StudyBAP
Unit 5 · Lesson 2b

2bAtomic Spectra and Energy Levels

Explore emission and absorption spectra, hydrogen's spectral series, and how spectroscopy reveals the quantum structure of atoms and the composition of distant stars.

Atomic spectra are the unique fingerprints of every element — spectroscopy lets us identify the composition of stars billions of light-years away, detect trace chemicals in the atmosphere, and probe the quantum structure of matter.

Lesson Overview

When atoms absorb or emit energy, they produce characteristic spectra — unique "fingerprints" of light. Emission spectra show bright colored lines on a dark background; absorption spectra show dark lines on a continuous rainbow. Both arise from electrons jumping between quantized energy levels: ΔE = hf. Hydrogen's spectral series are named by their discoverers — Lyman (UV), Balmer (visible), and Paschen (IR). Spectroscopy is one of the most powerful tools in science, used to identify elements in distant stars, analyze chemical composition, and probe quantum structure.

Key Concepts

Emission Spectrum

Bright lines on dark background — atoms emit photons when electrons drop to lower levels

Absorption Spectrum

Dark lines on continuous spectrum — atoms absorb photons to excite electrons to higher levels

Photon Energy Formula

ΔE = hf = hc/λ — photon energy equals the energy difference between levels

Lyman Series

Transitions to n=1; UV range (91–122 nm)

Balmer Series

Transitions to n=2; visible range (364–656 nm) — H-alpha at 656 nm (red)

Paschen Series

Transitions to n=3; infrared range (820 nm–1.87 μm)

Example 1

A hydrogen electron transitions from n = 3 to n = 2. Calculate the wavelength of the emitted photon and identify the spectral series.

Answer:ΔE = 13.6(1/2² − 1/3²) = 13.6(1/4 − 1/9) = 13.6 × 5/36 = 1.889 eV = 3.02×10⁻¹⁹ J. λ = hc/ΔE = (6.626×10⁻³⁴ × 3×10⁸) / 3.02×10⁻¹⁹ ≈ 658 nm. This is the H-alpha line in the Balmer series (visible red light).
Example 2

Calculate the wavelength of the photon emitted when a hydrogen electron falls from n = 2 to n = 1 (Lyman series).

Answer:ΔE = 13.6(1/1² − 1/2²) = 13.6(1 − 0.25) = 13.6 × 0.75 = 10.2 eV = 1.632×10⁻¹⁸ J. λ = hc/ΔE = (6.626×10⁻³⁴ × 3×10⁸) / 1.632×10⁻¹⁸ ≈ 122 nm. This is UV — the Lyman-alpha line.
Example 3

Explain the difference between an emission spectrum and an absorption spectrum. Why do they contain the same wavelengths?

Answer:Emission: hot gas emits photons at specific wavelengths when electrons drop to lower levels — bright lines on dark background. Absorption: cool gas in front of a continuous source absorbs the same wavelengths to excite electrons — dark lines on rainbow background. Same wavelengths because the same energy differences ΔE = hf are involved in both processes.
Example 4

A photon of wavelength 486 nm is emitted by hydrogen. Identify which transition produced it.

Answer:E_photon = hc/λ = (6.626×10⁻³⁴ × 3×10⁸) / (486×10⁻⁹) = 4.09×10⁻¹⁹ J = 2.55 eV. Check Balmer series (n→2): ΔE = 13.6(1/4 − 1/n²) = 2.55 → 1/n² = 0.25 − 2.55/13.6 = 0.25 − 0.1875 = 0.0625 → n² = 16 → n = 4. This is the n=4→n=2 transition (H-beta line, blue-green).
Example 5

How is spectroscopy used in astronomy to determine the composition of a star's atmosphere?

Answer:Starlight passes through the star's cooler outer atmosphere. Atoms there absorb specific wavelengths matching their energy level differences, creating dark absorption lines in the spectrum. By matching these dark lines to known laboratory spectra of elements, astronomers identify which elements are present. The Fraunhofer lines in the Sun's spectrum reveal hydrogen, helium, sodium, calcium, and many other elements.
Guided Problem 1

Calculate the energy and wavelength of the photon emitted when a hydrogen electron transitions from n = 4 to n = 3 (Paschen series).

Hint: ΔE = 13.6(1/3² − 1/4²) eV. Then λ = hc/ΔE.

Guided Problem 2

A hydrogen atom absorbs a photon of wavelength 97.2 nm. To which energy level does the electron jump from the ground state?

Hint: Find photon energy E = hc/λ. Then E_final = E₁ + E_photon = −13.6 + E_photon. Solve Eₙ = −13.6/n² for n.

Guided Problem 3

Why does the Balmer series appear in the visible range while the Lyman series is in the UV?

Hint: Compare the energy differences for transitions ending at n=1 (Lyman) vs n=2 (Balmer). Higher ΔE means shorter wavelength.

Guided Problem 4

A sodium street lamp emits bright yellow light at 589 nm. Calculate the energy difference between the two sodium energy levels responsible for this emission.

Hint: Use ΔE = hc/λ. Convert to eV.

Guided Problem 5

How many distinct spectral lines can be emitted by hydrogen atoms excited to the n = 4 level as they return to the ground state?

Hint: Count all possible downward transitions: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1. How many is that?

Key Vocabulary

Emission Spectrum

A spectrum of bright lines produced when excited atoms emit photons as electrons fall to lower energy levels. Each element has a unique emission spectrum.

Example: Hydrogen's visible emission spectrum shows four lines: red (656 nm), blue-green (486 nm), blue-violet (434 nm), and violet (410 nm).

Absorption Spectrum

A continuous spectrum with dark lines where atoms have absorbed photons, exciting electrons to higher energy levels. The dark lines match the emission lines of the same element.

Example: The Sun's absorption spectrum (Fraunhofer lines) reveals the elements in its atmosphere — dark lines at 589 nm indicate sodium.

Balmer Series

The series of hydrogen spectral lines produced by transitions ending at n=2. These lines fall in the visible range (364–656 nm).

Example: The H-alpha line (656 nm, red) is the n=3→n=2 transition; H-beta (486 nm, blue-green) is n=4→n=2.

Spectroscopy

The study of the interaction between matter and electromagnetic radiation, used to identify elements and compounds by their spectral fingerprints.

Example: Astronomers use spectroscopy to determine the chemical composition, temperature, and velocity of stars millions of light-years away.

Interactive Practice — 5 Questions

1

An emission spectrum consists of:

2

The Balmer series of hydrogen involves transitions ending at:

3

The energy of a photon emitted during an atomic transition equals:

4

Absorption spectra are produced when:

5

Spectroscopy is used in astronomy primarily to:

Independent Practice

1

Calculate the wavelengths of all four visible Balmer series lines of hydrogen (transitions from n=6, 5, 4, 3 to n=2). Identify the color of each.

2

Explain the difference between emission and absorption spectra. Why do they contain the same wavelengths for a given element?

3

A hydrogen atom in the n=4 state can emit photons as it returns to the ground state. List all possible transitions and calculate the wavelength of each photon.

4

The Lyman-alpha line of hydrogen is at 122 nm. Verify this using the energy level formula and calculate the photon's energy in eV.

5

★ A distant galaxy shows hydrogen absorption lines shifted to longer wavelengths (redshift). The H-alpha line (normally 656 nm) appears at 722 nm. Calculate the recession speed of the galaxy as a fraction of the speed of light using the Doppler formula: z = Δλ/λ₀ ≈ v/c.

Challenge
⚠️

Common Mistakes

❌

Confusing emission spectra (bright lines) with absorption spectra (dark lines).

✅

Emission: hot gas glows → bright lines on dark background. Absorption: cool gas in front of continuous source → dark lines on rainbow. Same wavelengths, opposite appearance.

❌

Thinking higher-energy transitions always produce visible light.

✅

Very high ΔE transitions (like Lyman series, ending at n=1) produce UV photons. Only Balmer series (ending at n=2) produces visible light for hydrogen.

💡

Math Tips

📌

For hydrogen spectral lines: E_photon = 13.6(1/n_low² − 1/n_high²) eV. Then λ = hc/E_photon = 1240 eV·nm / E_photon(eV). The shortcut hc = 1240 eV·nm saves calculation time.