C.5 Doppler Effect Questions | IB Physics SL 2025

1C-2-14

Wavefront diagrams·C.5 Doppler effect

Paper 2Easy11 marks

Short answer & extended response11 steps to full marksExplain

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.

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 point	Mark	Notes
Part (a)
circumference = 2π × 1.20 = 7.54 m	✓ 1
v = 7.54 × 2.0 = 15 m s⁻¹	✓ 1	Award [2] for CNA
Part (b)
H at C «where the buzzer moves directly towards L»	✓ 1	Accept a position on the circle near C
W at A «where the buzzer moves directly away from L»	✓ 1	Accept 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	✓ 1	OWTTE
Part (d)
2	✓ 1	The 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	✓ 1	MP2 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	✓ 1	OWTTE

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

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving (with A.2 — motion along a circular path) Command term: Explain

2C-2-15

Doppler for light·C.5 Doppler effect

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksDetermine

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.

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 point	Mark	Notes
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	✓ 1	OWTTE
Part (b)
Δλ = (490.0 − 482.2)/2 = 3.9 nm «half the width of the line»	✓ 1	Accept 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	✓ 1	Accept 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	✓ 1	OWTTE
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»	✓ 1	Accept 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	✓ 1	Allow ECF from (b)
Part (e)(ii)
d = r/θ = 8.3 × 10¹³/6.9 × 10⁻⁶	✓ 1	Allow ECF from (e)(i)
d = 1.2 × 10¹⁹ m	✓ 1	Accept 1.1–1.3 × 10¹⁹ m. Award [2] for CNA
Part (f)
L = 4πd²b = 4π × (1.2 × 10¹⁹)² × 6.0 × 10⁻⁹	✓ 1	Allow ECF from (e)(ii)
L = 1.1 × 10³¹ W	✓ 1	Accept 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	✓ 1	OWTTE

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)

Syllabus understandingC.5 — that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space; the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer, as given by Δf/f = Δλ/λ ≈ v/c (with B.1 — the concept of apparent brightness as given by b = L/4πd²) Command term: Determine

3C-1A-40

Doppler for light·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.5 — the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer, as given by Δf/f = Δλ/λ ≈ v/c Command term: Determine

4C-1A-41

Red shift & blue shift·C.5 Doppler effect

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.5 — the relative change in frequency or wavelength observed for a light wave due to the Doppler effect as given by Δf/f = Δλ/λ ≈ v/c Command term: Determine

5C-1A-42

Doppler effect for sound·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark1 step to full marksDescribe

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

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves Command term: Describe

6C-1A-43

Wavefront diagrams·C.5 Doppler effect

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDescribe

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?

Diagram NOT accurately drawn

Observer AObserver B

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.5 — the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving Command term: Describe

7C-1A-44

Red shift & blue shift·C.5 Doppler effect

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.5 — that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space Command term: Determine

8C-1B-18

Red shift & blue shift·C.5 Doppler effect

Paper 1BHard11 marks

Data-based question11 steps to full marksDetermine

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 / day	λ / nm ± 0.005 nm	v / km s⁻¹
0.0	656.223	−26
1.1	656.290	4.6
2.0	656.360	37
3.2	656.407
4.1	656.380	46
5.0	656.312	15
6.1	656.229	−23
7.0	656.209	−32
8.2	656.256	−11
9.0	656.315	16
10.1	656.389	50
11.0	656.401	55

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 point	Mark	Notes
Part (a)
Δv = cΔλ/λ = 3.00 × 10⁸ × 0.005/656.28 = 2290 m s⁻¹ ≈ 2 km s⁻¹	✓ 1	Must 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	✓ 1	Accept 58.1 km s⁻¹.
The wavelength is longer than the laboratory value (red shift), so S is moving away from the Earth	✓ 1	Must 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	✓ 1	Do 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	✓ 1	Accept the value of v on the curve's centre line.
≈ +12 km s⁻¹, i.e. the system is moving away from the Earth	✓ 1	Accept +8 to +16 km s⁻¹. Direction needed.
Part (d)(ii)
Half the peak-to-peak variation: (57 − (−33))/2 ≈ 45 km s⁻¹	✓ 1	Accept 42 to 48 km s⁻¹.
Part (e)
Period from the graph (e.g. time between successive maxima) ≈ 7.6 day	✓ 1	Accept 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	✓ 1	Accept 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	✓ 1	OWTTE.

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)

Syllabus understandingC.5 — the relative change in frequency or wavelength observed for a light wave due to the Doppler effect as given by Δf/f = Δλ/λ ≈ v/c; that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space Command term: Determine

9C-1B-20

Doppler for light·C.5 Doppler effect

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

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 = λ_west − λ_east. 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 / nm	Δλ / pm ± 0.3 pm
430.8	6.0
518.4	7.2
589.0	7.9
617.3	8.2
656.3	8.8
854.2	11.8

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 point	Mark	Notes
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	✓ 1	Uses Δλ/λ ≈ v/c.
The difference is therefore 2vλ/c, which is proportional to λ with gradient 2v/c	✓ 1	OWTTE.
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	✓ 1	Accept 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	✓ 1	Accept 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	✓ 1	Accept ±0.0003 pm nm⁻¹ on each reading.
Δv ≈ ±180 m s⁻¹, so v = (2.04 ± 0.18) km s⁻¹	✓ 1	Accept ±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	✓ 1	Allow 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	✓ 1	Must 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	✓ 1	Accept 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)

10C-2-33

Doppler radar·C.5 Doppler effect

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksExplain

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 point	Mark	Notes
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»	✓ 1	Accept 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	✓ 1	OWTTE.
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	✓ 1	Accept 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	✓ 1	Allow 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	✓ 1	OWTTE.
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»	✓ 1	Award [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	✓ 1	OWTTE.

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)

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer as given by Δf/f = Δλ/λ ≈ v/c Command term: Explain

11C-2-34

Red shift & blue shift·C.5 Doppler effect

Paper 2Easy12 marks

Short answer & extended response12 steps to full marksExplain

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.

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 point	Mark	Notes
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	✓ 1	OWTTE.
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	✓ 1	Accept 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»	✓ 1	Accept 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»	✓ 1	Accept '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	✓ 1	OWTTE.
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	✓ 1	Award [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)

Syllabus understandingC.5 — that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space; the relative change in frequency or wavelength observed for a light wave due to the Doppler effect as given by Δf/f = Δλ/λ ≈ v/c; E.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions Command term: Explain

12C-2-35

Wavefront diagrams·C.5 Doppler effect

Paper 2Medium10 marks

Short answer & extended response10 steps to full marksDeduce

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.

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 point	Mark	Notes
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	✓ 1	Accept 0.35–0.40 m. Measurement across several wavefronts expected.
Behind: λ = 0.63 m	✓ 1	Accept 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»	✓ 1	ALTERNATIVE: successive dots are 0.125 m apart, covered in one period. OR λ_behind − λ_ahead = 2uT.
u = 0.50/(4 × 1.47 × 10⁻³) = 85 m s⁻¹	✓ 1	Accept 80–90 m s⁻¹. Allow ECF from (b)(i) and (b)(ii).
Part (c)
Greater than 680 Hz	✓ 1	MP1 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»	✓ 1	Allow 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	✓ 1	OWTTE.
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	✓ 1	Accept 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)

Syllabus understandingC.5 — the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving; the nature of the Doppler effect for sound waves and electromagnetic waves Command term: Deduce

13C-2-36

Doppler for light·C.5 Doppler effect

Paper 2Hard11 marks

Short answer & extended response11 steps to full marksDetermine

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.

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 point	Mark	Notes
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»	✓ 1	OWTTE.
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	✓ 1	Award [2] for CNA.
Part (c)
3.5 days = 3.5 × 86 400 = 3.0 × 10⁵ s	✓ 1	Accept 3.4–3.6 days.
Part (d)
GMm/r² = m(2π/T)²r, so r³ = GMT²/4π²	✓ 1	Accept 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»	✓ 1	Must see full substitution OR answer to at least 3 s.f.
Part (e)
Speed of the planet v_p = 2πr/T = 2π × 7.0 × 10⁹/3.02 × 10⁵ = 1.45 × 10⁵ m s⁻¹	✓ 1	Allow ECF from (c).
m_pv_p = Mv_s, so m_p = 2.2 × 10³⁰ × 180/1.45 × 10⁵	✓ 1	Allow ECF from (b).
m_p = 2.7 × 10²⁷ kg	✓ 1	Accept 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	✓ 1	OWTTE.

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)

Syllabus understandingC.5 — that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space; the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer as given by Δf/f = Δλ/λ ≈ v/c; D.1 — Newton's universal law of gravitation as given by F = Gm1m2/r²; A.2 — circular motion is caused by a centripetal force acting perpendicular to the velocity; the conservation of linear momentum Command term: Determine

C.5 Doppler effect: IB Physics SL exam-style questions

13 practice questions on C.5 Doppler effect

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