Atomic Spectra: Emission, Absorption, and Electronic Transitions

Understand emission and absorption spectra

Atomic Spectra: Emission, Absorption, and Electronic Transitions

Atomic spectra arise from quantized energy transitions of electrons in atoms. When an atom absorbs energy (electrical discharge, flame, or photon), an electron can be promoted from a lower to a higher energy level. As the electron relaxes back down, the atom emits photons at discrete energies, producing line spectra that act as "fingerprints" for each element.

This topic connects quantum theory, electronic structure, and practical spectroscopy used in lab analysis and astrophysics. You should be able to describe how lines form, interpret simple spectra, and apply energy–wavelength–frequency relationships.

Key ideas

  • Electrons occupy discrete energy levels. Transitions between levels involve exact energy quanta.
  • Emission spectra: bright lines on a dark background (photons emitted as electrons relax).
  • Absorption spectra: dark lines on a continuous spectrum (photons absorbed to promote electrons).
  • Energy–light relationships: E = hν and c = λν, so E = hc/λ. Higher energy → higher frequency → shorter wavelength.
  • Each element has unique transitions → unique line positions; useful for identification.

Emission vs. absorption spectra

  • Emission: A hot, low-density gas emits light at specific wavelengths, giving bright lines. Example: Hydrogen discharge tube shows Balmer lines in the visible region.
  • Absorption: A continuous light source passing through a cool gas yields missing wavelengths (dark lines) where atoms absorbed light to excite electrons. The pattern of missing lines matches the element’s allowed transitions.

Both reflect the same quantized energy differences. If an absorption line occurs at 656 nm (Hα), the same transition can emit at 656 nm when the electron relaxes.

Hydrogen spectrum and energy levels

The hydrogen atom is the classic case where energy levels can be modeled simply. The allowed energies (n = 1, 2, 3, …) get closer together at higher n. Series of lines correspond to electrons ending in a specific level:

  • Lyman series: transitions to n = 1 (ultraviolet)
  • Balmer series: transitions to n = 2 (visible and near UV)
  • Paschen, Brackett, Pfund: transitions to n = 3, 4, 5 (infrared)

The Balmer lines (Hα ≈ 656 nm, Hβ ≈ 486 nm, Hγ ≈ 434 nm, Hδ ≈ 410 nm) are often seen in laboratory spectra and stellar spectra. As n increases, levels converge and the spectral lines crowd together.

Interpreting spectra qualitatively

  • Line positions: Indicate the energy gaps between quantized states; shorter wavelengths mean larger energy changes.
  • Line intensity: Related to transition probability and population of the upper level; more populated and more allowed transitions are brighter.
  • Line broadening: Collisions, Doppler motion, and instrument resolution broaden otherwise sharp lines.

Practical uses

  • Element identification: Flame tests (Na⁺ yields yellow; K⁺ lilac; Cu²⁺ green/blue) and emission spectroscopy use characteristic lines.
  • Quantitative analysis: Atomic absorption spectroscopy (AAS) and inductively coupled plasma optical emission spectroscopy (ICP-OES) measure concentrations from line intensities.
  • Astrophysics: Absorption lines in stellar spectra reveal composition, temperature, and motion (via Doppler shifts).

Problem walkthroughs

  1. Which has higher energy: a 400 nm photon or a 700 nm photon?
  • Use E = hc/λ. Smaller λ → larger E. So 400 nm photons carry more energy than 700 nm photons.
  1. Why do different elements have different line patterns?
  • Nuclear charge and electron–electron interactions create unique sets of energy levels; allowed transitions differ by element, so spectra are element-specific.
  1. In a discharge tube, bright lines at specific wavelengths appear. What process produces these lines?
  • Excited electrons relax to lower levels, emitting photons with energies equal to the level differences.

Common mistakes

  • Thinking atoms emit a continuous spectrum. Continuous spectra come from hot solids or dense gases; isolated atoms produce lines.
  • Mixing up emission and absorption diagrams; remember emission adds bright lines, absorption removes lines from a continuum.
  • Ignoring selection rules and relative intensities; not every transition is equally likely.

Quick reference

  • Emission: excited → lower level + photon (bright lines)
  • Absorption: lower level + photon → excited (dark lines)
  • E = hc/λ; higher E → shorter λ
  • Hydrogen: Balmer series visible; Lyman UV; others IR
  • Spectra identify elements and quantify amounts in samples

📚 Practice Problems

1Problem 1easy

Question:

What is the difference between an emission spectrum and an absorption spectrum?

💡 Show Solution

Emission Spectrum: • Produced when excited electrons fall to lower energy levels • Electrons emit photons of specific wavelengths • Appears as bright lines on a dark background • Example: Heated gas emitting light

Absorption Spectrum: • Produced when ground-state electrons absorb photons and jump to higher energy levels • Only photons of specific wavelengths are absorbed • Appears as dark lines on a bright continuous background • Example: White light passing through cool gas

Key insight: The wavelengths in emission and absorption spectra for the same element are identical - they represent the same energy transitions, just in opposite directions.

2Problem 2medium

Question:

Calculate the energy (in joules) of a photon with wavelength 486 nm (blue light in hydrogen's emission spectrum). Use h = 6.626 × 10⁻³⁴ J·s and c = 3.00 × 10⁸ m/s.

💡 Show Solution

Use the equation: E = hc/λ

Step 1: Convert wavelength to meters λ = 486 nm = 486 × 10⁻⁹ m = 4.86 × 10⁻⁷ m

Step 2: Calculate energy E = (6.626 × 10⁻³⁴ J·s)(3.00 × 10⁸ m/s) / (4.86 × 10⁻⁷ m) E = (19.878 × 10⁻²⁶) / (4.86 × 10⁻⁷) E = 4.09 × 10⁻¹⁹ J

The photon has energy of 4.09 × 10⁻¹⁹ joules.

This corresponds to an electron transition in hydrogen (n=4 to n=2).

3Problem 3medium

Question:

Why do different elements produce different line spectra (unique "fingerprints")?

💡 Show Solution

Each element has a unique electron configuration and energy level structure:

  1. Different nuclear charge (Z) • Affects the strength of attraction between nucleus and electrons • Creates element-specific energy levels

  2. Different number of electrons • Creates unique patterns of electron-electron repulsion • Results in different available transitions

  3. Unique energy level spacing • Each element has its own set of allowed energy states • Transitions between levels produce specific photon energies: ΔE = E_final - E_initial = hc/λ

  4. Result: Element-specific wavelengths • Each transition produces a photon of specific energy/wavelength • The collection of all possible transitions creates a unique pattern

This is why spectroscopy can identify elements - each has a unique "barcode" of spectral lines!

4Problem 4hard

Question:

An electron in a hydrogen atom drops from n=5 to n=2. Calculate the wavelength of the emitted photon using the Rydberg equation: 1/λ = R_H(1/n_f² - 1/n_i²), where R_H = 1.097 × 10⁷ m⁻¹.

💡 Show Solution

Given: n_i = 5 (initial) n_f = 2 (final) R_H = 1.097 × 10⁷ m⁻¹

Step 1: Apply Rydberg equation 1/λ = R_H(1/n_f² - 1/n_i²) 1/λ = (1.097 × 10⁷)(1/4 - 1/25) 1/λ = (1.097 × 10⁷)(0.25 - 0.04) 1/λ = (1.097 × 10⁷)(0.21) 1/λ = 2.304 × 10⁶ m⁻¹

Step 2: Solve for λ λ = 1/(2.304 × 10⁶) λ = 4.34 × 10⁻⁷ m = 434 nm

This is in the visible range (violet-blue), part of the Balmer series (transitions to n=2).

5Problem 5hard

Question:

A sodium lamp emits yellow light at 589 nm. If a sodium atom absorbs this photon, what happens to the electron? How much energy does it gain (in kJ/mol)?

💡 Show Solution

Step 1: Calculate energy per photon E = hc/λ λ = 589 nm = 5.89 × 10⁻⁷ m E = (6.626 × 10⁻³⁴)(3.00 × 10⁸)/(5.89 × 10⁻⁷) E = 3.37 × 10⁻¹⁹ J per photon

Step 2: What happens to the electron? The electron absorbs the photon and jumps to a higher energy level (excited state). The energy difference between the levels equals the photon energy.

Step 3: Convert to kJ/mol E = (3.37 × 10⁻¹⁹ J/photon)(6.022 × 10²³ photons/mol) E = 2.03 × 10⁵ J/mol = 203 kJ/mol

The electron gains 203 kJ/mol of energy and moves to an excited state. This is a reversible process - when the electron falls back, it emits the same yellow light.