a lower-energy one. Sorting transitions by their final principal quantum number produces the named Lyman, Balmer and Paschen series.

Core explanation

For hydrogen, Eₙ = −hcR H/n², with n = 1, 2, 3, … and R H the hydrogen Rydberg wavenumber. A transition from n high to n low, where n high > n low, releases photon energy E high − E low. Substitution gives hc/λ = hcR H(1/n low² − 1/n high²). Cancelling hc yields the line-position formula 1/λ = R H(1/n low² − 1/n high²). Both bracketed terms are positive, and the first is larger, so an emitted photon has a positive wavelength.

Every line in a series has the same n low. For the Lyman series, n low = 1; its lines lie in the ultraviolet. The Balmer series ends at n low = 2; its familiar visible lines include 3 → 2, although the series also extends toward the near ultraviolet. The Paschen series ends at n low = 3 and lies in the infrared. The spectral regions follow from the size of the energy releases, not merely from the labels. A fall to n = 1 usually releases more energy than a comparable fall to n = 3.

As n high grows without bound at fixed n low, 1/n high² tends to zero. The lines crowd toward a limiting wavenumber R H/n low². Physically this is the bound–free threshold for the lower level: reaching the continuum from that level takes hcR H/n low². The line spacing shrinks near the limit because adjacent high-n levels have increasingly similar energies. The Schrödinger treatment reproduces this formula while also predicting wavefunctions and angular-momentum states, which the elementary Bohr orbit picture does not provide.

Absorption uses the same energy differences in the opposite direction. An atom initially at n low can absorb a photon of the matching energy and reach n high, provided the transition is allowed and the population is present. Observing a particular line therefore depends on state populations and transition probabilities as well as the level energies. The Rydberg formula predicts positions, not intensities.

Step-by-step reasoning

Choose the initial and final n values and identify which is higher in energy. For an emission problem, use n high > n low and compute the positive energy difference. Insert the numbers into 1/λ = R H(1/n low² − 1/n high²), check that λ has units of length, and name the series using only n low. For an absorption problem, the direction reverses but the photon energy magnitude is the same.

Visual explanation

Draw horizontal levels at negative energies, with n = 1 lowest and successively closer levels approaching E = 0. Draw three downward arrows ending at n = 1, n = 2 and n = 3 in different colours. At the right, place a wavenumber axis showing each group of lines crowding toward its own series limit.

Real-world analogy

A staircase offers fixed landing heights. A person stepping down between two landings loses a definite height difference, and different pairs give different drops. The atom similarly has discrete energy differences. The analogy stops there: the electron is a quantum state, not a tiny person travelling along a visible stairway between stationary orbits.

Real-world example

The red H-alpha line in hydrogen emission is the Balmer 3 → 2 transition, near 656 nm. Astronomers and laboratory spectroscopists use its measured position to recognise hydrogen and investigate motion through Doppler shifts. Its presence and strength also depend on whether the emitting gas has enough excited atoms; the energy formula alone cannot tell how bright it will appear.

Why?

Quantisation of bound-state energy restricts the photon energies available in single-electron transitions. The subtraction of two negative level energies is crucial: a higher n has a less negative energy, so moving down releases positive energy. The common endpoint determines a series because all such transitions share one term in the energy-difference formula.

Common misconception

It is easy to call 3 → 2 a Lyman line because it starts from n = 3; a series is named by the final lower level, so it is Balmer. It is also wrong to treat every Balmer wavelength as visible: the high-n lines approach the ultraviolet series limit. Finally, a calculated line position does not imply every hydrogen sample emits that line at measurable intensity.

Worked example

For 3 → 2, the reciprocal wavelength is R H(1/2² − 1/3²) = R H(1/4 − 1/9) = 5R H/36. Taking R H ≈ 1.097 × 10⁷ m⁻¹ gives 1/λ ≈ 1.524 × 10⁶ m⁻¹, hence λ ≈ 6.56 × 10⁻⁷ m = 656 nm. The lower level is n = 2, so this is a Balmer line. It is in the visible red region. Rounding the Rydberg constant changes only the final digits.

Quick check

1. Which named series contains the hydrogen transition 5 → 3, and does it emit or absorb in that direction? Answer: It ends at n = 3, so it belongs to the Paschen series. The electron drops from higher to lower energy, so a photon is emitted; its wavelength is in the infrared region.

Exam focus

Write the energy difference before substituting into the wavelength equation to avoid a negative photon energy. Use reciprocal metres for R H if λ is required in metres. Identify a series by n low, state whether the arrow is up or down, and remember that the limit uses n high → ∞ rather than n high = 0.

Advanced insight

The simple formula assumes an ideal hydrogen atom and describes dominant gross-structure energies. Reduced electron–proton mass gives the hydrogen-specific R H rather than the infinite-nuclear-mass constant. Fine structure, hyperfine structure and external fields can split or shift observed lines on a smaller scale. These refinements do not remove the central inverse-square pattern.

Summary

Hydrogen's Eₙ = −hcR H/n² energies imply 1/λ = R H(1/n low² − 1/n high²). Transitions ending at n = 1, 2 and 3 form the Lyman, Balmer and Paschen series. Higher initial n values make lines converge toward an ionisation threshold. The formula determines line positions; intensity needs additional physics.

Practice questions

1. Calculate the wavelength of the Lyman 2 → 1 line using R H = 1.097 × 10⁷ m⁻¹. Answer: 1/λ = R H(1 − 1/4) = 3R H/4 ≈ 8.23 × 10⁶ m⁻¹. Thus λ ≈ 1.216 × 10⁻⁷ m = 121.6 nm, in the ultraviolet. 2. What is the limiting wavelength of the Balmer series, and what process does it represent? Answer: As n high → ∞, 1/λ limit = R H/4, so λ limit = 4/R H ≈ 365 nm. It corresponds to the bound–free threshold from n = 2, or the reverse of capture from the continuum into n = 2. 3. A hydrogen atom initially in n = 2 absorbs a photon matched to the 2 ↔ 4 difference. Which level is reached, and which series names the reverse emission? Answer: Absorption moves it upward to n = 4. The reverse 4 → 2 emission ends at n = 2 and is a Balmer line with the same photon-energy magnitude.