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IB Physics Standard Level · Topic C.5

C.5 Doppler effect

48 free, high-quality past-paper-style questions on Doppler effect for IB Physics SL, each with a full mark scheme. Covers: Wavefront diagrams; Doppler for light; Red shift & blue shift; and more. For when you have finished the past papers, or want to practise one topic at a time.C.5 at Higher Level →Read the C.5 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.258 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1C-1A-43Wavefront diagrams Paper 1AEasy1 mark

The diagram shows the wavefronts emitted by a source of sound that is moving to the right, towards observer A and away from observer B.

Which row describes what observers A and B hear, compared with the sound emitted?

Wavefront diagram for a source moving to the right: the wavefronts bunch up ahead of it and spread out behind it.sourceABthe source is moving towards observer A
Diagram NOT accurately drawn
Observer AObserver B
Show mark scheme
Marking pointMarkNotes
Step 1The source moves right between emitting successive wavefronts, so the wavefronts are closer together ahead of it (towards A) and further apart behind it (towards B).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2A therefore receives more wavefronts per second (higher frequency) and B fewer (lower frequency); the speed of sound in the air is unchanged.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe two observers are the wrong way round: the wavefronts are closer together on the side the source is moving towards.
  • BOnly the observer being approached hears a higher frequency; the observer being left behind hears a lower one.
  • CThe speed of sound is set by the medium and is the same in both directions; it is the wavelength and frequency that change.
  • DCorrect: the wavefronts are bunched ahead of the source, so A hears a higher frequency, and spread out behind it, so B hears a lower one.
2C-1A-90Doppler for light Paper 1AEasy1 mark

A distant star moves relative to the Earth with a speed much less than the speed of light c. Light from the star is observed on the Earth.

Which statement about the light received on the Earth is correct?

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Marking pointMarkNotes
Step 1The speed of light is c for every observer; the Doppler effect changes the wavelength and frequency, not the speed.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The shift depends on the relative velocity of the star and the Earth along the line joining them (Δλ/λ ≈ v/c); a velocity at right angles to that line gives no shift, so B is correct.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • ASpeeds are not added for light: the light arrives at c whatever the motion of the star. It is the wavelength that changes.
  • BCorrect: only the relative velocity along the line of sight produces a Doppler shift, so the line-of-sight component is what is measured.
  • CThis takes 'red shift' literally. Each line moves to a longer wavelength by the same small fraction v/c; a blue line stays blue.
  • DFor light only the relative velocity of the source and the observer matters, so the Earth moving towards the star gives a blue shift too.
3IBP2S-026-01-01Wavefront diagrams Paper 1AEasy1 mark

An ambulance with its siren sounding travels at constant speed along a straight road. A stationary pedestrian stands at the very edge of the road, so close that the ambulance moves effectively directly towards her, passes her and then moves directly away from her. The air is still. Which describes the frequency the pedestrian hears?

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Marking pointMarkNotes
The Doppler shift depends on the source's velocity towards or away from the observer, not on its distance. The ambulance moves directly towards the pedestrian at constant speed. Each wavefront is emitted closer to her than the last by a fixed amount, so the heard frequency f' = fv/(v − vs) is constant and higher than the emitted frequency. Once the ambulance has passed, it moves directly away at the same speed. The wavefronts are then stretched by a fixed amount, so f' = fv/(v + vs) is constant and lower than emitted. Option A mistakes the rising loudness for rising pitch. Option C ignores the Doppler effect. Option D reverses the direction of the shift.✓ 1Answer B

Answer: B

Common error: A source gets louder as it comes closer, but that is not the same as its frequency increasing. For a source moving directly towards or away from you at constant speed, the shifted frequency is steady.

4IBP2S-026-01-02Red shift & blue shift Paper 1AEasy1 mark

A line in the spectrum of a distant galaxy is observed at a longer wavelength than the same line measured in a laboratory on Earth. What can be deduced?

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Marking pointMarkNotes
A longer observed wavelength is a red shift. Light from a receding source is stretched, so the galaxy is moving away from Earth. Temperature changes the relative intensity of lines, not their positions. The speed of light in a vacuum is the same for all observers.✓ 1Answer B

Answer: B

Exam tip: Red shift means a longer wavelength and a lower frequency, so the source is receding. Blue shift means a shorter wavelength, so the source is approaching.

5IBP2S-026-01-06Doppler effect for sound Paper 1AEasy1 mark

A drone carrying a buzzer flies in still air directly towards a stationary microphone. Compared with measurements made when the drone is stationary, which quantity measured at the microphone is unchanged?

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Marking pointMarkNotes
The speed of sound depends only on the medium (the air), not on the motion of the source. The motion of the drone compresses the wavefronts, so the wavelength decreases. The frequency therefore increases and the period decreases.✓ 1Answer C

Answer: C

Exam tip: For sound, the wave speed is fixed by the medium. A moving source changes λ, and f then follows from f = v/λ.

6IBP2S-026-00-01Wavefront diagrams Paper 1AEasy1 mark

A small loudspeaker moves at constant speed through still air while emitting sound of constant frequency. Viewed from above, the circular wavefronts are crowded close together on the north side of the loudspeaker and are widely spaced on the south side. Which statement is correct?

NS
Wavefronts of the sound emitted by the loudspeaker (viewed from above)
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Marking pointMarkNotes
Each wavefront spreads from the point where the source was when it was emitted. A source moving north catches up with the wavefronts ahead of it, so they bunch together on the north side. The wave speed in air is unchanged, so the shorter wavelength to the north means more wavefronts pass a northern observer each second: a higher frequency. B reverses the motion. C and D wrongly assume that the motion of the source changes the wave speed.✓ 1Answer A

Answer: A

Common error: The wave speed depends only on the medium, not on the motion of the source. Bunched wavefronts mean a shorter wavelength, not faster waves.

7IBP2S-026-00-02Red shift & blue shift Paper 1AEasy1 mark

A hydrogen spectral line has a wavelength of 486 nm when measured in a laboratory. In the spectrum of a distant galaxy, the same line is observed at 492 nm. What can be concluded?

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Marking pointMarkNotes
The observed wavelength is longer than the laboratory wavelength, so the light is red-shifted. A red-shift means the source and observer are moving apart, so the galaxy is receding. B confuses red-shift with blue-shift. C is wrong because the speed of light in a vacuum is the same for all observers. D is wrong because temperature does not shift the position of a particular spectral line in this way.✓ 1Answer A

Answer: A

Exam tip: Identify the same line in both spectra and compare its wavelengths. Longer observed wavelength means red-shift, which means receding.

8IBP2S-026-00-05Wavefront diagrams Paper 1AEasy1 mark

A cyclist rides directly towards a stationary siren that emits sound of constant frequency. The air is still. Which quantity, as measured by the cyclist, has the same value as it would if the cyclist were at rest?

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Marking pointMarkNotes
The source is at rest in the air, so the wavefronts are evenly spaced and the wavelength in the air is unchanged. The moving cyclist meets the wavefronts at a greater relative speed (C changes). The cyclist therefore meets more wavefronts per second, so the detected frequency rises (A changes) and the period falls (D changes).✓ 1Answer B

Answer: B

Common error: For a moving observer, the wavelength in the medium does not change. The frequency changes because the speed of the waves relative to the observer changes.

9IBP2S-026-00-10Doppler for light Paper 1AEasy1 mark

The Sun rotates about an axis roughly perpendicular to the line of sight from Earth. Light is collected separately from its two opposite edges (limbs) on the solar equator. Which statement is correct?

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Marking pointMarkNotes
At the limbs, the rotational velocity points along the line of sight. One limb approaches Earth, so its light is blue-shifted. The other limb recedes, so its light is red-shifted. C overlooks that at the limbs the motion is towards or away from the observer. B confuses the direction the light travels with the motion of the source. D is wrong because the speed of light is the same for all observers.✓ 1Answer A

Answer: A

Exam tip: Only the component of velocity along the line of sight produces a Doppler shift. At the limbs of a rotating body, this component is at its largest.

10C-1A-40Doppler for light Paper 1AMedium1 mark

Hydrogen atoms in a distant gas cloud emit radio waves of frequency 1420.4 MHz. The cloud is moving directly away from the Earth at 2.1 × 105 m s−1.

What is the frequency of these radio waves when they are received on the Earth?

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Marking pointMarkNotes
Step 1Δf = f v/c = 1420.4 × (2.1 × 105/3.00 × 108) = 0.99 MHz.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The cloud is receding, so the radiation is red-shifted: its wavelength increases and its frequency decreases, giving 1420.4 − 1.0 = 1419.4 MHz.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AAdds the shift: assumes that a red shift raises the frequency as well as the wavelength.
  • BAssumes that the frequency of electromagnetic waves cannot change and only the wavelength is shifted.
  • CCorrect: Δf/f ≈ v/c gives a shift of about 1.0 MHz, and a receding source gives a lower frequency.
  • DDoubles the shift (uses Δf = 2fv/c, a factor of 2 that Δf/f ≈ v/c does not contain), giving 1420.4 − 2.0 = 1418.4 MHz.
11C-1A-42Doppler effect for sound Paper 1AMedium1 mark

A drone emitting a steady note flies at constant velocity in a straight line directly towards a stationary listener, passes just above her, and then flies directly away from her.

Ignoring the short interval as the drone passes overhead, which row describes the frequency she hears?

While the drone approachesWhile the drone moves away
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Marking pointMarkNotes
Step 1Ahead of a moving source the wavefronts are closer together and behind it they are further apart; at constant velocity along the line to the listener this spacing does not change, so the frequency heard is constant: higher while approaching, lower while receding.✓ 1Answer C

Answer: C  ·  1 stage of work, one mark

Every option, and why

  • AThis confuses pitch with loudness: the sound gets louder as the drone comes closer, but at constant velocity the frequency heard stays constant.
  • BThe speed of sound in the air is unchanged, but the motion of the source changes the spacing of the wavefronts, and so the frequency heard.
  • CCorrect: a constant approach velocity gives a steady higher frequency, and a constant recession velocity gives a steady lower frequency.
  • DThis is reversed: the wavefronts are bunched up ahead of the drone, so the frequency is higher while it approaches.
12C-1A-58Wavefront diagrams Paper 1AMedium1 mark

A loudspeaker S emits a note of constant frequency in still air. A cyclist O rides at constant speed directly towards S. The diagram shows some of the wavefronts emitted by S.

Which row describes the wavelength of the sound in the air between S and O, and the frequency heard by the cyclist, compared with those for a stationary listener?

Equally spaced circular wavefronts around a stationary loudspeaker S, with a cyclist O to the right moving towards SSOcyclist moving towards Swavefronts from S (still air)
Diagram NOT to scale
Wavelength in the airFrequency heard
Show mark scheme
Marking pointMarkNotes
Step 1The source is stationary, so the wavefronts are equally spaced circles: the wavelength in the air is unchanged.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Moving towards S, the cyclist meets successive wavefronts at a greater rate than a stationary listener, so the frequency heard is higher.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis treats the moving observer like a moving source. Only a moving source changes the spacing of the wavefronts in the air.
  • BThe wavelength is right, but moving towards the source means meeting more wavefronts per second, not fewer.
  • CCorrect: the wavefront spacing is set by the stationary source, but the cyclist's motion towards S makes the wavefronts arrive more often.
  • DThe Doppler effect also occurs when only the observer moves: the frequency heard depends on how often the wavefronts reach the observer.
13C-1A-91Wavefront diagrams Paper 1AMedium1 mark

A small source emits sound of constant frequency while moving along a straight line in still air. The diagram, drawn to scale, shows four wavefronts emitted at equal time intervals, seen from above at one instant. The small dots show where the source was when each wavefront was emitted and S is the position of the source at this instant.

Which row describes the motion of the source?

Plan view of four circular wavefronts whose centres lie on a horizontal line; the centres get closer together towards the left, where the source S isS• position of the source when each wavefront was emitted
Diagram drawn to scale
Direction of motionSpeed of the source
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Marking pointMarkNotes
Step 1The largest wavefront was emitted first. Its centre is on the right and S is on the left, so the source moves to the left (the wavefronts are also closest together on the left).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2In equal time intervals the source moves 0.85, 0.50, 0.30 and then 0.20 of the distance the sound travels: the dots get closer together towards S, so the speed of the source is decreasing.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • ABoth readings are reversed: the source moves towards the side where the wavefronts are bunched, and the dots get closer together towards S, which means it is slowing down.
  • BThis takes the source to move towards the side where the wavefronts are further apart. The oldest, largest wavefront is centred on the right, so the source is moving to the left.
  • CThe direction is correct, but the distance between successive dots decreases towards S: the source covers less distance in each equal time interval, so it is slowing down.
  • DCorrect: the source moves to the left (the oldest wavefront is centred furthest right) and the decreasing spacing of the dots shows its speed decreasing.
14C-1A-92Red shift & blue shift Paper 1AMedium1 mark

An observer on the Earth's equator measures the wavelength of an absorption line in the spectrum of sunlight. On the day of the measurements the Sun rises due east, is directly overhead at noon and sets due west. The Earth rotates from west to east. Ignore the orbital motion of the Earth.

Which row compares the wavelength measured at sunrise and at sunset with the wavelength measured at noon?

At sunriseAt sunset
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Marking pointMarkNotes
Step 1The rotation carries the observer eastwards at about 460 m s−1. At noon this velocity is at right angles to the direction of the Sun, so it gives no shift: the noon value is the reference.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2At sunrise the Sun is due east, so the observer moves directly towards it: the line is blue-shifted to a shorter wavelength. At sunset the Sun is due west, so the observer moves away from it: the line is red-shifted to a longer wavelength (Δλ/λ ≈ 1.5 × 10−6 in each case).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the observer approaches the Sun at sunrise (shorter wavelength) and recedes from it at sunset (longer wavelength).
  • BThis reverses the direction of rotation: an eastward-moving observer approaches the rising Sun in the east.
  • CThis confuses the reddening of the Sun near the horizon, caused by scattering in the atmosphere, with a Doppler red shift. Scattering does not change the wavelength of a spectral line.
  • DThis assumes that only motion of the source causes a Doppler shift. For light the relative velocity of the source and the observer is what matters, and the observer is moving.
15IBP2S-026-01-03Red shift & blue shift Paper 1AMedium1 mark

The hydrogen-alpha line has a laboratory wavelength of 656.28 nm. In the spectrum of a star it is observed at 656.06 nm. The speed of light is 3.00 × 108 m s−1. What is the velocity of the star along the line of sight?

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Marking pointMarkNotes
The shift is Δλ = 656.28 − 656.06 = 0.22 nm. Using v ≈ cΔλ/λ gives v = 3.00×108 × 0.22/656.28 ≈ 1.0×105 m s−1 = 1.0×102 km s−1. The observed wavelength is shorter than the laboratory value. This is a blue shift, so the star is approaching Earth, and A is correct. Option B gives the right speed but the wrong direction. Options C and D wrongly halve or double the shift, as though the light had been reflected.✓ 1Answer A

Answer: A

Common error: Always check the sign of the shift before stating a direction. A shorter observed wavelength means the source is approaching.

16IBP2S-026-01-07Red shift & blue shift Paper 1AMedium1 mark

A spiral galaxy is seen edge-on from Earth. Its centre recedes from Earth at 600 km s−1. Stars at its two opposite edges orbit the centre at 200 km s−1, moving along the line of sight. How do spectral lines from the two edges appear?

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Marking pointMarkNotes
The line-of-sight velocities of the two edges are 600 + 200 = 800 km s−1 away and 600 − 200 = 400 km s−1 away. Both edges are receding, so both are red-shifted, but by different amounts. Option A would be correct only if the rotation speed were greater than the recession speed.✓ 1Answer B

Answer: B

Exam tip: Add the velocities along the line of sight first. The sign of the total velocity decides whether the light is red-shifted or blue-shifted.

17IBP2S-026-01-09Red shift & blue shift Paper 1AMedium1 mark

A star emits two spectral lines with laboratory wavelengths of 400 nm and 600 nm. The star recedes from Earth at a speed much less than c. How does the wavelength shift Δλ of the 600 nm line compare with that of the 400 nm line?

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Marking pointMarkNotes
From Δλ/λ ≈ v/c, the fractional shift is the same for every line. Therefore Δλ is proportional to λ, and the ratio of the shifts is 600/400 = 1.5. Option A treats the shift as a fixed number of nanometres. Option D wrongly squares the ratio.✓ 1Answer B

Answer: B

Common error: A Doppler shift is a fractional change. Lines at longer wavelengths move by more nanometres.

18IBP2S-026-00-03Doppler radar Paper 1AMedium1 mark

A traffic radar unit emits microwaves of frequency 24 GHz. The microwaves reflect from a car travelling directly towards the unit at 30 m s−1. What is the difference between the emitted frequency and the frequency of the reflected microwaves received back at the unit?

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Marking pointMarkNotes
The car first acts as a moving observer and receives a shifted frequency. It then re-emits this frequency as a moving source, so the frequency is shifted a second time. The total shift is Δf ≈ 2fv/c = 2 × 24×109 × 30 / 3.0×108 = 4.8×103 Hz. A uses a single shift fv/c. C counts the factor 2 twice. D contains a power-of-ten slip.✓ 1Answer B

Answer: B

Common error: For a wave reflected from a moving object, the Doppler shift happens twice. Leaving out the factor 2 halves the answer.

19IBP2S-026-00-04Red shift & blue shift Paper 1AMedium1 mark

A star moves in a circular orbit in a binary system whose orbital plane contains the line of sight from Earth. The centre of mass of the system is at rest relative to Earth. A spectral line with a laboratory wavelength of 600.0 nm is observed to vary steadily between 599.8 nm and 600.2 nm. What is the orbital speed of the star?

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Marking pointMarkNotes
The largest shift from the rest wavelength is Δλ = 0.2 nm. This shift occurs when the star moves directly towards or directly away from Earth, so the full orbital speed is then along the line of sight. Then v = cΔλ/λ = 3.0×108 × 0.2/600.0 = 1.0×105 m s−1 = 100 km s−1.
A halves the shift (0.1 nm).
C uses the full range of 0.4 nm as the shift.
D doubles the full range (0.8 nm).
✓ 1Answer B

Answer: B

Exam tip: Measure the Doppler shift from the rest wavelength, not across the whole range between the extreme values.

20IBP2S-026-00-07Doppler effect for sound Paper 1AMedium1 mark

A medical Doppler ultrasound probe directs ultrasound along a blood vessel. The ultrasound reflects from blood cells moving away from the probe. The frequency of the reflected ultrasound differs from the emitted frequency by Δf. The probe frequency is changed from 4 MHz to 8 MHz, and the blood speed falls to half its original value. What is the new frequency difference?

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Marking pointMarkNotes
The shift for a reflected wave is Δf ≈ 2fv/c, so Δf is proportional to the product fv. Doubling f and halving v leaves fv unchanged, so the shift is still Δf. A comes from treating the shift as proportional to v/f. C considers only the change in frequency. D treats both changes as increases.✓ 1Answer B

Answer: B

Exam tip: Write down the proportionality Δf ∝ fv before scaling. This avoids errors when two quantities change at once.

21IBP2S-026-00-09Wavefront diagrams Paper 1AMedium1 mark

A source moves at constant speed u through a medium in which the wave speed is v. Directly ahead of the source, the spacing of the wavefronts is half the spacing directly behind it. What is u?

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Marking pointMarkNotes
In one period T, a wavefront travels vT while the source travels uT. So λahead = (v − u)T and λbehind = (v + u)T. Setting (v − u)/(v + u) = 1/2 gives 2v − 2u = v + u, so u = v/3. C compares the wavelength ahead with the wavelength from a stationary source, not with the wavelength behind. A and D come from algebraic slips.✓ 1Answer B

Answer: B

Exam tip: Think about how far a wavefront and the source each move in one period. This gives both wavelengths directly.

22C-1A-41Red shift & blue shift Paper 1AHard1 mark

Two absorption lines in the spectrum of a galaxy have laboratory wavelengths of 420 nm and 630 nm. The 420 nm line is observed to be red-shifted by Δλ.

What is (red shift of the 630 nm line)/Δλ?

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Marking pointMarkNotes
Step 1Δλ/λ ≈ v/c is the same for every line in the light from the same galaxy, so the shift Δλ ∝ λ.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The ratio of the shifts is 630/420 = 1.5.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis inverts the ratio of the wavelengths; the longer wavelength has the larger shift.
  • BThis assumes every line is shifted by the same number of nanometres; it is the fractional shift Δλ/λ that is the same.
  • CCorrect: Δλ ∝ λ at a given speed, so the ratio is 630/420 = 1.5.
  • DThis squares the ratio of the wavelengths; the shift is directly proportional to the wavelength.
23C-1A-44Red shift & blue shift Paper 1AHard1 mark

A star moves in a circular orbit in a binary star system whose orbit is seen edge-on from the Earth. The binary system as a whole is not moving relative to the Earth. A line of laboratory wavelength λ0 in the star's spectrum varies periodically between λ0 − δ and λ0 + δ.

What is the orbital speed of the star?

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Marking pointMarkNotes
Step 1The line is shifted most, by δ either side of λ0, when the star moves directly towards or directly away from the Earth.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Δλ/λ ≈ v/c gives v = cδ/λ0.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis treats δ as the full range of the variation; the full range is 2δ, and the shift either side of λ0 is δ.
  • BCorrect: the maximum shift from λ0 is δ, so v = cδ/λ0.
  • CThis uses the full range, 2δ, as the shift; the wavelength moves only δ either side of the laboratory value.
  • DThe ratio has been inverted, which gives a speed far greater than c.
24C-1A-59Doppler effect for sound Paper 1AHard1 mark

A motorboat whose engine emits a note of constant frequency f0 travels at constant speed along a straight course across a lake. A stationary observer stands at the end of a jetty, 40 m from the course. The boat approaches from far away, passes the observer and moves far away again.

Which graph shows how the frequency f heard by the observer varies with time t?

Four sketch graphs of the frequency heard f against time t, with the emitted frequency f0 dashed, labelled A to DA.ftf₀B.ftf₀C.ftf₀D.ftf₀
Sketch graphs, not drawn to scale; the dashed line shows f₀
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Marking pointMarkNotes
Step 1The frequency heard depends on the component of the boat's velocity along the line to the observer: far away this is almost the full speed, towards (higher than f0) or away (lower than f0), so f is nearly constant there.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2As the boat passes, this component falls smoothly through zero (heard frequency f0) and reverses, so f falls smoothly, most rapidly near the closest approach — graph A.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: f is almost constant above f0 while the boat is far away and approaching, falls smoothly through f0 as it passes, and levels off below f0.
  • BThis sudden drop would be heard only if the boat passed directly through the observer's position. At 40 m from the course the velocity component towards the observer changes gradually.
  • CThis assumes the frequency falls at a steady rate. Far from the observer the velocity component along the line of sight hardly changes, so f is almost constant there.
  • DThis confuses frequency with loudness: the sound is loudest at the closest approach, but the frequency there is about f0, between its higher and lower values.
25C-1A-93Red shift & blue shift Paper 1AHard1 mark

Two stars X and Y move in circular orbits about their common centre of mass, which is at rest relative to the Earth. The Earth lies in the plane of the orbits. A spectral line of laboratory wavelength λ0 appears in the spectra of both stars. The line from X varies between λ0 − 0.030 nm and λ0 + 0.030 nm, and the line from Y varies between λ0 − 0.090 nm and λ0 + 0.090 nm.

What is (mass of X)/(mass of Y)?

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Marking pointMarkNotes
Step 1At the same λ0, Δλ ∝ v, so the orbital speed of Y is 0.090/0.030 = 3 times that of X.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The centre of mass is at rest, so the total momentum is zero: mXvX = mYvY, giving mX/mY = vY/vX = 3.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis assumes the star with the larger shift is the more massive one. The faster star has the smaller mass, because the two momenta are equal in magnitude.
  • BThe two stars have the same orbital period, but that does not make their masses equal: their speeds, and so their shifts, differ.
  • CCorrect: the speeds are in the ratio 1 : 3 and the momenta are equal in magnitude, so X is 3 times as massive as Y.
  • DThis equates the kinetic energies of the stars (½mv² the same), which squares the speed ratio. It is the momenta that are equal.
26C-1B-20Doppler for light Paper 1BMedium10 marks

A published solar atlas gives the wavelengths of absorption lines in light from the east and west edges (limbs) of the Sun's equator. As seen from the Earth, the Sun's rotation carries the east limb towards the Earth and the west limb away from it. For six lines of wavelength λ, a student uses the atlas to find Δλ = λwest − λeast. Each value of Δλ has an uncertainty of ±0.3 pm (1 pm = 10−12 m).

The graph shows Δλ against λ with error bars. The dashed lines are the lines of maximum and minimum gradient.

λ / nmΔλ / pm ± 0.3 pm
430.86.0
518.47.2
589.07.9
617.38.2
656.38.8
854.211.8
Difference in wavelength between the two limbs against wavelength, with error bars and dashed lines of maximum and minimum gradient0100200300400500600700800900λ / nm02468101214Δλ / pm
Graph drawn to scale
(a)

The speed of the Sun's surface at the equator is v. Explain why the gradient of the graph is expected to be 2v/c.

(2)
(b)

Draw the line of best fit on the graph.

(1)
(c)

Determine v using your line of best fit.

(2)
(d)

Using the dashed lines, determine the absolute uncertainty in *v*.

(2)
(e)

The radius of the Sun is 6.96 × 108 m. Determine the time for one rotation of the Sun at the equator, in days, and compare it with the value of about 25 days found by tracking sunspots.

(2)
(f)

Suggest why the student uses the difference between the two limbs rather than the shift of one limb from the laboratory wavelength.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Each limb shows a Doppler shift of magnitude vλ/c: the west limb (receding) is red-shifted and the east limb (approaching) is blue-shifted✓ 1Uses Δλ/λ ≈ v/c.
The difference is therefore 2vλ/c, which is proportional to λ with gradient 2v/c✓ 1OWTTE.
Part (b)
Single straight line lying between the dashed lines and passing through all error bars✓ 1
Part (c)
gradient = 0.0136 pm nm−1 = 1.36 × 10−5✓ 1Accept 1.30 × 10−5 to 1.39 × 10−5. Must convert pm/nm (factor 10−3).
v = c × gradient/2 = 3.00 × 108 × 1.36 × 10−5/2 = 2.04 × 103 m s−1✓ 1Accept 1.95 to 2.09 km s−1. Allow ECF. Award [2] for CNA.
Part (d)
Gradients of the dashed lines ≈ 0.0150 and 0.0127 pm nm−1, giving v = 2250 m s−1 and 1900 m s−1✓ 1Accept ±0.0003 pm nm−1 on each reading.
Δv ≈ ±180 m s−1, so v = (2.04 ± 0.18) km s−1✓ 1Accept ±120 to ±230 m s−1.
Part (e)
T = 2πR/v = 2π × 6.96 × 108/2.04 × 103 = 2.14 × 106 s = 24.8 days✓ 1Allow ECF from (c).
With the uncertainty in v the period lies between about 22.8 and 27.2 days, which includes 25 days, so the two methods agree✓ 1Must refer to the uncertainty or a percentage difference.
Part (f)
The motion of the Earth relative to the Sun (and any other shift common to both limbs) adds the same shift to both limbs, so it cancels in the difference; the difference also doubles the effect to be measured✓ 1Accept either idea.

Answers: (c) v ≈ 2.04 × 103 m s−1  ·  (d) Δv ≈ ±180 m s−1  ·  (e) ≈ 24.8 days (the remaining parts are explanations — see the table above)

27C-1B-26Doppler radar Paper 1BMedium12 marks

A student uses a small microwave radar module, of the type used in automatic doors, to test the Doppler effect for electromagnetic waves. The module transmits microwaves of frequency f = 24.125 GHz along a straight track that slopes down towards it. A trolley released from rest on the track rolls directly towards the module and reflects the microwaves back to it. The module outputs a signal whose frequency fD is the difference between the transmitted and received frequencies. Theory predicts fD = 2fv/c, where v is the speed of the trolley and c is the speed of light.

A light gate measures the time t for a card of length (0.1000 ± 0.0005) m on the trolley to pass through it, with an uncertainty of ±0.2 ms, and v is calculated from the card length and t. A data logger records fD, with an uncertainty of ±4 Hz, as the card passes through the gate. The trolley is released from six different points. The graph shows fD against v with error bars.

runt / ms ± 0.2 msv / m s−1fD / Hz ± 4 Hz
1222.20.45073
2138.90.720115
3104.20.960154
482.0196
568.01.471237
657.51.739278
Doppler frequency f_D against trolley speed v: six points with error bars, no line drawn00.40.81.21.62v / m s⁻¹050100150200250300Doppler frequency / Hz
Graph drawn to scale
(a)

Complete the table for run 4.

(1)
(b)

Show that the absolute uncertainty in v for run 6 is about 0.015 m s−1.

(2)
(c)

Draw the line of best fit on the graph.

(1)
(d)

Determine the gradient of the line of best fit, including its unit.

(2)
(e)

Determine, using your gradient, a value for the speed of light c.

(2)
(f)

The lines of maximum and minimum gradient through the error bars have gradients of 167 m−1 and 151 m−1. Determine the absolute uncertainty in your value of c, and state whether the result is consistent with the accepted value of 3.00 × 108 m s−1.

(2)
(g)

Suggest why fD is recorded while the card is passing through the light gate, rather than at any other time during the run.

(1)
(h)

The module was in fact pointed at a small angle to the track. State and explain the effect of this on the value of c found in (e).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = 0.1000/0.0820 = 1.22 «m s−1»✓ 1Accept 1.220 or 1.2195.
Part (b)
Fractional uncertainty = 0.0005/0.1000 + 0.2/57.5 = 0.005 + 0.0035 = 0.0085 «0.85 %»✓ 1
Δv = 0.0085 × 1.739 = 0.015 «m s−1»✓ 1Must see the calculation AND an answer to at least 2 s.f.
Part (c)
Single straight line through all the error bars, passing close to the origin✓ 1Do not accept a line joining the first and last points.
Part (d)
Uses a triangle covering more than half the line✓ 1
gradient = 160 m−1 «Hz per m s−1»✓ 1Accept 157 to 163 m−1. The unit is required.
Part (e)
fD/v = 2f/c, so c = 2f/gradient✓ 1
c = 2 × 24.125 × 109/160 = 3.02 × 108 m s−1✓ 1Allow ECF from (d). Accept 2.96 × 108 to 3.07 × 108 m s−1. Award [2] for CNA.
Part (f)
2f/167 = 2.89 × 108 and 2f/151 = 3.20 × 108 m s−1, so Δc = ±0.15 × 108 m s−1✓ 1Accept the larger difference from the value in (e). Accept ±0.1 × 108 to ±0.2 × 108 m s−1.
c = (3.02 ± 0.15) × 108 m s−1: 3.00 × 108 m s−1 lies within this range, so the result is consistent✓ 1Allow ECF from (e). The conclusion must be based on the range.
Part (g)
The trolley accelerates down the slope, so v and fD change during the run: both must be measured at the same position «at the same time»✓ 1OWTTE.
Part (h)
The module detects only the component of the velocity along its beam, so every fD is too small: the gradient is too small and c is too large✓ 1Direction AND reason needed.

Answers: (a) 1.22 m s−1  ·  (b) 0.015 m s−1  ·  (d) 160 m−1  ·  (e) c = 3.02 × 108 m s−1  ·  (f) (3.02 ± 0.15) × 108 m s−1: consistent (the remaining parts are explanations — see the table above)

28C-1B-42Red shift & blue shift Paper 1BMedium10 marks

The Sun moves relative to the stars in its neighbourhood. A student uses a published catalogue of stellar spectra to determine the speed V of this motion. She selects seven groups of nearby stars in different directions in the sky. For each group the catalogue gives the mean shift Δλ = λobserved − λ0 of a calcium absorption line of laboratory wavelength λ0 = 422.67 nm, after correction for the motion of the Earth. θ is the angle between the direction to the group and the direction in which the Sun is moving. The stars in each group move randomly, so that on average each group is at rest in the neighbourhood of the Sun. Each value of Δλ has an uncertainty of ±3 pm (1 pm = 10−12 m).

The graph shows Δλ against cos θ, with error bars.

θ / °Δλ / pm ± 3 pm
10−26
35−24
60−13
85−4
11011
14020
17028
Mean shift of the line against cos θ for seven groups of stars, with error bars−1−0.500.51cos θ−40−30−20−10010203040Δλ / pm
Graph drawn to scale
(a)

The group at θ = 10° has a negative value of Δλ. State what this shows about the motion of the Sun relative to this group.

(1)
(b)

Explain why Δλ is expected to be proportional to cos θ.

(2)
(c)

Draw the line of best fit on the graph.

(1)
(d)

Determine V, using your line of best fit.

(3)
(e)

The lines of maximum and minimum gradient that pass through all the error bars have gradients of −29.7 pm and −25.5 pm. A published value of V is 19.5 km s−1. Determine whether the student's result is consistent with the published value.

(2)
(f)

Suggest why the student uses the mean shift for a group of many stars rather than the shift for a single star in each direction.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
the Sun and the group are moving towards each other «the distance between them is decreasing; the line is blue-shifted»✓ 1Accept 'the Sun is moving towards this group'. OWTTE.
Part (b)
the shift depends only on the component of the relative velocity along the line of sight, which is V cos θ «towards the group»✓ 1
Δλ/λ0 ≈ −V cos θ/c, so Δλ = −(λ0V/c) cos θ, which is proportional to cos θ «λ0, V and c constant»✓ 1Sign not required for MP2.
Part (c)
single straight line, passing through all the error bars and close to the origin✓ 1Do not accept a line joining the end points or dot-to-dot.
Part (d)
gradient read from a large triangle ≈ −27.7 pm✓ 1Accept −29.0 to −26.4 pm. The gradient has the unit of Δλ because cos θ has none.
V = −c × gradient/λ0 «with the gradient in m»✓ 1Allow the magnitude of the gradient.
V = 3.00 × 108 × 27.7 × 10−12/422.67 × 10−9 = 19.6 km s−1✓ 1Accept 18.7–20.6 km s−1. Allow ECF from their gradient. Award [3] for CNA.
Part (e)
V lies between 18.1 km s−1 and 21.1 km s−1, so ΔV ≈ ±1.5 km s−1 «V = (19.6 ± 1.5) km s−1»✓ 1Accept ±1 to ±2 km s−1. Accept the larger difference from the value in (d).
19.5 km s−1 lies within this range, so the result is consistent with the published value✓ 1Allow ECF from (d). The decision must be based on the range.
Part (f)
individual stars have their own random velocities «of similar size to V», which would hide the effect of the Sun's motion; these random velocities average out over many stars✓ 1OWTTE.

Answers: (d) V ≈ 19.6 km s−1  ·  (e) V = (19.6 ± 1.5) km s−1; consistent (the remaining parts are explanations — see the table above)

29C-1B-18Red shift & blue shift Paper 1BHard11 marks

An online spectroscopic archive lists the measured wavelength λ of the hydrogen absorption line whose laboratory wavelength is 656.28 nm, in the spectrum of a star S that orbits a companion star. The wavelengths are quoted with an uncertainty of ±0.005 nm. A student calculates the velocity v of S along the line of sight (positive when S is moving away from the Earth) and plots v against time t.

t / dayλ / nm ± 0.005 nmv / km s−1
0.0656.223−26
1.1656.2904.6
2.0656.36037
3.2656.407
4.1656.38046
5.0656.31215
6.1656.229−23
7.0656.209−32
8.2656.256−11
9.0656.31516
10.1656.38950
11.0656.40155
Radial velocity of the star against time with error bars, no curve drawn024681012t / day-40-20020406080v / km s⁻¹
Graph drawn to scale
(a)

Show that the uncertainty in each wavelength corresponds to an uncertainty in v of about 2 km s−1.

(1)
(b)

Calculate v for t = 3.2 day and state whether S is moving towards or away from the Earth at this time.

(2)
(c)

Draw the curve of best fit on the graph.

(1)
(d)
(i)

Determine the velocity of the whole system along the line of sight.

(2)
(ii)

Determine the orbital speed of S.

(1)
(e)

Estimate the orbital period of S and hence determine the radius of its orbit. Assume the orbit is circular and that the Earth lies in the plane of the orbit.

(3)
(f)

Suggest why, if the Earth does not lie in the plane of the orbit, the radius found in (e) is a minimum value.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δv = cΔλ/λ = 3.00 × 108 × 0.005/656.28 = 2290 m s−1 ≈ 2 km s−1✓ 1Must see the substitution AND an answer to at least 2 s.f.
Part (b)
Δλ = 656.407 − 656.280 = 0.127 nm, v = 3.00 × 108 × 0.127/656.28 = 58 km s−1✓ 1Accept 58.1 km s−1.
The wavelength is longer than the laboratory value (red shift), so S is moving away from the Earth✓ 1Must be consistent with the sign of Δλ.
Part (c)
Smooth, regular wave-shaped (sinusoidal) curve passing through the error bars, with maxima and minima of similar height✓ 1Do not accept dot-to-dot.
Part (d)(i)
Maximum and minimum of the curve read as about 57 km s−1 and −33 km s−1; system velocity = mean of these✓ 1Accept the value of v on the curve's centre line.
≈ +12 km s−1, i.e. the system is moving away from the Earth✓ 1Accept +8 to +16 km s−1. Direction needed.
Part (d)(ii)
Half the peak-to-peak variation: (57 − (−33))/2 ≈ 45 km s−1✓ 1Accept 42 to 48 km s−1.
Part (e)
Period from the graph (e.g. time between successive maxima) ≈ 7.6 day✓ 1Accept 7.2 to 8.0 day.
r = vP/2π, with P in seconds✓ 1
r = 45 × 103 × 7.6 × 86400/2π = 4.7 × 109 m✓ 1Accept 4.2 × 109 to 5.3 × 109 m. Allow ECF from (d)(ii) and the period.
Part (f)
The Doppler shift gives only the component of the orbital velocity along the line of sight, so the true orbital speed (and hence radius) is larger than the value measured✓ 1OWTTE.

Answers: (a) 2.3 km s−1  ·  (b) v ≈ 58 km s−1, moving away  ·  (d)(i) ≈ +12 km s−1  ·  (d)(ii) ≈ 45 km s−1  ·  (e) P ≈ 7.6 day; r ≈ 4.7 × 109 m (the remaining parts are explanations — see the table above)

30C-1B-43Doppler for light Paper 1BHard12 marks

A student uses archived spectra of sunlight reflected from Saturn's rings to test two models of the rings. In model 1 the rings are a solid disc rotating as a single body, so that the speed v of the ring material is proportional to its distance r from the centre of Saturn. In model 2 the rings consist of separate particles, each moving in its own circular orbit around Saturn, so that v² = GM/r, where M is the mass of Saturn.

At seven values of r, on the side of the rings where the ring material moves directly away from the Earth, the shift Δλ of a solar absorption line of laboratory wavelength 516.73 nm was measured, after correction for the motion of Saturn itself. Because the light is sunlight reflected by moving material, and the Sun and the Earth are in almost the same direction as seen from Saturn, Δλ/λ = 2v/c. Each Δλ has an uncertainty of ±0.3 pm. The uncertainty in r is negligible.

The graph shows lg(v/m s−1) against lg(r/m) for six of the measurements.

r / 107 mΔλ / pm ± 0.3 pmlg(r/m)lg(v/m s−1)
7.677.17.8814.350
8.572.67.9294.324
9.668.77.9824.300
10.665.2
11.462.78.0574.260
12.660.08.1004.241
13.557.58.1304.222
lg(v / m s−1) against lg(r / m) for six of the seven points in Saturn's rings7.857.907.958.008.058.108.15lg( r / m )4.204.244.284.324.36lg( v / m s⁻¹ )
Graph drawn to scale
(a)

Show that v for r = 7.6 × 107 m is about 22 km s−1.

(1)
(b)

Complete the table for r = 10.6 × 107 m and plot this point on the graph.

(2)
(c)

Draw the line of best fit on the graph.

(1)
(d)

Determine the gradient of the line.

(2)
(e)

Deduce, with reference to your answer to (d), which model the data support.

(2)
(f)

Assume that model 2 is correct.

(i)

Determine, using the data for r = 10.6 × 107 m, the mass M of Saturn and its absolute uncertainty.

(3)
(ii)

The accepted value of the mass of Saturn is 5.68 × 1026 kg. Comment on your answer to (f)(i).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = cΔλ/2λ = 3.00 × 108 × 77.1 × 10−12/(2 × 516.73 × 10−9) = 2.24 × 104 «m s−1»✓ 1Must see full substitution OR answer to 3 s.f. Do not award if the factor 2 is omitted.
Part (b)
lg(r/m) = 8.025 AND lg(v/m s−1) = 4.277 «v = 1.89 × 104 m s−1»✓ 1Accept 8.03 and 4.28. Do not penalise extra decimal places. Do not award if v is not in m s−1.
point plotted at (8.025, 4.277)✓ 1Accept within half a small square. Allow ECF from their table values.
Part (c)
single straight line with points scattered evenly on both sides✓ 1Do not accept dot-to-dot.
Part (d)
uses a triangle covering more than half the line «or two well-separated points on the line»✓ 1
gradient = −0.50✓ 1Accept −0.46 to −0.54. A negative sign is required.
Part (e)
model 1 predicts lg v = lg r + constant, a gradient of +1; model 2 predicts lg v = −½ lg r + ½ lg(GM), a gradient of −0.5✓ 1Both predicted gradients needed.
the gradient is about −0.5 «v decreases as r increases», so the data support model 2 «separate orbiting particles»✓ 1Allow ECF from (d). MP2 only scores if MP1 scores.
Part (f)(i)
M = v²r/G = (1.89 × 104)² × 1.06 × 108/6.67 × 10−11 = 5.69 × 1026 kg✓ 1Allow ECF from (b).
fractional uncertainty in v = 0.3/65.2 = 0.0046; fractional uncertainty in M = 2 × 0.0046 = 0.0092 «0.9 %»✓ 1Uncertainty in r negligible, as stated.
M = (5.69 ± 0.05) × 1026 kg✓ 1MP3 is for matching the precision of value and uncertainty. Award [3] for CNA.
Part (f)(ii)
5.68 × 1026 kg lies within the range of the result, so the result agrees with the accepted value «supporting model 2»✓ 1Allow ECF from (f)(i).

Answers: (b) 8.025, 4.277; point plotted  ·  (d) gradient ≈ −0.50  ·  (f)(i) M = (5.69 ± 0.05) × 1026 kg (the remaining parts are explanations — see the table above)

31C-2-14Wavefront diagrams Paper 2Easy11 marks

A student whirls a small buzzer on a string in a horizontal circle of radius 1.20 m, at a constant rate of 2.0 revolutions per second. The buzzer emits sound of constant frequency 2.40 kHz. A listener L stands in the same horizontal plane, a large distance from the circle.

Figure 1 is a plan view. The buzzer moves anticlockwise through the points A, B, C and D.

Plan view: a buzzer moves anticlockwise round a horizontal circle of centre O, through points A (top), B (right), C (bottom) and D (left); a listener L is far away to the rightOABCDmotion of buzzerLlistener
Figure 1 — Diagram NOT to scale
(a)

Calculate the speed of the buzzer.

(2)
(b)

On Figure 1, label with the letter H the position of the buzzer when it emits the sound that L hears with the highest frequency, and with the letter W the position when it emits the sound that L hears with the lowest frequency.

(2)
(c)

Explain, with reference to the wavefronts emitted by the buzzer, why the frequency heard by L is greater than 2.40 kHz for sound emitted at H.

(2)
(d)

State the number of times each second that the frequency heard by L reaches its maximum value.

(1)
(e)

State and explain the frequency heard by L for sound emitted when the buzzer is at B.

(2)
(f)

The student whirls the buzzer faster. State and explain the effect on the highest frequency heard by L.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
circumference = 2π × 1.20 = 7.54 m✓ 1
v = 7.54 × 2.0 = 15 m s−1✓ 1Award [2] for CNA
Part (b)
H at C «where the buzzer moves directly towards L»✓ 1Accept a position on the circle near C
W at A «where the buzzer moves directly away from L»✓ 1Accept a position on the circle near A
Part (c)
the buzzer moves towards L between emitting successive wavefronts, so the wavefronts travelling towards L are closer together «shorter wavelength»✓ 1
the speed of sound is unchanged, so more wavefronts reach L each second✓ 1OWTTE
Part (d)
2✓ 1The buzzer moves towards L once in each revolution
Part (e)
2.40 kHz «the emitted frequency»✓ 1
at B the velocity of the buzzer is perpendicular to the line joining it to L, so it is not moving towards or away from L✓ 1MP2 only scores if MP1 scores
Part (f)
the highest frequency heard increases✓ 1
the buzzer moves further between successive wavefronts, so the wavefronts ahead of it are even closer together✓ 1OWTTE

Answers: (a) v = 15 m s−1  ·  (d) 2  ·  (e) 2.40 kHz (the remaining parts are explanations — see the table above)

32C-2-34Red shift & blue shift Paper 2Easy12 marks

The figure shows four absorption lines of hydrogen in a laboratory spectrum, and the same four lines in the spectrum of a distant galaxy. The laboratory wavelengths are 410.2 nm, 434.0 nm, 486.1 nm and 656.3 nm. The wavelengths measured in the galaxy spectrum are marked on the figure.

Three spectrum strips on a wavelength scale from 400 nm to 700 nm: laboratory hydrogen absorption lines, the same four lines in a galaxy at 418.4, 442.7, 495.8 and 669.4 nm, and a blank strip for a star.laboratorygalaxystar418.4442.7495.8669.4400450500550600650700wavelength / nm
Diagram drawn to scale
(a)

State how the figure shows that the galaxy is moving away from the Earth.

(1)
(b)
(i)

Outline how astronomers can identify these galaxy lines as the hydrogen lines even though none of them is at its laboratory wavelength.

(1)
(ii)

Determine, using the 656.3 nm line, the speed of the galaxy relative to the Earth.

(2)
(iii)

Outline why the shift in wavelength is larger for the 656.3 nm line than for the 410.2 nm line.

(1)
(c)

A star in a close orbit around the black hole at the centre of our galaxy is, at one instant, moving directly towards the Earth at 3.0 × 106 m s−1. On the strip labelled 'star', draw the positions of the four hydrogen lines in the spectrum of this star.

(2)
(d)
(i)

Explain how the dark absorption lines in the spectrum of the galaxy are formed.

(3)
(ii)

Calculate, in eV, the energy of a photon of wavelength 656.3 nm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Every line is at a longer wavelength than in the laboratory «red-shifted»✓ 1
Part (b)(i)
The galaxy lines have the same pattern «relative spacing»: every wavelength is shifted by the same fraction Δλ/λ✓ 1OWTTE.
Part (b)(ii)
Δλ/λ = (669.4 − 656.3)/656.3 = 0.0200✓ 1
v = 0.0200 × 3.00 × 108 = 6.0 × 106 m s−1✓ 1Accept 5.99 × 106 m s−1. Award [2] for CNA.
Part (b)(iii)
Δλ = λv/c, so for the same speed the shift is proportional to the laboratory wavelength✓ 1
Part (c)
All four lines at shorter wavelengths than the laboratory lines «blue shift»✓ 1
Each shifted by about 1 % of its wavelength: about 4 nm for the 410.2 nm line, rising to about 7 nm for the 656.3 nm line «406, 430, 481, 650 nm»✓ 1Accept positions within ±3 nm.
Part (d)(i)
Light with a continuous spectrum from the hot interiors of the galaxy's stars passes through cooler gas «in the stars' outer layers»✓ 1Accept 'cooler gas between the source and the observer'.
Atoms of the gas absorb only photons whose energy equals the difference between two of their discrete energy levels «E = hf», so only certain wavelengths are absorbed✓ 1
The atoms re-emit this energy in all directions, so less light of these wavelengths reaches the observer, giving dark lines✓ 1OWTTE.
Part (d)(ii)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/656.3 × 10−9 = 3.03 × 10−19 J✓ 1
E = 3.03 × 10−19/1.60 × 10−19 = 1.89 eV✓ 1Award [2] for CNA.

Answers: (b)(ii) 6.0 × 106 m s−1  ·  (d)(ii) 1.89 eV (the remaining parts are explanations — see the table above)

33C-2-73Doppler effect for sound Paper 2Easy12 marks

A Doppler ultrasound probe P is used to study the flow of blood in an artery in the arm. P is held against the skin and emits ultrasound of frequency 5.0 MHz. Red blood cells moving in the artery scatter some of the ultrasound back to P, which detects it. The speed of ultrasound in soft tissue and blood is 1.54 × 103 m s−1.

Probe P on the left emitting equally spaced plane wavefronts travelling to the right; a red blood cell C on the right moves to the left towards PPwavefronts travel this wayCwavefronts from the probe (spacing greatly exaggerated)
Figure 1 — Diagram NOT to scale
Four positions C1 to C4 of a cell moving left towards probe P, at equal time intervals; one circular wavefront centred on C1 is drawnPC1C2C3C4direction of motion of the cell
Figure 2 — Diagram NOT to scale; the speed of the cell is greatly exaggerated
(a)
(i)

The ultrasound scattered back to P by a moving red blood cell travels through the tissue at the same speed as the ultrasound emitted by P. State why.

(1)
(ii)

Some of the ultrasound is reflected back to P by the wall of the artery, which is stationary. State and explain how the frequency of this reflected ultrasound compares with 5.0 MHz.

(2)
(b)

Figure 1 shows a red blood cell C moving towards P, and some of the wavefronts emitted by P.

(i)

Explain, with reference to Figure 1, why the frequency at which wavefronts reach C is greater than 5.0 MHz.

(2)
(ii)

The cell scatters each wavefront it meets in all directions, acting as a moving source. Figure 2 shows the positions C1, C2, C3 and C4 of the cell at the instants when it scattered four successive wavefronts. The wavefront scattered at C1 is shown at the instant the cell reaches C4. Draw, on Figure 2, the wavefronts scattered at C2 and C3 at this instant.

(2)
(iii)

Hence state how the frequency of the scattered ultrasound detected by P compares with the frequency at which wavefronts reach C.

(1)
(c)

The probe is turned so that its beam is perpendicular to the artery. Explain why no change in frequency is now detected.

(2)
(d)

During part of each heartbeat the blood in the artery flows away from the probe. State how the detected frequency shows this.

(1)
(e)

Suggest why a gel is spread between the probe and the skin.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
the speed of a sound wave is set by the medium «the tissue», not by the motion of the source✓ 1OWTTE. Do not accept 'all sound travels at the same speed'.
Part (a)(ii)
the same «5.0 MHz»✓ 1MP1 can be awarded only with a reason.
the wall does not move relative to P, so it meets the wavefronts at the rate they are emitted and re-emits them with the same spacing «wavelength»✓ 1OWTTE.
Part (b)(i)
the source is stationary, so the wavefronts are equally spaced and the wavelength in the tissue is unchanged✓ 1
C moves towards the wavefronts, so it meets them at a greater rate than a stationary cell «more than 5.0 × 106 per second»✓ 1OWTTE.
Part (b)(ii)
two circles, one centred on C2 and one centred on C3✓ 1Accept circular arcs on the P side only.
radii in the ratio 2 : 1 and about ⅔ and ⅓ of the radius of the C1 wavefront, so that the wavefronts are closer together on the side towards P than on the far side✓ 1Accept any drawing in which the spacing towards P is clearly less than on the far side.
Part (b)(iii)
greater «than the frequency at which wavefronts reach C», because the scattered wavefronts are closer together towards P✓ 1OWTTE.
Part (c)
the velocity of the cells is then perpendicular to the line joining them to the probe✓ 1
so the cells are not moving towards or away from P «the component of velocity along the beam is zero» and there is no Doppler effect✓ 1MP2 only scores if MP1 scores.
Part (d)
the detected frequency is less than 5.0 MHz «the shift reverses sign»✓ 1
Part (e)
without the gel there would be an air gap, and most of the ultrasound would be reflected at the air–skin boundary «so little enters the arm»✓ 1OWTTE.
34IBP2S-026-01-11Red shift & blue shift Paper 2Easy5 marks

Astronomers detect an exoplanet by observing a periodic Doppler shift in the light from its host star. A sodium line with a laboratory wavelength of 589.0000 nm is observed to vary repeatedly between a minimum of 588.9996 nm and a maximum of 589.0004 nm. The speed of light is c = 3.00 × 108 m s−1.

(a)

Explain why the observed wavelength of the line varies.

(2)
(b)

Calculate the maximum speed of the star along the line of sight.

(2)
(c)

State how the orbital period of the planet can be found from these observations.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The star orbits the common centre of mass of the star–planet system, so it moves alternately towards and away from Earth . When the star approaches, its light is blue-shifted to a shorter wavelength; when it recedes, its light is red-shifted to a longer wavelength .✓ 2
Part (b)
Δλ = 589.0004 − 589.0000 = 0.0004 nm, used in v = cΔλ/λ . v = 3.00×108 × 0.0004/589.0000 ≈ 204 m s−1 (accept 2.0×102 m s−1) .✓ 2
Part (c)
The period equals the time for one complete cycle of the wavelength variation, for example the time between successive maxima .✓ 1

Exam tip: Measure Δλ from the laboratory value, not across the full range of variation. Using the full range would double the speed.

35IBP2S-026-01-12Red shift & blue shift Paper 2Easy5 marks

A hydrogen line has a laboratory wavelength of 486.1 nm. The table gives the wavelength of this line observed in the spectra of three galaxies.

Galaxy | Observed wavelength / nm
A | 487.0
B | 485.6
C | 490.0

c = 3.00 × 108 m s−1.

(a)

Identify, with a reason, the galaxy that is moving towards Earth.

(1)
(b)

Calculate the speed of galaxy C relative to Earth.

(2)
(c)

Most galaxies show red shifts. Outline why the blue shift of galaxy B is consistent with an expanding universe.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B, because its observed wavelength is shorter than the laboratory value (blue shift) .✓ 1
Part (b)
Δλ = 490.0 − 486.1 = 3.9 nm . v = cΔλ/λ = 3.00×108 × 3.9/486.1 ≈ 2.4×106 m s−1 .✓ 2
Part (c)
Galaxy B is probably nearby, and its motion is dominated by local gravitational attraction, for example within a galaxy group . The recession due to expansion grows with distance, so for a nearby galaxy it can be smaller than this local motion .✓ 2

Exam tip: The formula Δλ/λ ≈ v/c is valid only when v is much less than c. Here v/c ≈ 0.008, so the approximation is fine.

36IBP2S-026-00-11Doppler radar Paper 2Easy6 marks

A radar speed gun at a tennis tournament measures the speed of a serve. The gun emits microwaves of frequency 10.5 GHz towards a ball moving directly towards it. The reflected microwaves return to the gun.

(a)

State why the frequency of the reflected microwaves received by the gun differs from the emitted frequency.

(2)
(b)

The frequency difference measured by the gun is 3.43 kHz. Calculate the speed of the ball.

(2)
(c)

Explain why the factor of 2 appears in the equation used in (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The ball is moving relative to the gun (towards it) . So the received frequency is Doppler-shifted, and here it is higher than the emitted frequency .✓ 2
Part (b)
Use Δf = 2fv/c, so v = cΔf/(2f) = 3.00×108 × 3430/(2 × 10.5×109) . v = 49.0 m s−1 .✓ 2
Part (c)
The ball acts as a moving observer and receives a frequency already shifted by about fv/c . It then re-emits (reflects) the waves as a moving source, adding a second shift of about fv/c, so the total is about 2fv/c .✓ 2

Common error: Using Δf = fv/c for a reflected signal doubles the calculated speed. Check whether the wave has been reflected from the moving object.

37C-2-35Wavefront diagrams Paper 2Medium10 marks

The engine of a racing car emits a note of frequency 680 Hz. The car moves in a straight line at constant speed. The figure is a computer simulation, drawn to scale, of four sound wavefronts emitted by the car, seen from above at one instant. The small dots show the positions of the car when the wavefronts were emitted; dot 1 is where the largest wavefront was emitted. D is the position of the car at the instant shown. The speed of sound in air is 340 m s−1.

Four circular sound wavefronts from a moving racing car, drawn to scale (0.50 m = 45 mm on the scale bar): small dots mark the centres of the wavefronts, 1 being the centre of the largest; the wavefronts are 0.375 m apart on the right and 0.625 m apart on the left; D is the car.1D0.50 m
Diagram drawn to scale
(a)

Draw an arrow on the figure to show the direction in which the car is moving.

(1)
(b)
(i)

Determine, using the scale of the figure, the wavelength of the sound directly ahead of the car and directly behind it.

(2)
(ii)

Calculate the time between the emission of two successive wavefronts.

(1)
(iii)

Deduce the speed of the car.

(2)
(c)

A spectator stands directly ahead of the car. State and explain whether the frequency she hears is greater than, equal to or less than 680 Hz.

(2)
(d)

Outline how the figure would differ if the car moved faster, but still more slowly than the sound.

(1)
(e)

State one way in which the Doppler effect for light differs from the Doppler effect for sound.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Arrow pointing to the right «towards the side where the wavefronts are closest together»✓ 1
Part (b)(i)
Ahead: λ = 0.38 m✓ 1Accept 0.35–0.40 m. Measurement across several wavefronts expected.
Behind: λ = 0.63 m✓ 1Accept 0.60–0.65 m.
Part (b)(ii)
T = 1/680 = 1.5 × 10−3 s «1.5 ms»✓ 1
Part (b)(iii)
Dot 1 to D is 0.50 m, covered in four periods «4 × 1.47 ms = 5.9 ms»✓ 1ALTERNATIVE: successive dots are 0.125 m apart, covered in one period. OR λbehind − λahead = 2uT.
u = 0.50/(4 × 1.47 × 10−3) = 85 m s−1✓ 1Accept 80–90 m s−1. Allow ECF from (b)(i) and (b)(ii).
Part (c)
Greater than 680 Hz✓ 1MP1 can be awarded only with a reason.
The sound travels at the same speed but the wavelength ahead is shorter, so more wavefronts reach her each second «f = v/λ»✓ 1Allow ECF from (b)(i).
Part (d)
The centres «dots» are further apart, so the wavefronts are closer together ahead of the car and further apart behind it✓ 1OWTTE.
Part (e)
Light needs no medium, so only the relative velocity of the source and the observer matters OR for light the relative speed is normally much less than c, so the shift is very small✓ 1Accept any one valid difference.

Answers: (b)(i) 0.38 m ahead, 0.63 m behind  ·  (b)(ii) 1.5 × 10−3 s  ·  (b)(iii) 85 m s−1 (the remaining parts are explanations — see the table above)

38C-2-47Red shift & blue shift Paper 2Medium12 marks

A distant star lies in the plane of the Earth's orbit. The Earth moves around the Sun in a circular orbit of radius 1.50 × 1011 m with a period of 365 days. An absorption line in the spectrum of the star has a laboratory wavelength of 589.000 nm. The table gives the wavelength of the line measured on two dates, six months apart, when the Earth was moving directly towards the star and directly away from it.

Motion of the EarthMeasured wavelength / nm
directly towards the star588.965
directly away from the star589.082
Plan view of the Earth's circular orbit around the Sun, with an arrow showing that the Earth moves anticlockwise. The distant star lies far to the right in the plane of the orbit.SunEarth's orbitto star (very far away)
Figure 1 — Diagram NOT to scale
(a)

Calculate the orbital speed of the Earth.

(1)
(b)

On Figure 1, label with the letter B the position of the Earth when the line is most blue-shifted and with the letter R the position when it is most red-shifted.

(2)
(c)
(i)

Determine the velocity of the star relative to the Sun along the line of sight. State whether the star is moving towards or away from the Sun.

(3)
(ii)

Determine whether the data in the table are consistent with your answer to (a).

(3)
(d)

Predict the wavelength of the line measured three months after the first date in the table. Explain your answer.

(2)
(e)

Astronomers search for planets by detecting periodic changes of a few metres per second in the velocity of a star. Suggest why measured velocities must first be corrected for the motion of the Earth.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = 2πr/T = 2π × 1.50 × 1011/(365 × 24 × 3600) = 2.99 × 104 m s−1✓ 1
Part (b)
B at the point of the orbit where the Earth's velocity is directed towards the star «the lowest point of the orbit in Figure 1»✓ 1
R at the opposite point of the orbit «the highest point»✓ 1Allow ECF for R diametrically opposite an incorrect B.
Part (c)(i)
The Earth's contributions are equal and opposite, so the star's own shift comes from the mean: (588.965 + 589.082)/2 − 589.000 = 0.0235 nm✓ 1
v = cΔλ/λ = 3.00 × 108 × 0.0235/589.000 = 1.2 × 104 m s−1✓ 1Accept 1.1–1.2 × 104 m s−1.
Away from the Sun «the mean line is red-shifted»✓ 1
Part (c)(ii)
Half the difference: (589.082 − 588.965)/2 = 0.0585 nm✓ 1Award [1 max] for using the whole difference «6.0 × 104 m s−1».
v = 3.00 × 108 × 0.0585/589.000 = 2.98 × 104 m s−1✓ 1
This agrees with (a) to within about 0.3 % «within the precision of the data», so yes✓ 1Allow ECF from (a). MP3 only if MP2 is scored.
Part (d)
The Earth's velocity is then perpendicular to the line joining it to the star, so it has no component along the line of sight «no Doppler shift from the Earth's motion»✓ 1OWTTE.
The measured wavelength is the mean of the two values, about 589.024 nm «shifted only by the star's own motion»✓ 1Allow ECF from (c)(i). Accept 589.023–589.024 nm.
Part (e)
The Earth's orbital speed «3 × 104 m s−1» is thousands of times larger than the change being searched for and also varies periodically, so it would hide «or mimic» the star's motion✓ 1OWTTE.

Answers: (a) 2.99 × 104 m s−1  ·  (c)(i) 1.2 × 104 m s−1, away from the Sun  ·  (c)(ii) 2.98 × 104 m s−1: consistent  ·  (d) 589.024 nm (the remaining parts are explanations — see the table above)

39C-2-75Doppler effect for sound Paper 2Medium13 marks

A student investigates the Doppler effect by releasing a small buzzer from rest at the top of a stairwell. The buzzer emits a note of constant frequency 3.20 kHz and falls 12.0 m vertically to a thick cushion on the floor, where it stops and continues to sound. Microphone M1 is at the point of release and microphone M2 is on the floor beside the cushion. Both microphones record the frequency of the sound they receive. Air resistance on the buzzer is negligible. The speed of sound in air is 340 m s−1.

Blank axes of frequency f in kHz against time t in seconds, with a dashed line at 3.20 kHz00.511.52t / s3.20f / kHz
Figure 1 — axes for (d)
(a)
(i)

Calculate the time taken by the buzzer to fall to the cushion.

(1)
(ii)

Calculate the speed of the buzzer just before it reaches the cushion.

(1)
(b)

Show that M1 first receives the sound emitted after the buzzer has stopped about 1.60 s after the buzzer is released.

(2)
(c)

State and explain how the frequency recorded by M2 while the buzzer is falling compares with 3.20 kHz.

(2)
(d)

Sketch, on Figure 1, the variation with time t of the frequency recorded by M1 from t = 0 to t = 2.0 s.

(3)
(e)

The experiment is repeated with the buzzer inside a light foam ball, so that air resistance is no longer negligible and the ball reaches its terminal speed well before it lands. Describe how the frequency recorded by M1 would differ from your sketch in (d).

(2)
(f)

A space probe accelerates directly away from the Earth. It transmits radio waves of constant frequency. Outline how measurements on the Earth of the frequency received at known times could be used to determine the acceleration of the probe.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
t = √(2h/g) = √(2 × 12.0/9.81) = 1.56 s✓ 1
Part (a)(ii)
v = gt OR √(2gh) = 15.3 m s−1✓ 1Allow ECF from (a)(i).
Part (b)
time for the sound to travel 12.0 m back up to M1 = 12.0/340 = 0.035 s✓ 1
t = 1.564 + 0.035 = 1.599 «s» ≈ 1.60 s✓ 1Must see both times added OR an answer to at least 4 s.f. «1.599 s».
Part (c)
higher than 3.20 kHz «and increasing during the fall»✓ 1
the buzzer moves towards M2 between emitting successive wavefronts, so the wavefronts below it are closer together «shorter wavelength»; the speed of sound is unchanged, so more wavefronts reach M2 each second✓ 1MP1 can be awarded only with a reason.
Part (d)
starts at 3.20 kHz at t = 0 and is below 3.20 kHz throughout the fall✓ 1
decreases continuously «the buzzer's speed away from M1 increases», with the lowest frequency just before the change in (b)✓ 1
returns abruptly to 3.20 kHz at t ≈ 1.60 s «not at 1.56 s» and stays constant until 2.0 s✓ 1Allow ECF from (b).
Part (e)
the frequency decreases less rapidly «the acceleration is smaller» and then becomes constant, below 3.20 kHz, once the ball reaches terminal speed✓ 1
the return to 3.20 kHz happens later, because the ball takes longer to fall✓ 1OWTTE.
Part (f)
for each time, the speed of the probe is found from v = cΔf/f «Δf = transmitted frequency − received frequency»✓ 1
the acceleration is the gradient of a graph of v against t «or Δv/Δt between two measurements»✓ 1OWTTE.

Answers: (a)(i) 1.56 s  ·  (a)(ii) 15.3 m s−1  ·  (b) 1.60 s (the remaining parts are explanations — see the table above)

40IBP2S-026-01-15Doppler radar Paper 2Medium6 marks

A deep-space probe transmits a radio signal of frequency 8.40 GHz. A ground station receives the signal with a frequency differing from the transmitted frequency by 280 kHz. The probe is moving directly away from Earth. c = 3.00 × 108 m s−1.

(a)

State and explain whether the received frequency is greater or smaller than 8.40 GHz.

(2)
(b)

Calculate the speed of the probe relative to Earth.

(2)
(c)

Later the probe moves at an angle to the line joining it to Earth. Explain why the speed calculated from the Doppler shift may then be less than the probe's actual speed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Smaller . The probe is receding, so the signal is red-shifted: the wavelength is stretched and the frequency lowered .✓ 2
Part (b)
v = cΔf/f . v = 3.00×108 × 2.80×105/8.40×109 = 1.0×104 m s−1 .✓ 2
Part (c)
The Doppler shift depends only on the component of velocity along the line of sight (the radial component) . The component perpendicular to the line of sight produces no shift, so the calculated speed is smaller than the actual speed .✓ 2

Exam tip: Convert GHz and kHz to Hz before substituting. Only the ratio Δf/f matters, so the units must match.

41IBP2S-026-00-12Red shift & blue shift Paper 2Medium7 marks

An astronomer measures three absorption lines in the spectrum of a galaxy. The table gives their laboratory and observed wavelengths.

Line | Laboratory wavelength / nm | Observed wavelength / nm
H-alpha | 656.3 | 669.4
H-beta | 486.1 | 495.8
Calcium K | 393.4 | 401.2

(a)

State and explain whether the galaxy is moving towards or away from Earth.

(2)
(b)

Show that the fractional change in wavelength Δλ/λ for the H-beta line is about 0.020.

(2)
(c)

Determine the speed of the galaxy relative to Earth, using all three lines.

(2)
(d)

Outline why the equation Δλ/λ ≈ v/c is valid for this galaxy.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
All observed wavelengths are longer than the laboratory values, so the light is red-shifted . Therefore the galaxy is moving away from Earth .✓ 2
Part (b)
Δλ = 495.8 − 486.1 = 9.7 nm . Δλ/λ = 9.7/486.1 = 0.01995 ≈ 0.020 .✓ 2
Part (c)
Fractional shifts are 13.1/656.3 = 0.01996, 9.7/486.1 = 0.01995 and 7.8/393.4 = 0.01983, with mean ≈ 0.0199 . v = c × Δλ/λ = 3.00×108 × 0.0199 ≈ 6.0×106 m s−1 .✓ 2
Part (d)
The speed found is only about 2% of c, so v ≪ c and the non-relativistic approximation holds .✓ 1

Exam tip: Using several lines and averaging their fractional shifts reduces the effect of random uncertainty in reading any single line.

42IBP2S-026-00-15Doppler radar Paper 2Medium6 marks

A space probe transmits a radio signal of frequency 8.400 000 GHz. A ground station receives the signal at 8.399 720 GHz. Assume the probe moves along the line joining it to the ground station.

(a)

State whether the probe is moving towards or away from Earth.

(1)
(b)

Calculate the speed of the probe relative to the ground station.

(2)
(c)

The rotation of the Earth gives the ground station a speed of up to 460 m s−1 along the line to the probe. Estimate the largest change in received frequency caused by the Earth's rotation alone.

(2)
(d)

Outline why the method gives only part of the probe's velocity if the probe is not moving along the line of sight.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Away, because the received frequency is lower (red-shift) .✓ 1
Part (b)
Δf = 2.80×105 Hz and v = cΔf/f = 3.00×108 × 2.80×105/8.40×109 . v = 1.0×104 m s−1 .✓ 2
Part (c)
Δf = fv/c = 8.40×109 × 460/3.00×108 . Δf ≈ 1.3×104 Hz, or about 13 kHz .✓ 2
Part (d)
Only the component of velocity along the line of sight causes a Doppler shift, so any motion perpendicular to it is not detected .✓ 1

Exam tip: When a frequency changes only in the fifth or sixth significant figure, subtract carefully to find Δf before dividing.

43C-2-15Doppler for light Paper 2Hard13 marks

A nova is an explosion on the surface of a white dwarf star. The explosion throws off a thin spherical shell of hot gas that expands in all directions. Assume that the speed of expansion is constant. Hydrogen in the shell emits light whose wavelength measured in the laboratory is 486.13 nm. Figure 1 shows how the intensity of this emission line, observed from the Earth, varies with wavelength. The white dwarf is at rest relative to the Earth.

Intensity of the emission line (arbitrary units) against wavelength from 480 nm to 492 nm: flat-topped between 482.24 nm and 490.02 nm; dashed line at the laboratory wavelength 486.13 nm480481482483484485486487488489490491492wavelength / nm0intensity / arbitrary units486.13 nm
Figure 1 — drawn to scale
(a)

Explain why the line is spread over a range of wavelengths rather than appearing at 486.13 nm only.

(2)
(b)

Determine the speed at which the shell expands.

(2)
(c)

State which part of the shell emits the light observed at the shortest wavelength.

(1)
(d)

Another nova, with a shell expanding at the same speed, is moving away from the Earth at 6.0 × 105 m s−1. Sketch, on Figure 1, the line that would be observed from this nova.

(2)
(e)

A telescope image taken 400 days after the explosion shows the shell as a ring of angular radius 6.9 × 10−6 rad. The radius of the shell is equal to its angular radius, in radians, multiplied by its distance from the Earth.

(i)

Calculate the radius of the shell 400 days after the explosion.

(1)
(ii)

Determine the distance from the Earth to the nova.

(2)
(f)

At its brightest the nova had an apparent brightness of 6.0 × 10−9 W m−2. Determine the luminosity of the nova at that time.

(2)
(g)

Suggest one reason why the distance found in (e)(ii) may be inaccurate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
different parts of the expanding shell have different components of velocity along the line of sight, from directly towards the Earth «near side» to directly away from it «far side»✓ 1
gas moving towards the Earth gives a blue shift and gas moving away gives a red shift, with Δλ/λ ≈ v/c for its line-of-sight component, so every shift between the two extremes is observed✓ 1OWTTE
Part (b)
Δλ = (490.0 − 482.2)/2 = 3.9 nm «half the width of the line»✓ 1Accept Δλ = 490.0 − 486.13 or 486.13 − 482.2
v = cΔλ/λ = 3.00 × 108 × 3.9/486.13 = 2.4 × 106 m s−1✓ 1Accept 2.3–2.5 × 106 m s−1. Award [1] max for using the full width «4.8 × 106 m s−1»
Part (c)
the part of the shell on the near side, moving directly towards the Earth✓ 1OWTTE
Part (d)
same flat-topped shape and the same width «about 7.8 nm»✓ 1
whole line shifted to longer wavelengths by about 1.0 nm «edges at about 483.2 nm and 491.0 nm»✓ 1Accept a shift of 0.8–1.2 nm. Do not award MP2 for a shift to shorter wavelengths
Part (e)(i)
r = vt = 2.4 × 106 × 400 × 86 400 = 8.3 × 1013 m✓ 1Allow ECF from (b)
Part (e)(ii)
d = r/θ = 8.3 × 1013/6.9 × 10−6✓ 1Allow ECF from (e)(i)
d = 1.2 × 1019 m✓ 1Accept 1.1–1.3 × 1019 m. Award [2] for CNA
Part (f)
L = 4πd²b = 4π × (1.2 × 1019)² × 6.0 × 10−9✓ 1Allow ECF from (e)(ii)
L = 1.1 × 1031 W✓ 1Accept 9.1 × 1030–1.3 × 1031 W or about 2.8 × 104 L☉. Award [2] for CNA
Part (g)
any one, e.g. the shell may slow down as it sweeps up surrounding gas, so its mean speed over 400 days is less than measured / the shell may not be spherical / the edge of the ring in the image is not sharp✓ 1OWTTE

Answers: (b) 2.4 × 106 m s−1  ·  (d) shift ≈ 1.0 nm  ·  (e)(i) 8.3 × 1013 m  ·  (e)(ii) 1.2 × 1019 m  ·  (f) 1.1 × 1031 W (the remaining parts are explanations — see the table above)

44C-2-33Doppler radar Paper 2Hard14 marks

A weather radar transmits pulses of microwaves of frequency 2.80 GHz and detects the microwaves reflected by raindrops. When the raindrops have a velocity component v along the radar beam, the frequency of the reflected microwaves received at the radar differs from the transmitted frequency f by Δf, where Δf = 2fv/c for v ≪ c.

(a)
(i)

Calculate the wavelength of the microwaves.

(1)
(ii)

Explain why Δf is twice the shift fv/c that the radar would detect if the raindrops were themselves sources of microwaves of frequency f.

(3)
(b)
(i)

Raindrops in one region of a storm move directly towards the radar at 18.0 m s−1. Calculate Δf, and state whether the received frequency is higher or lower than 2.80 GHz.

(2)
(ii)

In another region the received frequency is 210 Hz lower than the transmitted frequency. Determine the velocity component of these raindrops along the beam.

(2)
(c)

In a third region the wind carries the raindrops at 25.0 m s−1 in a direction at 60° to the radar beam, approaching the radar. Determine Δf, and explain why the speed deduced from Δf is less than the wind speed.

(3)
(d)

A pulse reflected from the storm returns 0.40 ms after it was transmitted. Calculate the distance from the radar to the storm.

(2)
(e)

Suggest how meteorologists could determine the full velocity of the wind in the storm.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = 3.00 × 108/2.80 × 109 = 0.107 m✓ 1
Part (a)(ii)
Moving towards the radar, a raindrop meets the wavefronts more often than f per second, so it receives a frequency shifted by about fv/c «moving observer»✓ 1Accept the argument for a receding drop.
The raindrop then re-radiates «reflects» the microwaves while moving, so the reflected wavefronts are bunched up in the direction of the radar «moving source»✓ 1
Each stage shifts the frequency by about fv/c, so the total shift is about 2fv/c✓ 1OWTTE.
Part (b)(i)
Δf = 2 × 2.80 × 109 × 18.0/3.00 × 108 = 336 Hz✓ 1
Higher, because the raindrops are approaching the radar✓ 1
Part (b)(ii)
v = cΔf/2f = 3.00 × 108 × 210/(2 × 2.80 × 109) = 11 m s−1✓ 1Accept 11.3 m s−1.
Directed away from the radar «because the received frequency is lower»✓ 1
Part (c)
Component along the beam = 25.0 cos 60° = 12.5 m s−1✓ 1
Δf = 2 × 2.80 × 109 × 12.5/3.00 × 108 = 233 Hz✓ 1Allow ECF from MP1.
Only the component of velocity along the line of sight produces a Doppler shift; the perpendicular component «21.7 m s−1» is not detected✓ 1OWTTE.
Part (d)
The pulse travels to the storm and back: d = ct/2✓ 1
d = 3.00 × 108 × 0.40 × 10−3/2 = 6.0 × 104 m «60 km»✓ 1Award [1 max] for 1.2 × 105 m.
Part (e)
Use a second radar at a different location to measure a second velocity component, and combine the two✓ 1OWTTE.

Answers: (a)(i) 0.107 m  ·  (b)(i) 336 Hz, higher  ·  (b)(ii) 11 m s−1 away from the radar  ·  (c) 233 Hz  ·  (d) 6.0 × 104 m (the remaining parts are explanations — see the table above)

45C-2-36Doppler for light Paper 2Hard11 marks

A planet orbits a star of mass 2.2 × 1030 kg. The gravitational pull of the planet makes the star move in a small circular orbit with the same period as the planet. Astronomers measure the wavelength of an absorption line in the spectrum of the star, whose laboratory wavelength is 500.00 nm. The graph shows how the change Δλ in the measured wavelength varies with time t. The Earth lies in the plane of the orbits.

Change in wavelength Δλ of a stellar absorption line against time over 7 days: a sinusoid of amplitude 3.0 × 10^-13 m and period 3.5 days, starting at zero and rising.01234567time t / days−4−3−2−101234Δλ / 10⁻¹³ m
Graph drawn to scale
(a)

Explain why the wavelength of the line varies periodically.

(2)
(b)

Determine the maximum speed of the star along the line of sight.

(2)
(c)

State, in seconds, the orbital period of the planet.

(1)
(d)

The planet moves in a circular orbit around the star. Show that the radius of the orbit of the planet is about 7.0 × 109 m.

(2)
(e)

The momentum of the planet is equal in magnitude to the momentum of the star. Determine the mass of the planet.

(3)
(f)

Suggest why the mass in (e) would be an underestimate if the Earth were not in the plane of the orbits.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The star's velocity component along the line of sight alternates between towards and away from the Earth as it moves round its orbit✓ 1
When it moves away the line is red-shifted «Δλ positive» and when it moves towards the Earth the line is blue-shifted «Δλ/λ ≈ v/c»✓ 1OWTTE.
Part (b)
Δλ = 3.0 × 10−13 m read from the graph✓ 1
v = cΔλ/λ = 3.00 × 108 × 3.0 × 10−13/500.00 × 10−9 = 180 m s−1✓ 1Award [2] for CNA.
Part (c)
3.5 days = 3.5 × 86 400 = 3.0 × 105 s✓ 1Accept 3.4–3.6 days.
Part (d)
GMm/r² = m(2π/T)²r, so r³ = GMT²/4π²✓ 1Accept v = 2πr/T with GMm/r² = mv²/r.
r = (6.67 × 10−11 × 2.2 × 1030 × (3.02 × 105)²/4π²)1/3 = 6.98 × 109 «m»✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (e)
Speed of the planet vp = 2πr/T = 2π × 7.0 × 109/3.02 × 105 = 1.45 × 105 m s−1✓ 1Allow ECF from (c).
mpvp = Mvs, so mp = 2.2 × 1030 × 180/1.45 × 105✓ 1Allow ECF from (b).
mp = 2.7 × 1027 kg✓ 1Accept 2.7–2.8 × 1027 kg. Award [3] for CNA.
Part (f)
Only a component of the star's velocity would then lie along the line of sight, so the measured speed «and so the mass» would be too small✓ 1OWTTE.

Answers: (b) 180 m s−1  ·  (c) 3.0 × 105 s  ·  (d) 6.98 × 109 m  ·  (e) 2.7 × 1027 kg (the remaining parts are explanations — see the table above)

46C-2-48Doppler for light Paper 2Hard13 marks

A small satellite moves in a circular orbit at a height of 500 km above the surface of the Earth. It transmits a radio signal of frequency 437.100 MHz. A ground station directly below the path of the satellite records the difference Δf between the received frequency and 437.100 MHz as the satellite passes overhead. Figure 1 shows how Δf varies with time t while the satellite is above the horizon; the satellite is directly overhead at t = 0. Ignore the rotation of the Earth.

Mass of the Earth = 5.97 × 1024 kg; radius of the Earth = 6.37 × 106 m.

Graph of received frequency minus 437.100 MHz, in kHz, against time from −400 s to +400 s. The curve starts at about +10.3 kHz at −346 s, falls slowly, passes steeply through zero at t = 0 and levels off near −10.3 kHz at +346 s.-400-300-200-1000100200300400time t / s-12-8-404812Δf / kHz
Figure 1 — drawn to scale
(a)

Show that the orbital speed of the satellite is about 7.6 × 103 m s−1.

(2)
(b)

State why Δf is positive at the start of the pass.

(1)
(c)

Calculate the value of Δf that would be recorded if the satellite moved directly towards the ground station at its orbital speed.

(1)
(d)
(i)

Explain why Δf is zero when the satellite is directly overhead.

(2)
(ii)

When the satellite is on the horizon, the angle θ between its velocity and the line from the satellite to the ground station is given by cos θ = R/(R + h), where R is the radius of the Earth and h is the height of the orbit. Determine whether Figure 1 is consistent with this.

(3)
(e)

A second ground station is 800 km to one side of the path of the satellite. Sketch, on Figure 1, the variation of Δf with t that this station records. The satellite is closest to this station at t = 0.

(3)
(f)

Suggest one reason why the frequencies recorded by a real ground station differ slightly from Figure 1.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
GMm/r2 = mv2/r with r = 6.37 × 106 + 5.00 × 105 = 6.87 × 106 m✓ 1Award [1 max] if r = 5.00 × 105 m or r = 6.37 × 106 m is used.
v = √(6.67 × 10−11 × 5.97 × 1024/6.87 × 106) = 7.61 × 103 «m s−1»✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (b)
The satellite is approaching the station «the distance between them is decreasing»✓ 1
Part (c)
Δf = fv/c = 437.1 × 106 × 7.61 × 103/3.00 × 108 = 11.1 kHz✓ 1Accept 11.0–11.1 kHz. Allow ECF from (a).
Part (d)(i)
Directly overhead, the velocity of the satellite is perpendicular to the line joining it to the station✓ 1
so the component of its velocity along the line of sight is zero «the distance is not changing» and there is no Doppler shift✓ 1OWTTE.
Part (d)(ii)
cos θ = 6.37/6.87 = 0.927, so the component along the line of sight = 7.61 × 103 × 0.927 = 7.06 × 103 m s−1✓ 1
Δf = f × 7.06 × 103/c = 10.3 kHz✓ 1ALTERNATIVE: (c) × cos θ = 11.1 × 0.927.
Figure 1 starts at about +10.3 kHz «and ends near −10.3 kHz», so yes, it is consistent✓ 1Accept a graph reading of 10.0–10.5 kHz. MP3 only if MP2 is scored.
Part (e)
Curve of the same shape passing through Δf = 0 at t = 0, positive before and negative after✓ 1
Smaller maximum |Δf| «less than 10.3 kHz» at the start and end✓ 1
Starting later and ending earlier «shorter pass» OR a less steep gradient at t = 0✓ 1Award MP3 for either feature.
Part (f)
The rotation of the Earth moves the ground station OR the transmitter frequency drifts «with temperature» OR the orbit is not exactly circular✓ 1Accept any one valid reason.

Answers: (a) 7.61 × 103 m s−1  ·  (c) 11.1 kHz  ·  (d)(ii) 10.3 kHz: consistent (the remaining parts are explanations — see the table above)

47C-2-74Doppler for light Paper 2Hard14 marks

In a tokamak fusion reactor, a plasma of deuterium and tritium nuclei is held at a very high temperature by magnetic fields. The plasma also contains a small number of carbon ions, which emit light of laboratory wavelength 529.05 nm. A spectrometer views the plasma. Figure 1 shows how the intensity of this line, as observed, varies with wavelength.

Mass of a carbon ion = 1.99 × 10−26 kg; mass of a deuterium ion = 3.34 × 10−27 kg.

Relative intensity against wavelength for the carbon line: a symmetric peak centred at 529.05 nm525526527528529530531532533wavelength / nm00.20.40.60.811.2relative intensity
Figure 1 — Graph drawn to scale
(a)

Explain why a very high temperature is needed for deuterium and tritium nuclei to fuse.

(2)
(b)

Explain why the random motion of the carbon ions spreads the line over a range of wavelengths.

(2)
(c)

The half-width w of the line is half its width at half the maximum intensity. Assume that w is the Doppler shift produced by an ion moving directly towards the spectrometer with speed v, where v is the speed of an ion that has the average kinetic energy of the carbon ions.

(i)

Show, using Figure 1, that v is about 4.5 × 105 m s−1.

(1)
(ii)

Determine the temperature of the plasma.

(2)
(d)

The deuterium ions are at the same temperature as the carbon ions. Deduce the ratio (speed of a deuterium ion with the average kinetic energy)/(speed of a carbon ion with the average kinetic energy).

(2)
(e)

The temperature of the plasma is increased to four times the value in (c)(ii). The total power emitted in the line is unchanged. Sketch, on Figure 1, the line that would now be observed.

(2)
(f)

The plasma also rotates around the tokamak. When the spectrometer views the plasma along a line tangential to this rotation, the centre of the line is at 529.110 nm. Determine the speed of rotation of the plasma and state its direction relative to the spectrometer.

(2)
(g)

Suggest why the temperature determined in (c)(ii) is only an estimate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
the nuclei are positively charged and repel each other «electrostatic repulsion»✓ 1
they must have very high kinetic energies to come close enough for the strong nuclear force to act «to fuse»; the average kinetic energy is proportional to the temperature✓ 1OWTTE.
Part (b)
ions moving towards the spectrometer emit light observed at shorter wavelengths «blue shift» and ions moving away at longer wavelengths «red shift»✓ 1
the ions have a range of velocity components along the line of sight, so Δλ = λv/c takes a range of values on both sides of 529.05 nm✓ 1OWTTE.
Part (c)(i)
w = 0.80 nm read from the graph AND v = cw/λ = 3.00 × 108 × 0.80/529.05 = 4.54 × 105 «m s−1»✓ 1Must see the reading of w AND full substitution OR answer to 3 s.f. Accept w = 0.78–0.82 nm.
Part (c)(ii)
½mv² = (3/2)kBT, so T = mv²/3kB✓ 1
T = 1.99 × 10−26 × (4.5 × 105)²/(3 × 1.38 × 10−23) = 9.7 × 107 K✓ 1Accept 8.7 × 107 to 1.1 × 108 K. Allow use of the unrounded speed «9.9 × 107 K». Award [2] for CNA.
Part (d)
same temperature, so the same average kinetic energy: ½mDvD² = ½mCvC²✓ 1
vD/vC = √(1.99 × 10−26/3.34 × 10−27) = 2.4✓ 1Accept 2.44. Award [1 max] for the inverse «0.41».
Part (e)
line still centred on 529.05 nm with about twice the half-width «w ≈ 1.6 nm, half-maximum points near 527.4 nm and 530.6 nm»✓ 1v ∝ √T, so doubling of width required for MP1. Accept a half-width of 1.4–1.8 nm.
maximum about half of the original height «same area under the curve»✓ 1Accept a peak between 0.4 and 0.6.
Part (f)
v = cΔλ/λ = 3.00 × 108 × 0.060/529.05 = 3.4 × 104 m s−1✓ 1Accept 3 × 104 m s−1.
away from the spectrometer «the line is red-shifted»✓ 1
Part (g)
only the component of each ion's velocity along the line of sight shifts the wavelength, whereas (3/2)kBT is the average kinetic energy for motion in three dimensions OR the ions have a wide range of speeds, so w does not correspond exactly to the speed of an ion with the average kinetic energy✓ 1Accept any one valid reason, e.g. the temperature is not the same everywhere along the line of sight.

Answers: (c)(ii) T ≈ 9.7 × 107 K «9.9 × 107 K with the unrounded speed»  ·  (d) 2.4  ·  (f) 3.4 × 104 m s−1 away from the spectrometer (the remaining parts are explanations — see the table above)

48IBP2S-026-00-16Doppler radar Paper 2Hard8 marks

A Doppler weather radar emits microwaves of frequency 2.80 GHz towards a rotating thunderstorm. Two regions of rain, P and Q, lie on opposite sides of the storm's centre, each the same distance from it. They are 10.0 km apart, on a line through the centre that is perpendicular to the radar beam. The reflected signal from region P has a frequency 560 Hz higher than the emitted frequency. The reflected signal from region Q has a frequency 373 Hz lower than the emitted frequency.

radarbeamPQcentre10.0 km
Diagram not to scale. Rotating storm viewed from above.
(a)

Explain why the frequency shift of the reflected signal is approximately 2fv/c, where v is the speed of the rain along the beam.

(2)
(b)

Calculate the speed of the rain along the beam in region P and in region Q, stating the direction of each.

(2)
(c)

The storm both moves towards the radar as a whole and rotates. Determine the speed of the storm as a whole and the rotational speed of the rain at P and Q.

(2)
(d)

Estimate the time for one rotation of the storm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The raindrops act as moving receivers, so they detect a frequency shifted by about fv/c . They then re-emit as moving sources, which adds a further shift of about fv/c, giving a total of 2fv/c for v ≪ c .✓ 2
Part (b)
P: v = cΔf/(2f) = 3.00×108 × 560/(2 × 2.80×109) = 30.0 m s−1, towards the radar . Q: v = 3.00×108 × 373/(5.60×109) = 20.0 m s−1, away from the radar .✓ 2
Part (c)
Take towards the radar as positive. With storm speed u and rotational speed w: u + w = 30.0 and u − w = −20.0, so u = 5.0 m s−1 towards the radar . w = 25 m s−1 . Allow ECF from (b) for both marks.✓ 2
Part (d)
The radius is 5.0 km, so T = 2πr/w = 2π × 5000/25 . T ≈ 1.26×103 s, about 21 minutes . Allow ECF from (c).✓ 2

Exam tip: When two opposite regions share a common motion plus a rotation, half the sum and half the difference of their line-of-sight velocities separate the two motions.