A spectral line at 500\ nm is redshifted by 10\ nm. Find the recession speed (c = 3.0 × 10^8\ m s^-1). [2 marks]

WJEC Eduqas Eduqas 2020-style practice: A spectral line of laboratory wavelength 656\ nm is observed from a distant galaxy at 689\ nm. Calculate the galaxy's recession speed. Take c = 3.0 × 10^8\ m s^-1.

Change in wavelength: λ = 689 - 656 = 33\ nm. Doppler redshift for v c: λλ = vc, so v = cλλ. v = (3.0 × 10^8)33656 = (3.0 × 10^8)(0.0503) = 1.5 × 10^7\ m s^-1. Markers reward λ = 33\ nm, (λ)/(λ) = (v)/(c), and the recession speed about 1.5 × 10^7\ m s^-1.

EnglandPhysicsSyllabus dot point

What evidence shows the universe is expanding, and how does Hubble's law lead to the Big Bang?

Orbits and the wider universe: the Doppler effect and redshift, Hubble's law, the age of the universe, and the evidence for the Big Bang including the cosmic microwave background.

A focused answer to the Eduqas A-Level Physics Component 2 cosmology content, covering the Doppler effect and the redshift of galaxies, Hubble's law, estimating the age of the universe, and the evidence for the Big Bang including the cosmic microwave background radiation.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Eduqas wants you to explain the Doppler effect and redshift, state and use Hubble's law, estimate the age of the universe from the Hubble constant, and describe the evidence for the Big Bang, including the cosmic microwave background radiation.

The answer

The Doppler effect and redshift

Galactic redshift and Hubble's law

The age of the universe

Evidence for the Big Bang

Examples in context

Redshift and Hubble's law are the foundations of modern cosmology, giving the scale, age and expansion history of the universe. The same Doppler reasoning measures the orbital speeds of binary stars, detects exoplanets from the wobble of their host stars, and is used in radar speed cameras and medical Doppler ultrasound. The cosmic microwave background, mapped in fine detail by space telescopes, is the strongest single piece of evidence for the Big Bang.

Try this

Q1. State Hubble's law. [1 mark]

Cue. The recession speed of a galaxy is proportional to its distance, $v = H_0 d$ .

Q2. A spectral line at $500\ \text{nm}$ is redshifted by $10\ \text{nm}$ . Find the recession speed ( $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$ ). [2 marks]

Cue. $v = c\frac{\Delta\lambda}{\lambda} = (3.0 \times 10^{8})\frac{10}{500} = 6.0 \times 10^{6}\ \text{m s}^{-1}$ .

Q3. State one piece of evidence for the Big Bang. [1 mark]

Cue. The cosmic microwave background radiation (or the observed hydrogen-to-helium abundance).

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20204 marksA spectral line of laboratory wavelength

656\ \text{nm}

is observed from a distant galaxy at

689\ \text{nm}

. Calculate the galaxy's recession speed. Take

c = 3.0 \times 10^{8}\ \text{m s}^{-1}

Show worked answer →

Change in wavelength: $\Delta\lambda = 689 - 656 = 33\ \text{nm}$ .

Doppler redshift for $v \ll c$ : $\dfrac{\Delta\lambda}{\lambda} = \dfrac{v}{c}$ , so $v = c\dfrac{\Delta\lambda}{\lambda}$ .

$v = (3.0 \times 10^{8})\dfrac{33}{656} = (3.0 \times 10^{8})(0.0503) = 1.5 \times 10^{7}\ \text{m s}^{-1}$ .

Markers reward $\Delta\lambda = 33\ \text{nm}$ , $\frac{\Delta\lambda}{\lambda} = \frac{v}{c}$ , and the recession speed about $1.5 \times 10^{7}\ \text{m s}^{-1}$ .

Eduqas 20224 marksState Hubble's law and use a Hubble constant of

2.3 \times 10^{-18}\ \text{s}^{-1}

to estimate the age of the universe in years. (1 year =

3.15 \times 10^{7}\ \text{s}

Show worked answer →

Hubble's law: the recession speed of a galaxy is proportional to its distance, $v = H_0 d$ , where $H_0$ is the Hubble constant.

The age of the universe is estimated as the reciprocal of the Hubble constant (assuming constant expansion): $t \approx \dfrac{1}{H_0} = \dfrac{1}{2.3 \times 10^{-18}} = 4.35 \times 10^{17}\ \text{s}$ .

In years: $t = \dfrac{4.35 \times 10^{17}}{3.15 \times 10^{7}} = 1.4 \times 10^{10}\ \text{years}$ , about 14 billion years.

Markers reward stating $v = H_0 d$ , $t \approx \frac{1}{H_0}$ , and the age about $1.4 \times 10^{10}$ years.

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)