Light from most distant galaxies has longer wavelengths than it had when emitted. This is redshift, and it is evidence that the galaxies are moving away from us. Where a question supplies a relationship, your job is to choose the right quantities, keep the units consistent, and say what the answer means.
This lesson sits in space physics and uses the wave ideas in wave behaviour.
What relationships might be supplied?
Two relationships are typical in this topic. Your paper states which ones are given, so check the Cambridge syllabus and the question.
- Change in wavelength and speed: Δλ ÷ λ = v ÷ c, where Δλ is the increase in wavelength, λ is the emitted wavelength, v is the speed of the galaxy away from us and c is the speed of light, 3.0 × 108 m/s. This holds when v is much smaller than c.
- A Hubble-type relationship: v = H0 × d, where d is the distance and H0 is a constant given in the question.
Steps for any supplied relationship:
- Write the relationship, then list each symbol with its value and unit.
- Convert units so they match, for example nanometres with nanometres.
- Substitute and calculate.
- Give the answer with a unit and a sensible number of significant figures.
- Say what the result means.
Worked example
All values below are invented for practice.
A spectral line has emitted wavelength 656.0 nm. From a distant galaxy, it is observed at 660.0 nm.
The constant H0 is given as 2.2 × 10-18 per second. Find the speed of the galaxy and its distance.
Step 1, change in wavelength: Δλ = 660.0 − 656.0 = 4.0 nm.
Step 2, ratio: Δλ ÷ λ = 4.0 ÷ 656.0 = 0.00610. Both wavelengths are in nm, so the units cancel.
Step 3, speed: v = 0.00610 × 3.0 × 108 = 1.83 × 106 m/s, about 1.8 × 106 m/s.
Step 4, distance: d = v ÷ H0 = 1.83 × 106 ÷ 2.2 × 10-18 = 8.3 × 1023 m.
Check the unit: m/s divided by 1/s gives metres.
Meaning: the wavelength increased, so the galaxy is moving away. In light-years, using 9.5 × 1015 m per light-year, the distance is about 8.8 × 107 light-years.
The mistake to watch for
A common slip is to use the observed wavelength as if it were the change.
Mistaken answer: Δλ ÷ λ = 660.0 ÷ 656.0 = 1.006, so v is about 3.0 × 108 m/s.
That would make the galaxy move at the speed of light. A result that large is a signal to re-check the step.
The correction: Δλ is the difference between observed and emitted wavelength, 4.0 nm. Always subtract first, then divide by the emitted wavelength. A second slip is mixing units, such as putting one wavelength in nm and the other in m.
Check yourself
Use c = 3.0 × 108 m/s and H0 = 2.2 × 10-18 per second, both supplied here.
1. Emitted wavelength 500 nm, observed 505 nm. Find the speed of the source.
Show answer
Δλ = 5 nm. Ratio = 5 ÷ 500 = 0.010. v = 0.010 × 3.0 × 108 = 3.0 × 106 m/s.
2. A galaxy moves away at 1.5 × 106 m/s. Find its distance using v = H0d.
Show answer
d = 1.5 × 106 ÷ 2.2 × 10-18 = 6.8 × 1023 m, so about 6.8 × 1023 m.
3. Two galaxies have different redshifts. The one with the larger redshift is moving away faster. Which is likely to be further from us, and why?
Show answer
The one with the larger redshift is further away. Observations show that galaxies with greater redshift are at greater distances, which is what the relationship v = H0d describes.
Where this leads next
The next step is to separate what is observed from what is concluded, in distinguishing observational evidence from a model conclusion. The bounds and rounding explainer helps you decide how many significant figures an answer built from given data deserves.
If substitution and units keep costing marks, online one-to-one Physics tuition allows a teacher to watch your working as you do it.