Line Spectra

Key idea: Distinguish emission and absorption line spectra, explain why spectra are discrete, and use ΔE = hf = hc/λ for transitions (A Level Physics).

GCE A-Level H2 Physics 2027

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Learning objectives

Analyse atomic energy levels and emission or absorption spectra.

1. Definitions (Must Know)

A. Emission line spectrum

An emission line spectrum is a set of discrete bright lines on a dark background, produced when excited atoms emit photons as electrons drop to lower energy levels.

B. Absorption line spectrum

An absorption line spectrum is a continuous spectrum with discrete dark lines removed, produced when atoms absorb photons of specific energies that match allowed transitions.

C. Transition energy

Photon energy in a transition is:

Δ E = hf = hc/λ

2. Key Ideas (What Earns Marks)

Scroll diagram horizontally to read all labels.

Absorption raises an electron only when the photon energy matches an allowed gap. Emission releases a photon whose energy equals the downward energy-level difference.

Line spectra exist because atomic energy levels are discrete.
Emission: photon energies correspond to downward transitions.
Absorption: only photons with exactly the right energies are absorbed (matching energy gaps).
Emission and absorption line positions match for the same gas (same energy gaps).

This lesson targets learning outcomes 19k–19m: discrete levels, emission vs absorption spectra, and photon energies in transitions.

3. Detailed Explanations

A. Emission spectra (how they form)

If a low-pressure gas is excited (e.g. in a discharge tube), electrons in atoms can move to higher energy levels.

When they return to lower levels, they emit photons with energies equal to the differences between levels. Because only certain energy differences exist, only certain photon frequencies/wavelengths appear.

Hydrogen's prominent visible Balmer lines are labelled with approximate wavelengths. Iron illustrates a much denser element-specific line pattern; absorption removes selected wavelengths from a continuous spectrum.

Hydrogen line data used in the figure

Approximate wavelength	Visible colour region
410.2 nm	violet
434.0 nm	violet-blue
486.1 nm	blue-green
656.3 nm	red

B. Absorption spectra (how they form)

If white light passes through a cooler gas, atoms absorb photons whose energies match allowed upward transitions.

Those wavelengths are missing from the transmitted light, so dark lines appear.

The bottom row of the comparison figure shows the corresponding absorption idea: dark lines appear where photons are absorbed for upward transitions.

C. Emission vs absorption (comparison)

Feature	Emission spectrum	Absorption spectrum
Background	dark	continuous
Lines	bright	dark
Process	electron drops and emits photon	electron absorbs photon and rises
Photon energies	equal to energy gaps	equal to the same energy gaps

4. Common Mistakes

Saying “any photon can be absorbed” (only photons matching energy gaps are absorbed).
Mixing up emission and absorption diagrams (remember: bright vs dark lines).
Forgetting Δ E = hf applies to each transition.

5. Exam Tips

If asked “why lines are discrete”, say: “energy levels are discrete so only certain ΔE exist”.
If asked for a photon wavelength, use: λ = hc/(Δ E) and keep units consistent.

6. Worked Examples

Problem

A spectrum shows a continuous rainbow background crossed by thin dark lines. Identify the spectrum and explain how it forms.

Study the worked solution

Read the visual evidence
Method
It is an absorption line spectrum.
Reason
A continuous background with selected dark wavelengths is the defining appearance of absorption.
Working
continuous background + dark lines → absorption
Explain the missing wavelengths
Method
Atoms absorb photons whose energies match allowed upward transitions.
Reason
Discrete atomic energy levels permit only particular energy gaps.
Working
Δ E = hf = hc/λ

Problem

An atomic transition has energy difference

Δ E = 2.0 eV. Estimate the photon wavelength using

hc = 1240 eV nm.

Try this before viewing the solution

Photon wavelength

Unit: nm

Hints

Hint 1: match the units

Because both

Δ E and

hc use eV-based units, use

λ = hc/Δ E directly in nanometres.

View solution step by step

Match photon and transition energy
Method
Δ E = hc/λ.
Reason
An allowed transition emits or absorbs one photon with the matching energy.
Working
λ = hc/(Δ E)
Evaluate
Method
λ = 620 nm.
Reason
The supplied hc unit returns wavelength in nanometres.
Working
λ = 1240/2.0 nm = 620 nm

Learner claim

Hydrogen emits a red line of wavelength

656 nm. A learner substitutes

656 directly into

Δ E = hc/λ while using SI values of

h and

c. Diagnose the error and find the photon energy. Take

h = 6.63 × 10⁻³⁴ J s and

c = 3.00 × 10⁸ m s⁻¹.

Try this before viewing the solution

Photon energy

Unit: J

View solution step by step

Repair the wavelength unit
Method
656 nm = 656 × 10⁻⁹ m.
Reason
SI values for h and c require wavelength in metres.
Working
λ = 6.56 × 10⁻⁷ m
Calculate photon energy
Method
Δ E = 3.03 × 10⁻¹⁹ J.
Reason
The energy of the emitted photon equals the downward level difference.
Working
Δ E = ((6.63 × 10⁻³⁴)(3.00 × 10⁸))/(656 × 10⁻⁹) = 3.03 × 10⁻¹⁹ J

Examination question

An atom emits a photon of energy

Δ E = 3.0 × 10⁻¹⁹ J. Find the photon frequency. Take

h = 6.63 × 10⁻³⁴ J s. [2 marks]

Try this before viewing the solution

Frequency calculation

View solution step by step

Use photon energy
1 mark
Method
f = Δ E/h.
Reason
The emitted photon carries the transition energy.
Working
f = (3.0 × 10⁻¹⁹)/(6.63 × 10⁻³⁴)
Evaluate
1 mark
Method
f = 4.5 × 10¹⁴ Hz.
Reason
Energy divided by J s gives s⁻¹.
Working
f = 4.5 × 10¹⁴ Hz

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark the photon relation and frequency.

Mark points

1 mark: Uses ΔE = hf or equivalent.
1 mark: Obtains 4.5 × 10¹⁴ Hz.

Independent transfer

An atom has energy levels at

0 eV,

2.0 eV and

3.5 eV. The atom is initially in the ground state. Determine which photon wavelengths it can absorb. Use

hc = 1240 eV nm.

Try this before viewing the solution

Allowed transitions and wavelengths

Hints

Hint 1: start from the occupied level

Only upward transitions beginning at the stated initial level are available.

View solution step by step

Select allowed gaps
Method
The available upward gaps are 2.0 eV and 3.5 eV.
Reason
The atom begins at 0 eV; the 2.0 → 3.5 eV gap is not initially accessible.
Working
0 → 2.0 eV and 0 → 3.5 eV
Convert the first gap
Method
The 2.0 eV gap absorbs 620 nm photons.
Reason
Photon energy must match the gap exactly in the idealised isolated-atom model.
Working
λ = 1240/2.0 = 620 nm
Convert the second gap
Method
The 3.5 eV gap absorbs 354 nm photons.
Reason
The larger energy gap corresponds to the shorter wavelength.
Working
λ = 1240/3.5 = 354 nm

Next step

Attempt why low pressure sharpens line spectra independently.

7. Mind Stretchers

Mind stretcher 1: Why low pressure helps line spectraExtension

Suggest why low-pressure gases give sharper line spectra than high-pressure gases.

Show Answer

At low pressure, atoms collide less often, so emitted/absorbed photons are less disturbed and lines are sharper.

At higher pressure, frequent collisions broaden and smear the lines.

Mind stretcher 2: Why do emission and absorption lines “match”?Extension

Explain why the wavelengths of emission lines for a gas match the wavelengths of the dark absorption lines for the same gas.

Show Answer

Both processes involve the same discrete energy gaps between atomic levels.

Emission occurs when electrons drop and emit photons of energy Δ E, while absorption occurs when electrons rise by absorbing photons of the same Δ E, so the same photon wavelengths appear.

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H2 Physics
Edition: GCE A-Level H2 Physics 2027

Line Spectra

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Learning objectives

1. Definitions (Must Know)

A. Emission line spectrum

B. Absorption line spectrum

C. Transition energy

2. Key Ideas (What Earns Marks)

3. Detailed Explanations

A. Emission spectra (how they form)

B. Absorption spectra (how they form)

C. Emission vs absorption (comparison)

4. Common Mistakes

5. Exam Tips

6. Worked Examples

Emission vs absorption identification

Read the visual evidence

Explain the missing wavelengths

Photon wavelength from energy gap

Match photon and transition energy

Evaluate

Photon energy from wavelength (SI units)

Repair the wavelength unit

Calculate photon energy

Frequency from transition energy

Use photon energy

Evaluate

Which photons can be absorbed? (energy levels)

Select allowed gaps

Convert the first gap

Convert the second gap

7. Mind Stretchers

Mind stretcher 1: Why low pressure helps line spectraExtension

Mind stretcher 2: Why do emission and absorption lines “match”?Extension