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Edexcel A Level Physics · Topic 5
Waves and the Particle Nature of Light
FREE Online Physics A Level Course130minutes
112marks
16questions
How it works
- Answer each question. Type into the boxes or pick an option, and use paper for working and drawings.
- Press Check my answers at the end of the question. Tick-box and number answers are marked for you.
- For written answers, compare yours with the mark scheme and tick each point you made. We highlight the key words we spot, but you decide, just like an examiner.
You will need: paper, a pen, a ruler and a calculator. Use c = 3.00 × 10⁸ m s⁻¹, h = 6.6 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C. It’s a long test: your answers save, so you can do it in two sittings. Your answers save automatically in this browser.
1
Question 1
Multiple choice12 marksTwelve quick questions from across the topic. Each part is worth 1 mark.
(a)
A sound is investigated with a microphone and an oscilloscope. On the x-axis, 1 division represents 5 ms. The frequency of the sound wave is
Oscilloscope trace
(1 mark)
Mark scheme
Key equationf = 1 ÷ Tone cycle = 4 divisions = 20 ms = 0.020 sA and B come from using milliseconds as if they were seconds.
Answer: C 50 Hz
(b)
The diagram represents a stationary wave on a string. Which diagram correctly shows the positions of the nodes N and antinodes A?
The stationary wave
(1 mark)
Mark scheme
Both fixed ends and the crossing point in the middle have zero amplitude, so all three are nodes. The antinodes are at the middle of each loop.
Answer: C
(c)
Which of the following would increase the amount of detail (resolution) in an ultrasound scan?
(1 mark)
Mark scheme
Shorter pulses let the echoes from close-together boundaries be told apart. A, B and C all mean a longer wavelength, which gives less detail.
Answer: D shorter pulses
(d)
The diagram shows how an image is formed of an object a small distance from a thin converging lens. Which label represents the focal point of the lens?
Ray diagram
(1 mark)
Mark scheme
A ray through the focal point leaves the lens parallel to the axis: that’s the ray through A. B is the centre of the lens; D is the image.
Answer: A
(e)
Two polarising filters X and Y sit between a source of unpolarised light and a detector. Their planes of polarisation start perpendicular. Which rotations give the maximum intensity at the detector?
The filters
(1 mark)
Mark scheme
In C the planes end up 270° (the same as 90°) further apart than they started, so they are now parallel. In A, B and D they are still crossed.
Answer: C X: 180° clockwise · Y: 90° anticlockwise
(f)
The effect of diffraction is more noticeable, in everyday life, with sound than with light. This is because
(1 mark)
Mark scheme
Diffraction is greatest when the gap is about the same size as the wavelength. Doorways are close to the wavelength of sound, but huge compared with light.
Answer: A sound has a much longer wavelength than light.
(g)
A beam of electrons is directed at a sheet of crystalline material, and the pattern shown appears on the screen. This pattern demonstrates
The arrangement
The pattern
(1 mark)
Mark scheme
Electrons passing through the gaps between atoms diffract: evidence that particles behave as waves.
Answer: A diffraction.
(h)
Which experiment provides evidence for the particle nature of light?
(1 mark)
Mark scheme
A shows electrons behaving as waves; B and D show light behaving as a wave.
Answer: C The ejection of photoelectrons from zinc when ultraviolet light falls on it
(i)
Which characteristic of the photoelectric effect cannot be explained by the wave theory of light?
(1 mark)
Mark scheme
Key equationhf = φ + ½mv²maxbelow the threshold, one photon never has enough energyWaves would deliver energy continuously, so brighter light should eventually free electrons. It doesn’t.
Answer: C Increasing the intensity of light below the threshold frequency does not result in the emission of photoelectrons.
(j)
Which of the following can be explained only by the wave nature of electromagnetic radiation?
(1 mark)
Mark scheme
B is about electrons, which aren’t electromagnetic radiation. A and C need photons.
Answer: D X-ray diffraction
(k)
Which row correctly matches properties of light to the model that best explains them?
(1 mark)
Mark scheme
Diffraction needs waves; line spectra need photons with set energies (ΔE = hf).
Answer: B Wave: diffraction · Particle: atomic line spectra
(l)
The diagram shows five energy levels in an atom. Which transition would be caused by the absorption of the lowest frequency of radiation?
Energy levels
(1 mark)
Mark scheme
Key equationΔE = hfsmallest gap → lowest frequencyD is the same gap, but going down is emission, not absorption.
Answer: C 4 to 5
2
Question 2
4 marks(a)
Some radio signals have a frequency of 218.6 MHz. Calculate their wavelength.
m
(2 marks)
Mark scheme
Key equationv = fλ
Wavelength: 1.37 m (3.00 × 10⁸ ÷ 2.186 × 10⁸)
💡 Mega means × 10⁶: use 218.6 × 10⁶ Hz.
💡 Upside down: λ = c ÷ f.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
(b)(i)
State what is meant by frequency.
(1 mark)
Mark scheme
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(b)(ii)
State what is meant by wavelength.
(1 mark)
Mark scheme
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3
Question 3
14 marksA string is fixed at one end and attached to a vibration generator at the other. When the generator is switched on, the wave shown is produced.
The string
*(a)
Name the type of wave produced on the string and explain how it has been formed.
Marks are given for a clear, logically organised answer using technical wording.
(4 marks)
Mark scheme
Key equationsuperposition: y = y1 + y2
Tick each marking point your answer makes:
(b)
The string between the generator and the fixed end is 1.8 m long and vibrates at 330 Hz. Calculate the speed of the waves on the string.
m s⁻¹
(3 marks)
Mark scheme
Key equationv = fλone loop is half a wavelength
Speed: 1200 m s⁻¹ (λ = 2 × 1.8 = 3.6 m; 330 × 3.6 = 1188)
💡 One loop is only half a wavelength, so λ = 2 × 1.8 m = 3.6 m.
💡 You divided. v = f × λ.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(c)(i)
The frequency is changed and the new wave is shown. The student can touch the string at X without disturbing the pattern. Explain why.
The new wave
(2 marks)
Mark scheme
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(c)(ii)
Calculate the new frequency of the vibration generator.
Hz
(1 mark)
Mark scheme
New frequency: 660 Hz (two loops: half the wavelength, double the frequency)
💡 Two loops means a shorter wavelength, so a higher frequency.
(c)(iii)
The string is lit by a strobe lamp, without changing the generator’s frequency. A section of the string now looks like two separate strands. Calculate the maximum possible strobe frequency that gives two strands.
Under the strobe lamp
Hz
(1 mark)
Mark scheme
Maximum strobe frequency: 1320 Hz (the string is caught twice per cycle: 2 × 660; error carried forward from (ii))
💡 At 660 Hz the string is caught once per cycle, in the same place: one strand. Two strands means twice per cycle.
💡 Use the new frequency from (ii), and the string must be caught twice per cycle.
Explain why this is a maximum.
(1 mark)
Mark scheme
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(d)
The student reads the frequency from the scale on the dial shown. Explain a disadvantage of this method of measuring the frequency.
The dial (Hz)
(2 marks)
Mark scheme
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4
Question 4
7 marksAn ultrasound source and two detectors are set up as shown, with the detectors connected to an oscilloscope. Photograph 1 shows the traces. Detector 2 is then moved slightly further from the source, giving photograph 2.
The arrangement
Photograph 1
Photograph 2
(a)
Explain the change in the traces between photograph 1 and photograph 2.
(3 marks)
Mark scheme
Key equationpath difference = (n + ½)λantiphase
Tick each marking point your answer makes:
(b)
Detector 2 is moved back, then steadily away from the source. Photograph 2 then photograph 1 appear alternately until nine cycles have been seen. Photographs 3 and 4 show the start and finish on a metre rule. The ultrasound frequency is 40.0 kHz. Calculate the speed of sound in air.
Photograph 3: start
Photograph 4: finish
m s⁻¹
(4 marks)
Mark scheme
Key equationv = fλ
Speed of sound: 329 m s⁻¹ (7.4 cm ÷ 9 = 0.822 cm; 40 000 × 0.00822)
💡 That used 7.4 cm as one wavelength. Nine cycles were seen, so λ = 7.4 ÷ 9.
💡 Your wavelength is still in cm: change it to metres.
💡 You may have read 15.4 cm and 8.0 cm the wrong way round or used kHz as Hz. λ = 0.00822 m, f = 40 000 Hz.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 3):
5
Question 5
4 marksThe photograph shows a kidney stone removed from a patient. It was detected using pulses of 2.5 MHz ultrasound. The speed of sound in body tissue is 1500 m s⁻¹.
The kidney stone
Justify whether this was a suitable frequency for detecting this kidney stone. First, calculate the wavelength.
m
(2 marks)
Mark scheme
Key equationv = fλ
Wavelength: 6 × 10⁻⁴ m (1500 ÷ 2.5 × 10⁶)
💡 Mega means × 10⁶: 1500 ÷ 2 500 000.
💡 Upside down: λ = v ÷ f.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
Now use the photograph and your wavelength to justify whether 2.5 MHz was suitable.
(2 marks)
Mark scheme
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6
Question 6
4 marksA small camera uses a thin converging lens. When it focuses on a distant scene, the distance from the lens to the detector is 4.0 cm.
(a)
Calculate the power of the lens in dioptres.
D
(1 mark)
Mark scheme
Key equationP = 1 ÷ ff in metres
Power: 25 D (1 ÷ 0.040 m)
💡 The focal length must be in metres: 4.0 cm = 0.040 m.
(b)
How far must the lens be able to move if it can focus on objects at any distance between 2.0 m and infinity?
mm
(3 marks)
Mark scheme
Key equation1/u + 1/v = 1/f
Distance the lens moves: 0.8 mm (1/v = 25 − 0.5, v = 40.8 mm; 40.8 − 40.0)
💡 Right, but in metres. The box wants mm: × 1000.
💡 That’s the new image distance. The lens moves by the difference from 40.0 mm.
💡 That’s for an object at 1.5 m. Use 2.0 m.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
7
Question 7
3 marksA child’s nature observation kit has a lens in its lid and an optional extra lens. The photographs show a fly with no lens, one lens and both lenses. A student measured these focal lengths:
| Lens | Focal length / cm |
|---|---|
| Lens in the lid | 12 |
| Optional lens | 17.5 |
| Both lenses together | 7.0 |
The kit
No lens, one lens, both lenses
A fly is 5.0 cm from the combined lenses (focal length 7.0 cm). Calculate the magnification.
(3 marks)
Mark scheme
Key equation1/u + 1/v = 1/freal is positiveKey equationm = v ÷ u
Magnification: 3.5 (1/v = 1/7.0 − 1/5.0, v = −17.5 cm; 17.5 ÷ 5.0)
💡 Right size: the minus sign just says the image is virtual. Tick the box.
💡 That used 1/v = 1/7 + 1/5. Rearrange: 1/v = 1/f − 1/u.
💡 That’s f ÷ u. Find the image distance first, then m = v ÷ u.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
8
Question 8
2 marksA telescope’s objective is a converging lens of power 0.820 D. It forms an image at P, which the eyepiece turns into a final image at Q.
The telescope (not to scale)
Show that the focal length of the objective lens is about 1200 mm.
mm
(2 marks)
Mark scheme
Key equationP = 1 ÷ f
Focal length: 1220 mm (1 ÷ 0.820 = 1.22 m). A “show that” needs more figures than the 1200 given.
💡 Right, in metres. The box wants mm: 1220.
💡 For a “show that”, give more figures than the question: 1220 mm.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
9
Question 9
8 marks(a)
State what is meant by the principle of superposition of waves.
(2 marks)
Mark scheme
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(b)(i)
A monochromatic light source lights a narrow single slit, then a double slit. Dark and light bands form on the screen.
Double slit
* Explain how the dark and light bands are formed by light from the two slits.
Marks are given for a clear, logically organised answer using technical wording.
(3 marks)
Mark scheme
Key equationpath difference = nλ (bright) (n + ½)λ (dark)
Tick each marking point your answer makes:
(b)(ii)
Polarising filters are placed behind the slits. With their planes parallel, the bands are still seen. When one filter is turned through 90°, there are no dark bands and the screen is lit evenly. Explain why.
With polarising filters
(3 marks)
Mark scheme
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10
Question 10
15 marksSome solutions rotate the plane of polarisation of polarised light, by an amount that depends on their concentration. A polarimeter measures that rotation.
(a)
Describe what is meant by the plane of polarisation of polarised light.
(2 marks)
Mark scheme
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(b)(i)
In step 1 there is no solution in the tube. In step 2 the analyser is rotated until light from the source can no longer be seen. Explain why.
A polarimeter
(2 marks)
Mark scheme
Key equationI = I0 cos² θθ = 90° gives zero
Tick each marking point your answer makes:
(b)(ii)
The solution is added (step 3) and the analyser rotated until no light is seen again (step 4). The angle turned is 55°. The plane of polarisation may have been rotated by more than 55°: suggest another possible angle of rotation.
°
(1 mark)
Mark scheme
Another possible angle: 235° (55 + 180; 415° also scores: use the “means the same” box)
💡 That’s 180 − 55. The plane repeats every 180° added on: 55 + 180.
Explain why the plane of polarisation may have been rotated by more than 55°.
(2 marks)
Mark scheme
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(b)(iii)
The experiment would give the same results using maximum intensity in steps 2 and 4. Suggest the advantage of turning the analyser until no light is seen.
(2 marks)
Mark scheme
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(c)
A sodium lamp is often used as the light source, because its two bright lines at 589.0 nm and 589.6 nm make it effectively monochromatic. Explain how the spectral lines are created with specific wavelengths.
(6 marks)
Mark scheme
Key equationE = hf = hc ÷ λphoton energy = difference between two levels
Tick each marking point your answer makes:
11
Question 11
10 marksIn a microwave oven, microwaves are absorbed by water molecules, increasing the internal energy of the food.
(a)
A leakage detector checks that no microwaves escape from the oven. Suggest why microwaves leaking to the surroundings could be dangerous to people.
Leakage detector
(2 marks)
Mark scheme
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(b)(i)
The door is glass with a layer of metal full of small holes, so the food can be seen but microwaves are kept in. It has been suggested that, because of diffraction, light can pass through the holes but microwaves cannot. Explain what is meant by diffraction.
The oven door
(2 marks)
Mark scheme
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(b)(ii)
Calculate the wavelength of the microwaves. Microwave frequency = 2.5 GHz.
m
(2 marks)
Mark scheme
Key equationv = fλ
Wavelength: 0.12 m (3.0 × 10⁸ ÷ 2.5 × 10⁹)
💡 Giga means × 10⁹: 2.5 GHz = 2.5 × 10⁹ Hz.
💡 Upside down: λ = c ÷ f.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
(b)(iii)
Use the close-up of the door and the ruler to determine the diameter of the holes.
Close-up of the door
mm
(1 mark)
Mark scheme
Diameter: 2 mm
💡 Right, in cm. The box wants mm.
💡 That’s the distance from one hole to the next. Measure across one hole.
(b)(iv)
* Discuss the suggestion that, because of diffraction, light can pass through the holes but microwaves cannot.
Marks are given for a clear, logically organised answer using technical wording.
(3 marks)
Mark scheme
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12
Question 12
5 marksIn a spectrometer, light from hot gas passes through a diffraction grating, giving the line spectrum shown.
The line spectrum
(a)
A line of wavelength 650 nm is seen in the first order at 19.9° from the zero order. Calculate the number of lines per metre of the grating.
m⁻¹
(3 marks)
Mark scheme
Key equationnλ = d sin θ
Lines per metre: 5.2 × 10⁵ m⁻¹ (d = 650 × 10⁻⁹ ÷ sin 19.9° = 1.91 × 10⁻⁶ m; 1 ÷ d)
💡 That’s the slit spacing d. Lines per metre = 1 ÷ d.
💡 You used cos. The grating equation uses sin θ.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(b)
Explain one precaution that could make the measurement of the angle more accurate.
(2 marks)
Mark scheme
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Either:
Or:
13
Question 13
4 marksThe diagram shows an arrangement used to demonstrate the diffraction of electrons.
Electron diffraction tube
(a)
On paper, sketch the pattern seen on the fluorescent screen, then compare it with the answer.
(1 mark)
Mark scheme
Compare your sketch with the answer below
Describe the pattern: what causes the bright and dark rings?
(1 mark)
Mark scheme
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(b)
State and explain how the pattern would change if the accelerating voltage were increased.
(2 marks)
Mark scheme
Key equationλ = h ÷ pfaster electrons, more momentum, shorter wavelength
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⚠️ Faster electrons have a shorter wavelength, so they diffract less: the rings shrink.
14
Question 14
6 marksExperiments on the photoelectric effect show that the kinetic energy of the photoelectrons depends on the frequency of the light, not its intensity, and that light below a threshold frequency cannot release photoelectrons.
How do these conclusions support a particle theory, but not a wave theory, of light?
(6 marks)
Mark scheme
Key equationhf = φ + ½mv²maxone photon, one electron
Any 6 of these, 1 mark each. Tick what you wrote (max 6):
15
Question 15
9 marksThe table gives the work functions of several metals. h = 6.6 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C, c = 3.0 × 10⁸ m s⁻¹.
| Metal | Work function / eV |
|---|---|
| Potassium | 2.2 |
| Zinc | 4.2 |
| Iron | 4.5 |
| Silver | 4.7 |
(a)
Explain what is meant by the work function.
(1 mark)
Mark scheme
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(b)
Convert the work function of zinc to joules.
J
(1 mark)
Mark scheme
Work function of zinc: 6.7 × 10⁻¹⁹ J (4.2 × 1.6 × 10⁻¹⁹)
💡 You divided. eV → J: multiply by 1.6 × 10⁻¹⁹.
(c)
Which of the metals has the highest threshold frequency?
(1 mark)
Mark scheme
Metal: Silver
Explain your answer.
(1 mark)
Mark scheme
Key equationφ = hf0
Tick each marking point your answer makes:
(d)(i)
A potassium photocell is lit by violet light of wavelength 420 nm. Calculate the energy of one photon, in J and in eV.
The photocell circuit
J
eV
(2 marks)
Mark scheme
Key equationE = hf = hc ÷ λ
Photon energy: 4.71 × 10⁻¹⁹ J (6.6 × 10⁻³⁴ × 3.0 × 10⁸ ÷ 420 × 10⁻⁹)
💡 Nano means × 10⁻⁹: 420 nm = 420 × 10⁻⁹ m.
Photon energy: 2.95 eV (4.71 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹)
(d)(ii)
Calculate the maximum kinetic energy of the photoelectrons emitted from the potassium (work function 2.2 eV).
eV
(2 marks)
Mark scheme
Key equationhf = φ + ½mv²max
Maximum kinetic energy: 0.75 eV (1.2 × 10⁻¹⁹ J) (2.95 − 2.2)
💡 You added. KEmax = hf − φ.
💡 Right, in joules. The box wants eV: 0.75.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
(d)(iii)
Calculate the stopping voltage for this experiment.
V
(1 mark)
Mark scheme
Key equationeVs = KEmaxin eV, the number is the same
Stopping voltage: 0.75 V
16
Question 16
5 marksThe energy of the nth level of the hydrogen atom is En = −13.6 eV ÷ n².
(a)
Calculate the energies of the n = 1 and n = 3 levels.
eV
eV
(2 marks)
Mark scheme
E1: −13.6 eV (ignore the sign)
💡 The sign is ignored here. Tick the box.
E3: −1.51 eV (−13.6 ÷ 9)
💡 The sign is ignored here. Tick the box.
💡 Square n: divide by 9, not 3.
(b)
Calculate the wavelength of the line emitted when electrons fall from n = 3 to n = 1.
m
(3 marks)
Mark scheme
Key equationE = hc ÷ λ
Wavelength: 1.0 × 10⁻⁷ m (12.1 eV = 1.93 × 10⁻¹⁸ J; λ = hc ÷ E = 1.02 × 10⁻⁷ m)
💡 Convert the 12.1 eV to joules first: × 1.6 × 10⁻¹⁹.
💡 That’s 13.6 eV on its own. Use the difference: 13.6 − 1.51 = 12.1 eV.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
Your results
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| Question | Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Q9 | Q10 | Q11 | Q12 | Q13 | Q14 | Q15 | Q16 | Total |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Out of | 12 | 4 | 14 | 7 | 4 | 4 | 3 | 2 | 8 | 15 | 10 | 5 | 4 | 6 | 9 | 5 | 112 |
| Your mark | – | – | – | – | – | – | – | – | – | – | – | – | – | – | – | – | 0 |
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