Emission and Absorption Spectra - AP Physics 2

SECTION 1

Historical Context & Motivation

The study of light emitted and absorbed by matter ranks among the most consequential experimental programs in the history of physics. When Isaac Newton first dispersed sunlight through a prism in the seventeenth century, he revealed a continuous rainbow of colors — but it was the discovery of dark lines interrupting that rainbow, and bright lines emitted by heated elements, that ultimately demanded an entirely new theory of matter. These spectral features could not be explained by classical wave optics, and their resolution required the birth of quantum mechanics.

1814 • Fraunhofer Lines

Joseph von Fraunhofer catalogued over 570 dark lines in the solar spectrum, labeling the most prominent with letters (e.g., the D-lines near 589 nm). He could not explain their origin but demonstrated they were reproducible features of sunlight.

1859 • Kirchhoff & Bunsen

Gustav Kirchhoff and Robert Bunsen showed that each chemical element produces a unique set of spectral lines when heated, and that the dark Fraunhofer lines corresponded to absorption by the same elements in the solar atmosphere. Spectroscopy became a tool for chemical identification.

1885 • Balmer Series

Johann Balmer discovered an empirical formula that predicted the visible wavelengths of hydrogen's emission lines with remarkable accuracy, suggesting an underlying mathematical structure to spectra.

1913 • Bohr Model

Niels Bohr proposed that electrons in hydrogen occupy quantized orbits with discrete energies. Transitions between these orbits produce or absorb photons of specific wavelengths, finally providing a theoretical basis for Balmer's formula.

1925–1926 • Quantum Mechanics

Heisenberg's matrix mechanics and Schrödinger's wave equation generalized Bohr's model to multi-electron atoms, explaining the spectra of all elements and the fine structure of spectral lines.

The central question that spectroscopy posed — why does each element emit and absorb only certain wavelengths of light? — became the gateway to quantum physics. Understanding emission and absorption spectra is therefore essential not only for the AP Physics 2 exam but for grasping the physical basis of modern atomic theory.

SECTION 2

Core Principles & Definitions

Spectral phenomena arise directly from the quantized energy levels of atoms. When an atom transitions between two energy states, it either releases or absorbs a photon whose energy exactly equals the difference between those states. This discrete nature of atomic energy is the origin of the line patterns that distinguish emission spectra and absorption spectra from the continuous spectrum produced by incandescent solids or dense gases.

Continuous Spectrum

Produced by hot, dense objects (solids, liquids, or high-pressure gases). Contains all wavelengths across a band and appears as a smooth rainbow. Described by blackbody radiation curves, it carries no line features.

Emission (Bright-Line) Spectrum

Produced by hot, low-pressure gases. Consists of discrete bright lines on a dark background. Each line corresponds to a specific electron transition from a higher to a lower energy level, releasing a photon.

Absorption (Dark-Line) Spectrum

Produced when continuous-spectrum light passes through a cooler, low-pressure gas. The gas absorbs photons at precisely the wavelengths it would emit, leaving dark lines in an otherwise continuous spectrum.

Quantized Energy Levels

Electrons in an atom can only occupy certain allowed energy states (n = 1, 2, 3, …). The ground state (n = 1) is the lowest. All higher states are called excited states. The energy difference between levels determines photon wavelength.

Kirchhoff's Three Laws of Spectroscopy

(1) A hot dense object emits a continuous spectrum. (2) A hot, diffuse gas emits a bright-line spectrum. (3) A cool gas in front of a continuous source produces an absorption spectrum. These empirical laws unify all three spectral types.

✦

KEY TAKEAWAY

Think of an atom's energy levels like the floors of a building with no ramps — only stairs. An electron can stand on any floor but never between floors. Emission is like tossing a ball from a high floor to a lower one: the ball (photon) carries exactly the energy corresponding to the height difference. Absorption is the reverse — catching a ball thrown from below and jumping up to a higher floor. Because the floors are at fixed heights, only balls of specific energies (wavelengths) can be thrown or caught.

SECTION 3

Visual Explanation — Energy-Level Transitions

Hydrogen energy-level diagram showing three transitions. Downward arrows represent emission: an electron drops to a lower level and a photon is released. The upward arrow represents absorption: the atom absorbs a photon and the electron is promoted to a higher level. Note that the spacing between levels decreases as n increases, reflecting the 1/n² dependence of the energy.

The diagram above illustrates several critical features. First, the energy levels are not evenly spaced — they converge as n increases, bunching together near the ionization limit at 0 eV. This means transitions involving the ground state (n = 1) release or require much more energy than transitions among higher levels. Second, the emission transition from n = 3 to n = 2 (the Hα line at 656.3 nm) falls in the visible red portion of the spectrum, while the n = 3 → 1 transition produces an ultraviolet photon — its larger energy gap yields a shorter wavelength. Third, absorption is the mirror process: the photon energy must match the gap precisely, or the atom will not absorb it. This selectivity is why absorption spectra consist of sharp dark lines at exactly the same wavelengths as the corresponding emission lines.

SECTION 4

Mathematical Framework

The quantitative treatment of emission and absorption spectra rests on two foundational relationships: the Bohr energy-level equation for hydrogen-like atoms and the photon energy equation. Together, they allow us to predict every spectral line wavelength from first principles.

ENERGY OF THE nTH LEVEL (HYDROGEN)

Eₙ = −13.6 eV / n²

E_n = energy of the electron in the nth level (eV); n = principal quantum number (1, 2, 3, …); −13.6 eV = the ground-state energy, equal to the ionization energy of hydrogen. The negative sign indicates a bound state.

PHOTON ENERGY FOR A TRANSITION

E_photon = |E_final − E_initial| = hf = hc / λ

h = Planck's constant = 6.626 × 10⁻³⁴ J·s = 4.136 × 10⁻¹⁵ eV·s; f = photon frequency (Hz); c = speed of light = 3.00 × 10⁸ m/s; λ = photon wavelength (m). The photon energy equals the magnitude of the energy difference between the two levels involved.

COMBINED WAVELENGTH EQUATION

1/λ = R_H (1/n_f² − 1/n_i²)

R_H = Rydberg constant for hydrogen = 1.097 × 10⁷ m⁻¹; n_f = final (lower) energy level; n_i = initial (higher) energy level. This equation is valid for emission when n_i > n_f. For AP Physics 2, you will most commonly use E = hf or the energy-level formula directly.

💡

AP Exam Tip

The AP Physics 2 equation sheet provides E = hf and the energy-level diagram for hydrogen. You are expected to calculate photon energies, wavelengths, and frequencies from given or computed energy differences. Practice converting between eV and joules: 1 eV = 1.602 × 10⁻¹⁹ J.

A key conceptual point in the mathematical framework is that the photon energy is always positive — it is the absolute value of the energy difference between levels. For emission, the electron moves from a higher (less negative) energy to a lower (more negative) energy, and the difference is carried away by the photon. For absorption, the incoming photon must supply exactly the right energy to promote the electron upward. If the photon energy does not match any available transition, the atom is transparent to that wavelength.

SECTION 5

Spectral Series of Hydrogen

The emission lines of hydrogen are organized into named spectral series, each defined by the lower energy level (n_f) to which the electron transitions. Every series converges to a series limit — the shortest wavelength produced when n_i → ∞ — and spans a particular region of the electromagnetic spectrum.

Series Name	Final Level ( $n_f$ )	Spectral Region	Key Wavelengths
Lyman	1	Ultraviolet	121.6 nm (Ly-α), 102.6 nm (Ly-β), series limit 91.2 nm
Balmer	2	Visible / near UV	656.3 nm (Hα, red), 486.1 nm (Hβ, cyan), 434.0 nm (Hγ, violet), limit 364.6 nm
Paschen	3	Infrared	1875 nm, 1282 nm, limit 820.4 nm
Brackett	4	Infrared	4051 nm, 2625 nm, limit 1458 nm

Hydrogen spectral series with representative wavelengths

The top bar shows a continuous spectrum for reference. The middle bar shows hydrogen's emission spectrum — bright colored lines on a dark background. The bottom bar shows the absorption spectrum — dark lines at the same wavelengths superimposed on a continuous rainbow. The four visible Balmer lines (Hα through Hδ) are labeled.

For the AP Physics 2 exam, the Balmer series is the most frequently tested because its lines fall in the visible range. However, you should understand conceptually why the Lyman series involves higher-energy photons (UV) — the transitions terminate at the deeply bound ground state — and why the Paschen and Brackett series involve lower-energy photons (IR), since the upper levels are closely spaced. Each element has its own unique set of energy levels, so each element's spectrum serves as a spectral fingerprint used in chemical analysis and astrophysics.

SECTION 6

Worked Example — Calculating a Spectral Line Wavelength

A hydrogen atom in the n = 4 excited state emits a photon and transitions to the n = 2 state. Determine the energy, wavelength, and color of the emitted photon.

Photon from the n = 4 → n = 2 Transition (Hβ line)

1
Step 1 — Find the Energy of Each Level

Using E_n = −13.6 eV / n²: E₄ = −13.6 / 4² = −13.6 / 16 = −0.850 eV E₂ = −13.6 / 2² = −13.6 / 4 = −3.40 eV

E₄ = −0.850 eV, E₂ = −3.40 eV
2
Step 2 — Calculate the Photon Energy

The photon carries the energy difference between the two levels: E_photon = |E₂ − E₄| = |−3.40 − (−0.850)| = |−2.55| = 2.55 eV Converting to joules: 2.55 eV × 1.602 × 10⁻¹⁹ J/eV = 4.09 × 10⁻¹⁹ J

E_photon = 2.55 eV = 4.09 × 10⁻¹⁹ J
3
Step 3 — Calculate the Wavelength

Using E = hc / λ, solve for λ: λ = hc / E = (6.626 × 10⁻³⁴ J·s)(3.00 × 10⁸ m/s) / (4.09 × 10⁻¹⁹ J) λ = 1.988 × 10⁻²⁵ / 4.09 × 10⁻¹⁹ = 4.86 × 10⁻⁷ m = 486 nm

λ = 486 nm
4
Step 4 — Identify the Color

A wavelength of 486 nm falls in the blue-cyan portion of the visible spectrum. This is the Hβ (H-beta) line of the Balmer series — one of the four visible hydrogen emission lines. You can verify this matches the second brightest line shown in the Balmer emission diagram from Section 5.

The emitted photon is cyan (blue-green) visible light.

SECTION 7

Emission vs. Absorption — Strengths, Limitations, and Applications

Feature	Emission Spectrum	Absorption Spectrum
Appearance	Bright colored lines on a dark background	Dark lines superimposed on a continuous spectrum
Physical Process	Electron drops from higher to lower energy level; photon emitted	Photon absorbed by electron, promoting it from lower to higher energy level
Source Conditions	Hot, low-pressure gas (Kirchhoff's 2nd law)	Cooler gas in front of a continuous source (Kirchhoff's 3rd law)
Line Positions	Identical wavelengths for a given element	Identical wavelengths for a given element
Astrophysical Use	Identifying composition of nebulae, stellar atmospheres	Identifying elements in stellar atmospheres (Fraunhofer lines); measuring Doppler shifts for radial velocity
Laboratory Use	Flame tests, gas discharge tubes, elemental analysis	UV-Vis spectrophotometry, atmospheric gas analysis

Comparison of emission and absorption spectra

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KEY TAKEAWAY

Emission and absorption spectra are two views of the same underlying physics, much like a photographic negative and positive of the same scene. A gas discharge tube and a gas cloud backlit by a bright source both encode the same set of energy-level spacings — one as bright lines, the other as dark gaps. Recognizing this complementarity is the single most important conceptual point for the AP exam.

Three interchangeable optical source modules—a dense source, discharge tube and continuum behind a gas cell—beside one spectrometer. A dense thermal source produces a continuum, an excited dilute gas produces emission lines, and cooler gas in front of a continuum can produce absorption lines. Photon energies match allowed atomic energy differences.

Three source conditions at one spectrometer. A dense thermal source produces a continuum, an excited dilute gas produces emission lines, and cooler gas in front of a continuum can produce absorption lines. Photon energies match allowed atomic energy differences. A dense thermal source produces a continuum, an excited dilute gas produces emission lines, and cooler gas in front of a continuum can produce absorption lines. Photon energies match allowed atomic energy differences.

SECTION 8

Connection to Advanced Theory

The Bohr model and its associated energy-level formula work beautifully for hydrogen but fail for multi-electron atoms. Full quantum mechanics — the Schrödinger equation — extends the concept of quantized energy levels to all elements by introducing additional quantum numbers (ℓ, m_ℓ, m_s) that describe orbital shape, orientation, and electron spin. These additional degrees of freedom lead to fine structure — closely spaced doublets and triplets that are invisible in the simple Bohr picture but detectable with high-resolution spectrometers.

Feature	Bohr Model (AP Physics 2)	Quantum Mechanical Model
Applicable to	Hydrogen and hydrogen-like ions (He⁺, Li²⁺)	All atoms and molecules
Electron description	Circular orbits with definite radius	Probability clouds (orbitals)
Energy quantization	Depends only on n	Depends on n, ℓ, and electron-electron interactions
Selection rules	Any n → n′ transition allowed	Δℓ = ±1 required for electric dipole transitions
Fine structure	Not predicted	Predicted via spin-orbit coupling

Bohr model vs. full quantum mechanics

For AP Physics 2, you need not solve the Schrödinger equation, but you should be aware that the simple Bohr picture is an approximation. The essential takeaway that survives the transition to quantum mechanics is the concept of quantized energy levels and the rule that photon energy equals the gap between levels. These principles are universal and apply to all atoms, even though the specific level spacings become more complex than the 1/n² pattern of hydrogen. Modern applications like lasers, LEDs, and astronomical spectroscopy all depend on precisely these ideas.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student observes the spectrum of a low-pressure gas discharge tube and sees four distinct bright lines on a dark background. A second student passes white light through a cool sample of the same gas and observes dark lines on a continuous background. Which of the following best explains why the dark lines appear at the same wavelengths as the bright lines?

A. The cool gas scatters all wavelengths of light equally, but the bright lines are simply brighter than the rest of the spectrum.
B. The atoms in the cool gas can only absorb photons whose energies match the differences between the atoms' quantized energy levels, which are the same energy differences that produce the emission lines.
C. The dark lines are caused by destructive interference between the white light and the light re-emitted by the gas.
D. The gas absorbs photons of all wavelengths, but the dark lines appear only where the gas re-emits light in a direction away from the observer.

Show answer (B)

B is correct. The same quantized energy-level spacings that determine which photon energies an atom can emit also determine which photon energies it can absorb. Choice A is wrong because scattering is not wavelength-selective in the way observed. Choice C incorrectly invokes wave interference — absorption is a quantum process, not a classical interference effect. Choice D is wrong because atoms do not absorb all wavelengths; they absorb only photons matching their energy-level gaps. While re-emission does occur in random directions, the selectivity itself is the key explanation.

Question 2: BASIC CALCULATION

A hydrogen atom transitions from the n = 3 state to the n = 1 state. What is the wavelength of the emitted photon? (Use E_n = −13.6 eV / n², h = 4.136 × 10⁻¹⁵ eV·s, c = 3.00 × 10⁸ m/s)

A. 656 nm
B. 122 nm
C. 103 nm
D. 486 nm

Show answer (C)

C is correct. E₃ = −13.6/9 = −1.51 eV and E₁ = −13.6/1 = −13.6 eV. The photon energy is |−13.6 − (−1.51)| = 12.09 eV. The wavelength is λ = hc/E = (4.136 × 10⁻¹⁵ eV·s)(3.00 × 10⁸ m/s) / 12.09 eV = 1.024 × 10⁻⁷ m ≈ 102.6 nm, which rounds to approximately 103 nm (Lyman-β, UV). Choice A (656 nm) is the Balmer-α line (n = 3 → 2). Choice B (122 nm) is the Lyman-α line (n = 2 → 1). Choice D (486 nm) is the Balmer-β line (n = 4 → 2). These are common distractors from other well-known transitions.

Question 3: INTERMEDIATE

An atom has three energy levels: the ground state at −8.0 eV, a first excited state at −5.0 eV, and a second excited state at −2.0 eV. A photon with energy 6.0 eV is incident on this atom when the atom is in its ground state. Which of the following correctly describes the result?

A. The atom absorbs the photon and the electron transitions to the second excited state.
B. The atom absorbs the photon and the electron transitions to the first excited state, with the remaining 3.0 eV emitted as a second photon.
C. The atom cannot absorb the photon because 6.0 eV does not match the energy difference to the first excited state.
D. The atom is ionized because 6.0 eV exceeds the energy of the second excited state.

Show answer (A)

A is correct. The energy difference between the ground state (−8.0 eV) and the second excited state (−2.0 eV) is exactly 6.0 eV, so the photon energy matches a valid transition. Choice B is wrong because partial absorption does not occur — the photon must be absorbed entirely and its energy must match a level difference exactly. Choice C incorrectly focuses on the ground-to-first-excited gap (3.0 eV) while ignoring the ground-to-second-excited gap (6.0 eV). Choice D is wrong because ionization would require at least 8.0 eV (to reach 0 eV from −8.0 eV), and 6.0 eV is insufficient.

Question 4: APPLIED

4 pts

An astronomer observes a distant star and records its absorption spectrum. She notices that the hydrogen Balmer-α line, which has a laboratory wavelength of 656.3 nm, appears at 659.6 nm in the star's spectrum. She wants to determine the star's radial velocity and also identify its atmospheric composition. (a) Explain how the shift in the Balmer-α line can be used to determine the star's radial velocity. (b) The astronomer also identifies dark lines at 589.0 nm and 589.6 nm in the absorption spectrum. Explain what these lines reveal about the star's atmospheric composition. (c) Explain why the absorption lines appear dark rather than bright in the observed spectrum. (d) The astronomer claims that if she observed the same star's atmosphere in isolation (without the bright stellar surface behind it), she would see bright emission lines at the same wavelengths. Justify her claim.

Show answer

Point 1 (Part a): The observed wavelength (659.6 nm) is longer than the laboratory wavelength (656.3 nm), indicating a redshift. By the Doppler effect, Δλ/λ₀ = v/c, so v = c × (659.6 − 656.3)/656.3 = c × 3.3/656.3 ≈ 0.00503c ≈ 1.51 × 10⁶ m/s, directed away from the observer. Point 2 (Part b): The closely spaced doublet at 589.0 nm and 589.6 nm corresponds to the sodium D-lines. Each element has a unique spectral fingerprint; therefore, the presence of these lines indicates that sodium is present in the star's outer atmosphere. Point 3 (Part c): The star's hot, dense interior emits a continuous spectrum. The cooler, low-pressure gas in the star's outer atmosphere absorbs photons at wavelengths matching its atoms' energy-level transitions, removing those photons from the continuous background. This produces dark lines (Kirchhoff's third law). Point 4 (Part d): The same quantized energy levels that permit absorption also permit emission. If the gas is observed without a bright continuum source behind it, atoms that have been excited (by thermal collisions or other means) will de-excite and emit photons at the same characteristic wavelengths, producing a bright-line emission spectrum. This is Kirchhoff's second law.

Question 5: CRITICAL THINKING

4 pts

A group of students is investigating the emission spectrum of an unknown gas. They use a diffraction grating spectrometer to measure the wavelengths of emitted light from a discharge tube containing the gas. (a) Describe an experimental procedure the students could use to measure the wavelengths of the emission lines. Include what equipment is needed and what measurements should be taken. (b) The students measure the following emission line wavelengths: 410 nm, 434 nm, 486 nm, and 656 nm. They hypothesize that the gas is hydrogen. Describe how they could use the Bohr model energy-level equation to test this hypothesis. (c) Explain a potential source of systematic error in their measurements and how it would affect their results. (d) The students also notice faint spectral lines at wavelengths not predicted by the Bohr model for hydrogen. Propose a plausible explanation for these additional lines.

Show answer

Point 1 (Part a): Equipment: gas discharge tube with power supply, diffraction grating with known slit spacing d, meter stick or angular scale, and a dark room. Procedure: Position the diffraction grating in front of the discharge tube. For each bright emission line visible in the first-order diffraction pattern (m = 1), measure the angle θ between the central maximum and the line. Calculate the wavelength using d sin θ = mλ. Repeat for each visible line and record all wavelengths. Point 2 (Part b): For each measured wavelength, calculate the photon energy using E = hc/λ. Then, for the hypothesis that the gas is hydrogen, check whether each energy matches a predicted Bohr model transition: E = 13.6(1/

n_f

² − 1/

n_i

²) eV with

n_f

= 2 (Balmer series). For example, 656 nm gives E ≈ 1.89 eV; the Bohr model predicts 13.6(1/4 − 1/9) = 1.89 eV for n = 3 → 2. If all four lines match Balmer series predictions (

n_i

= 3, 4, 5, 6 to

n_f

= 2), the hypothesis is supported. Point 3 (Part c): A systematic error could arise from miscalibration of the angular scale or incorrect knowledge of the grating spacing d. If d is slightly larger than assumed, all calculated wavelengths (λ = d sin θ / m) would be systematically too large. This would cause all measured energies to be systematically too small, shifting all predicted quantum numbers and potentially leading the students to reject a correct hypothesis. Point 4 (Part d): The additional faint lines could be due to trace impurities in the discharge tube (other gases contaminating the sample). Alternatively, they could be second-order diffraction lines (m = 2) from the grating, appearing at different angles and thus different apparent wavelengths. A third possibility is that they represent fine-structure splitting not predicted by the simple Bohr model but arising from spin-orbit coupling effects described by quantum mechanics.

SUMMARY

Summary — Emission and Absorption Spectra

Summary

Atoms possess quantized energy levels described by the principal quantum number n. When an electron transitions from a higher to a lower energy state, the atom releases a photon whose energy equals the gap between levels: E = hf = hc/λ. This produces an emission spectrum — bright lines at discrete wavelengths on a dark background. The reverse process, absorption, occurs when a photon of the correct energy is absorbed by the atom, promoting the electron upward and creating dark lines in an otherwise continuous spectrum. For hydrogen, the Bohr model gives energy levels as E_n = −13.6 eV / n², predicting all observed spectral series.

The three types of spectra — continuous, emission, and absorption — are summarized by Kirchhoff's three laws. Each element's unique set of energy levels gives it a distinct spectral fingerprint, enabling chemical identification in laboratories and across the cosmos. On the AP Physics 2 exam, focus on calculating photon energies and wavelengths from energy-level diagrams, recognizing the complementary relationship between emission and absorption, and applying the Rydberg/Balmer framework to hydrogen.