Measure redshift from stellar spectral lines
Compare observed and rest wavelengths to estimate the recession speed of a distant source.
Section 25 · Astronomy and cosmology
A Level · Cambridge International Physics 9702, 2025–2027 syllabus. Topic 25 was checked against Cambridge’s 2028–2030 syllabus; a full reconciliation of every remaining syllabus-year change is still outstanding. Online study supports the theory but cannot replace supervised practical work, full timed assessment or human subject-teacher review.
Before you begin
- Use powers of ten, inverse-square relationships, proportional reasoning and SI units; distinguish wavelength, frequency, speed, flux and luminosity.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does a redshift tell you about a known spectral line?
What you will learn
- Identify wavelength increases in distant-object spectra
- Calculate redshift from a known spectral line
- Use the low-speed relation between wavelength shift and relative speed
Known spectral lines reveal a wavelength shift
Atoms and ions produce emission or absorption lines at known wavelengths in the laboratory. In spectra from distant objects, these lines can appear at longer wavelengths than their known rest values.
For a line with rest wavelength λ and observed wavelength λobs, define redshift z = Δλ/λ = (λobs − λ)/λ. A positive redshift means the line is shifted to a longer wavelength.
For small speeds, the fractional shift estimates relative speed
For motion along the line of sight at speeds much smaller than c, the magnitude of the fractional wavelength shift is approximately v/c: Δλ/λ ≈ v/c. Thus v ≈ cz.
For a receding source the wavelength increases and frequency decreases. These shifts have opposite signs; do not interpret the approximate relation as saying wavelength and frequency both increase together.
Pause and explain
For low speeds, which relation estimates recession speed from redshift?
Worked example
A spectral line has rest wavelength 600 nm and is observed at 606 nm. Estimate the source’s recession speed using c = 3.00 × 10⁸ m s⁻¹.
Show the worked solution
- Δλ = 606 − 600 = 6 nm.
- z = Δλ/λ = 6/600 = 0.010.
- v ≈ cz = 0.010 × 3.00 × 10⁸ = 3.0 × 10⁶ m s⁻¹.
Answer 3.0 × 10⁶ m s⁻¹ away from the observer.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A rest line at 500 nm is observed at 505 nm. Find z and estimate v using c = 3.0 × 10⁸ m s⁻¹.
Give me a hint
Use the rest wavelength in the denominator.
Compare my reasoning
- Δλ = 505 − 500 = 5 nm.
- z = 5/500 = 0.010.
- v ≈ cz = 3.0 × 10⁶ m s⁻¹.
z = 0.010 and v ≈ 3.0 × 10⁶ m s⁻¹ away.
Look for these in your work
- Finds the shift
- Uses rest wavelength
- Uses v ≈ cz
A distant object’s known absorption lines all appear at longer wavelengths than in the laboratory. What does this indicate, and what additional information would estimate its speed?
Give me a hint
Separate the qualitative shift from the quantitative calculation.
Compare my reasoning
- The lines are redshifted because their observed wavelengths are longer.
- For a small line-of-sight speed, measure a rest wavelength and its observed wavelength.
- Calculate z = Δλ/λ, then estimate v ≈ cz.
The source is receding in the low-speed model; measuring the fractional wavelength shift gives v ≈ cz.
Look for these in your work
- Identifies redshift
- Uses a known rest wavelength
- Connects the measured shift to speed
A line shifts from 400 nm to 404 nm. Calculate the fractional frequency change approximately and explain its sign for a receding source.
Give me a hint
For small shifts, the fractional frequency change has the opposite sign to the fractional wavelength change.
Compare my reasoning
- Δλ/λ = (404 − 400)/400 = 0.010.
- For a fixed wave speed, the fractional frequency shift is approximately −Δλ/λ.
- The frequency decreases by about 0.010, or 1.0%.
The frequency change is approximately −1.0%; the frequency decreases as the wavelength is redshifted.
Look for these in your work
- Calculates the fractional wavelength shift
- Uses the opposite sign for frequency
- Explains the decrease for redshift
Common mistakes
- Dividing the shift by the observed wavelength instead of the rest wavelength
- Calling a shorter observed wavelength a redshift
- Forgetting that the frequency shift has the opposite sign to the wavelength shift
What can you explain now?
What does a redshift tell you about a known spectral line?
Compare with the explanation
Its observed wavelength is longer than its rest wavelength
Redshift moves a line towards longer wavelength.
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Make it stick
For a receding source, why does wavelength increase while frequency decreases?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.