Edexcel International A Level Physics · Unit 2 · Waves and Electricity
Diffraction and the wave nature of electrons
40 free, high-quality past-paper-style questions on Diffraction and the wave nature of electrons for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 135 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U2-4-A-01Unit 2 · ASEasy1 mark
Plane water waves in a ripple tank pass through a gap in a barrier. The waves spread out only slightly beyond the gap. Which change would make the waves spread out more?
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Answer
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Notes
Spreading is greatest when the gap width is similar to the wavelength. The gap is at present much wider than λ, so narrowing the gap towards λ increases the spreading. A higher frequency would shorten λ and reduce spreading; amplitude has no effect on the angle of spreading.
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Answer D
Answer: D
Exam tip: The amount of diffraction depends on the ratio of wavelength to gap width. Ask whether a change brings the gap closer to λ.
Huygens' construction is used to predict the position of a wavefront a short time later. Which statement describes the construction?
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Huygens' construction treats every point on a wavefront as a source of secondary wavelets spreading at the wave speed; the new wavefront is the line that touches all of them (their envelope).
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Answer A
Answer: A
Exam tip: A complete Huygens statement has three parts: points on the wavefront are sources of wavelets; the wavelets spread at the wave speed; the new wavefront is their envelope.
In a school laboratory, microwaves of wavelength 2.8 cm from a transmitter pass through a gap of width 3.0 cm between two metal plates. A receiver is moved around the far side of the plates. c = 3.00 × 108 m s−1.
Figure: microwaves approaching the gap between two metal plates (not to scale)
(a)
Calculate the frequency of the microwaves.
(1)
(b)
Use Huygens' construction to explain what happens to the microwaves beyond the gap.
(3)
(c)
The gap is widened to 30 cm. State and explain how the signal detected well to one side of the gap changes.
(1)
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Part (a)
f = c/λ = 3.00 × 108 ÷ 0.028 = 1.1 × 1010 Hz
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Part (b)
each point on the wavefront in the gap acts as a source of secondary wavelets
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the new wavefront is the envelope of the wavelets / the line touching all the wavelets
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the gap is similar in size to the wavelength, so the wavefronts become (nearly) semicircular and the microwaves spread into the region behind the plates
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Part (c)
the signal to the side becomes weaker, because the gap is now much wider than the wavelength so there is little diffraction / spreading
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Exam tip: Finish a Huygens explanation by comparing the gap with the wavelength — that comparison decides how much spreading there is.
An electron travels at a speed of 3.2 × 106 m s−1. e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s.
(a)
Calculate the momentum of the electron.
(1)
(b)
Calculate the de Broglie wavelength of the electron.
(2)
(c)
The atoms in a nickel crystal are about 2.5 × 10−10 m apart. State whether a beam of these electrons would be diffracted noticeably by the crystal, giving a reason.
(1)
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Part (a)
p = mv = 9.11 × 10−31 × 3.2 × 106 = 2.9 × 10−24 kg m s−1 (2.92 × 10−24)
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Part (b)
λ = h/p = 6.63 × 10−34 ÷ 2.92 × 10−24
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= 2.3 × 10−10 m (allow error carried forward)
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Part (c)
yes, because the wavelength is similar to (about the same as) the atomic spacing
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Exam tip: Keep the unrounded momentum in your calculator for the next step, and give the final wavelength to the same s.f. as the data.
Figure 1 shows four ripple tank set-ups drawn to scale on a grid of 1.0 cm squares. In each one, straight wavefronts travel towards a gap in a barrier. In which set-up will the waves spread out most after passing through the gap?
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Answer
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Notes
Read the wavelength (spacing of the wavefronts) and the gap width from the grid, then compare them. λ ÷ gap: set-up 1, 1.0 ÷ 2.0 = 0.50; set-up 2, 2.0 ÷ 8.0 = 0.25; set-up 3, 3.0 ÷ 7.0 = 0.43; set-up 4, 2.0 ÷ 3.0 = 0.67. The gap is closest to the wavelength in set-up 4, so the spreading is greatest there.
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Answer D
Answer: D
Common error: Judging by the gap width alone, or by the wavelength alone. The amount of spreading depends on how the gap compares with λ, so work out λ ÷ gap for each case.
Light of wavelength 589 nm is incident normally on a diffraction grating. Figure 1 shows the beams travelling from the grating to the central maximum and to the two first-order maxima. How many lines per mm does the grating have?
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The 48° angle is between the two first-order beams, so θ = 48° ÷ 2 = 24°. d = λ ÷ sin θ = 589 × 10−9 ÷ sin 24° = 1.45 × 10−6 m. 1/d = 6.91 × 105 lines per m = 691 lines per mm.
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Answer B
Answer: B
Common error: Using the angle between the two first-order beams as θ. In nλ = d sin θ, θ is measured from the straight-through direction, so halve it.
Figure 1 shows the patterns produced when a beam of X-rays, and then a beam of electrons, pass through the same thin metal foil. The two patterns are drawn to the same scale. Which conclusion can be drawn from the figure?
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Both patterns have rings of radius 12 mm and 21 mm, so both beams are spread through the same angles by the same atoms. The angles of a diffraction pattern depend on the wavelength compared with the atomic spacing, so the electrons have the same wavelength as the X-rays. The electron pattern is a diffraction pattern, showing wave behaviour.
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Answer A
Answer: A
Exam tip: Identical ring sizes from the same foil mean identical wavelengths. Comparing with X-rays, already known to be waves, shows that the electron rings are a diffraction effect.
Figure 1 shows a plan view of straight water waves in a ripple tank travelling towards a gap in a barrier. The bar that makes the waves vibrates with a frequency of 12 Hz.
(a)
Use the figure to determine the wavelength of the water waves.
(1)
(b)
Calculate the speed of the waves.
(1)
(c)
Describe the wavefronts beyond the gap and explain their shape using Huygens' construction.
(2)
(d)
The frequency is increased to 24 Hz. The speed of the waves does not change. State and explain how the spreading beyond the gap changes.
(1)
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Answer
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Part (a)
4.8 cm spans four wavelengths (five wavefronts), so λ = 4.8 ÷ 4 = 1.2 cm
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Part (b)
v = fλ = 12 × 0.012 = 0.14 m s−1 (0.144 m s−1; allow error carried forward)
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Part (c)
each point on the wavefront in the gap acts as a source of secondary wavelets, and the new wavefront is the envelope of (the line touching) these wavelets
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the gap (1.5 cm) is similar to the wavelength (1.2 cm), so the wavefronts beyond the gap are (almost) semicircular with the same spacing, spreading into the regions behind the barrier
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Part (d)
λ halves to 0.60 cm, so the gap is now 2.5 wavelengths wide and the waves spread out less
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Common error: Dividing the marked distance by the number of wavefronts. Five wavefronts enclose only four wavelengths.
Figure 1 is a plan view of a room and a corridor separated by a wall with a doorway. A loudspeaker S in the room emits a steady note. A listener stands at L in the corridor. The speed of sound in air is 340 m s−1.
(a)
Use the figure to determine the width of the doorway.
(1)
(b)
The loudspeaker emits a note of frequency 425 Hz and then a note of frequency 4250 Hz. Calculate the wavelength of each note.
(2)
(c)
Explain which note the listener at L hears more loudly.
(1)
(d)
The listener cannot see the loudspeaker. Explain why light from S does not reach L by diffraction at the doorway.
(1)
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Part (a)
the doorway runs from 1.6 m to 2.4 m (four squares): width = 0.80 m
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Part (b)
λ = v/f = 340 ÷ 425 = 0.80 m
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λ = 340 ÷ 4250 = 0.080 m
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Part (c)
the 425 Hz note: its wavelength equals the doorway width, so it diffracts (spreads) strongly through the doorway towards L, whereas the 4250 Hz note has a wavelength much smaller than the doorway and stays mostly in the shaded region
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Part (d)
the wavelength of light (about 5 × 10−7 m) is very much smaller than the doorway width, so the diffraction is negligible and the light travels in straight lines through the doorway
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Exam tip: Compare the wavelength with the gap width every time: a wavelength similar to the gap gives strong spreading; a much smaller wavelength passes almost straight through.
Figure 1 shows an electron diffraction tube and a face-on view of its screen. e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s.
(a)
Name the part of the tube that diffracts the electrons, and explain why it is suitable.
(2)
(b)
Explain why the tube must be evacuated.
(1)
(c)
Use the figure to calculate the de Broglie wavelength of the electrons.
(2)
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Answer
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Part (a)
the (thin) graphite film
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its atoms are arranged regularly, like a grating, with a spacing of about 10−10 m; this is larger than the electrons' de Broglie wavelength (about 2 × 10−11 m) but within a factor of about ten, so the electrons are diffracted through angles large enough to observe
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Part (b)
so that the electrons do not collide with (are not scattered by) air molecules before reaching the film and the screen
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Part (c)
p.d. = 4.0 kV, so Ek = eV = 1.60 × 10−19 × 4000 = 6.40 × 10−16 J; with Ek = ½mv2, p = √(2mEk) = 3.41 × 10−23 kg m s−1
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λ = h/p = 1.9 × 10−11 m
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Exam tip: The accelerating p.d. is the one between cathode and anode. Find Ek = eV, then p = √(2mEk), then λ = h/p.
Blue light of wavelength 470 nm is incident normally on a grating with 600 lines per mm. What is the angle between the second-order maximum and the straight-through direction?
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Answer
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d = 1 ÷ 6.00 × 105 = 1.67 × 10−6 m. sin θ = nλ/d = 2 × 470 × 10−9 × 6.00 × 105 = 0.564, so θ = 34.3°.
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Answer C
Answer: C
Exam tip: Find sin θ first and check it is less than 1, then take the inverse sine with the calculator in degree mode.
Red light of wavelength 650 nm is incident normally on a grating with 520 lines per mm. What is the total number of maxima produced, including the central maximum?
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d = 1 ÷ 5.20 × 105 = 1.92 × 10−6 m and d/λ = 2.96. sin θ cannot exceed 1, so the highest order is 2. Maxima: orders 0, ±1, ±2, a total of 5.
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Answer C
Answer: C
Common error: Rounding d/λ = 2.96 up to 3. A third order would need sin θ greater than 1, which is impossible — always round down.
A student shines a laser normally onto a grating with 100 lines per mm. On a screen 2.40 m from the grating, the distance between the two first-order maxima is 0.310 m. What is the wavelength of the laser light?
Figure: laser, grating and screen (not to scale)
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x = 0.310 ÷ 2 = 0.155 m, tan θ = 0.155 ÷ 2.40, θ = 3.70°. d = 1 ÷ 1.00 × 105 = 1.00 × 10−5 m. λ = d sin θ = 6.4 × 10−7 m.
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Answer A
Answer: A
Exam tip: The measured distance spans both first-order maxima, so halve it before finding tan θ = x/D.
Which observation provides evidence that electrons have wave properties?
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Answer
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Rings on the screen form a diffraction pattern. Diffraction is a wave property, so electron diffraction through graphite is evidence that electrons behave as waves. The other observations are explained by treating electrons as charged particles.
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Answer D
Answer: D
Exam tip: Evidence for wave behaviour must be a wave effect: diffraction or interference. Deflection by fields and emission from surfaces are particle-type explanations.
A mercury lamp emits a yellow line of wavelength 577 nm. Light from the lamp falls normally on a diffraction grating, and the first-order maximum for this line is observed at an angle of 20.3° to the straight-through direction.
Figure: first-order maxima for the 577 nm line (not to scale)
(a)
Explain, in terms of path difference, why bright maxima are formed only at certain angles.
(2)
(b)
Calculate the slit spacing of the grating.
(2)
(c)
Calculate the number of lines per mm on the grating.
(1)
(d)
Calculate the angle of the second-order maximum for the same line.
(1)
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Answer
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Part (a)
light from adjacent slits has a path difference of d sin θ to a distant point
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where this path difference equals a whole number of wavelengths the waves from all slits arrive in phase and superpose constructively; at other angles they cancel
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Part (b)
d = nλ/sin θ = 577 × 10−9 ÷ sin 20.3°
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= 1.66 × 10−6 m
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Part (c)
1 ÷ 1.663 × 10−6 = 6.01 × 105 lines per m = 601 lines per mm (allow 600)
A student uses a grating with 250 lines per mm to find the wavelength of a red laser. The screen is 0.850 m from the grating. She measures the distance x from the central maximum to the maxima of orders 1, 2 and 3 on one side: n = 1: x = 0.140 m; n = 2: x = 0.292 m; n = 3: x = 0.475 m.
Figure: grating and screen, drawn to scale, showing the maxima on one side; θ is the angle for the second-order maximum
(a)
Calculate the angle θ for the second-order maximum.
(1)
(b)
The values of sin θ for n = 1, 2 and 3 are 0.163, 0.325 and 0.488. Explain why a graph of sin θ against n should be a straight line through the origin, and state what its gradient represents.
(2)
(c)
The gradient of the best-fit line is 0.163. Calculate the wavelength of the laser light.
(2)
(d)
The student measures each distance with a metre rule. Suggest why the third-order reading gives the most reliable single value of the wavelength.
(1)
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Answer
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Part (a)
tan θ = 0.292 ÷ 0.850, θ = 19.0° (18.96°)
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Part (b)
sin θ = (λ/d) × n, which has the form y = mx with constant λ/d, so a straight line through the origin
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gradient = λ/d
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Part (c)
d = 1 ÷ 2.50 × 105 = 4.00 × 10−6 m
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λ = 0.163 × 4.00 × 10−6 = 6.52 × 10−7 m
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Part (d)
x is largest for the third order, so the same reading uncertainty gives the smallest percentage uncertainty
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Model sketch
Figure: sin θ against n, a straight line through the origin with gradient λ/d
Exam tip: Plotting sin θ against n uses all the orders at once; the gradient gives λ/d, and a line through the origin checks the grating equation.
In an electron diffraction tube, electrons are accelerated through a potential difference of a few kilovolts and then pass through a very thin film of graphite before hitting a fluorescent screen.
(a)
Describe the pattern seen on the screen.
(1)
(b)
Explain why this pattern is evidence that electrons behave as waves.
(2)
(c)
The accelerating potential difference is increased. State and explain what happens to the rings.
(2)
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Part (a)
a bright central spot surrounded by concentric bright rings (with dark regions between them)
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Part (b)
the pattern is a diffraction pattern / is produced by diffraction and superposition
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diffraction is a property of waves; the atomic spacing in graphite is similar to the electrons' wavelength
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Part (c)
the electrons have greater speed / momentum, so their de Broglie wavelength λ = h/p is shorter
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so they are diffracted through smaller angles and the rings become smaller
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Common error: Explaining the rings by electrons colliding with or being repelled by atoms. The rings must be explained as diffraction, using the electrons' wavelength.
In an electron gun, electrons are accelerated from rest through a potential difference of 1800 V. e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s.
(a)
Calculate the kinetic energy gained by each electron, in joules.
(1)
(b)
Show that the speed of the electrons is about 2.5 × 107 m s−1.
(2)
(c)
Calculate the de Broglie wavelength of the electrons.
(2)
(d)
Suggest why electron beams of this kind are suitable for studying the arrangement of atoms in thin metal films, where the atoms are about 1 × 10−10 m apart.
(1)
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Answer
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Notes
Part (a)
Ek = eV = 1.60 × 10−19 × 1800 = 2.9 × 10−16 J (2.88 × 10−16 J)
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Part (b)
½mv2 = 2.88 × 10−16 J, so v = √(2 × 2.88 × 10−16 ÷ 9.11 × 10−31)
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v = 2.51 × 107 m s−1
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Part (c)
p = mv = 9.11 × 10−31 × 2.51 × 107 = 2.29 × 10−23 kg m s−1
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λ = h/p = 2.9 × 10−11 m
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Part (d)
the wavelength is of a similar order of magnitude to the atomic spacing (about a third of it), so the electrons are diffracted noticeably by the regular arrangement of atoms
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Exam tip: In a “show that”, give the answer to at least one more significant figure than the value quoted in the question.
Figure 1 shows Huygens' construction for a straight wavefront in a gap in a barrier. Secondary wavelets (dashed) have been drawn from points on the wavefront in the gap. In each of the panels P, Q, R and S, a different thick line has been drawn as the new wavefront. In which panel is the new wavefront drawn correctly?
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The new wavefront is the envelope of the wavelets: a line that touches every wavelet and cuts through none. Across the gap it is straight; beyond each edge it follows the outside of the wavelet from the edge point, curving round into the region behind the barrier. Only R does this.
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Answer C
Answer: C
Exam tip: A correct Huygens wavefront touches every wavelet and never cuts through one. At the edges of a gap it follows the wavelets from the edge points round into the region behind the barrier.
A student shines a laser of wavelength 600 nm normally onto a diffraction grating and measures the angle θ of the maxima of orders n = 1 to 4. Figure 1 shows her graph of sin θ against n. How many lines per mm does the grating have?
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Answer
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From nλ = d sin θ, the gradient of sin θ against n is λ/d. Gradient = 0.48 ÷ 4 = 0.12, so d = 600 × 10−9 ÷ 0.12 = 5.0 × 10−6 m. Lines per metre = 1/d = 2.0 × 105, so lines per mm = 200.
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Answer B
Answer: B
Exam tip: The gradient of sin θ against n is λ/d. Use a point far up the line and divide by its value of n.
Figure 1 shows how sin θ for the first-order maximum produced by a diffraction grating depends on the wavelength λ of light incident normally on the grating. Light of wavelength 700 nm is incident normally on the grating. What is the highest order of maximum that can be observed?
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Answer
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At 700 nm the graph gives sin θ1 = 0.28 = λ/d. For order n, sin θ = 0.28n, which cannot exceed 1, so n ≤ 1 ÷ 0.28 = 3.57. The highest order is 3.
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Answer A
Answer: A
Exam tip: The line has gradient 1/d. Read sin θ1 at the wavelength given; the highest order is the whole-number part of 1 ÷ sin θ1.
Laser light is incident normally on a grating with 150 lines per mm. A screen is placed parallel to the grating, 1.50 m from it. Figure 1 shows the spots on the screen, with a metre rule placed alongside. What is the wavelength of the laser light?
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Answer
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Notes
The second-order spots are at 26.0 cm and 74.0 cm, so x = 74.0 − 50.0 = 24.0 cm. tan θ = 0.240 ÷ 1.50 = 0.160, θ = 9.09°. d = 1 ÷ 1.50 × 105 = 6.67 × 10−6 m, so λ = d sin θ ÷ 2 = 5.3 × 10−7 m.
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Answer B
Answer: B
Common error: Using the outermost spots with n = 1. Count the orders out from the central spot and use the matching n in nλ = d sin θ.
Figure 1 shows an electron gun. Electrons leave the cathode with negligible speed and are accelerated to the anode. The potentials of the cathode and the anode relative to earth are shown. e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s. What is the de Broglie wavelength of the electrons that reach the graphite film?
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Answer
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The p.d. between cathode and anode is +0.30 − (−1.20) = 1.50 kV. Kinetic energy eV = 1.60 × 10−19 × 1500 = 2.40 × 10−16 J. With eV = ½mv2, p = √(2meeV) = 2.09 × 10−23 kg m s−1, so λ = h/p = 3.2 × 10−11 m.
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Answer C
Answer: C
Common error: Subtracting the sizes of the two potentials. The accelerating p.d. is the difference between them: +0.30 kV − (−1.20 kV) = 1.50 kV.
A diffraction grating with 450 lines per mm is mounted on a spectrometer. Light from a lamp that emits two visible spectral lines, X and Y, falls normally on the grating. Figure 1 shows the angle scale with the readings for the straight-through position and for the first-order images of X and Y on both sides.
(a)
Use the figure to determine the first-order diffraction angle θ for each line.
(2)
(b)
Calculate the wavelength of each line.
(2)
(c)
Suggest why the angle between the two first-order images is measured and halved, rather than the angle from the straight-through position to one image.
(1)
(d)
Show that the second-order image of X is further from the straight-through position than the first-order image of Y.
d = 1 ÷ 4.50 × 105 = 2.22 × 10−6 m; λX = d sin 11.0° = 4.24 × 10−7 m
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λY = d sin 17.0° = 6.50 × 10−7 m (allow error carried forward)
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Part (c)
the measured angle is twice as large, so its percentage uncertainty is halved / it cancels the error if the grating is not exactly normal to the beam or the straight-through reading is slightly off
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Part (d)
sin θ = 2 × 4.24 × 10−7 ÷ 2.22 × 10−6 = 0.382, θ = 22.4°, which is greater than 17.0°
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Exam tip: Read each pointer to the nearest 1° division, take the difference between the two images of the same order and halve it to get θ.
Figure 1 shows part of a diffraction grating seen through a microscope. Light of wavelength 532 nm from a laser is incident normally on the grating.
(a)
Use the figure to determine the slit spacing d of the grating.
(1)
(b)
Calculate the number of lines per mm.
(1)
(c)
Calculate the angle between the second-order maximum and the straight-through direction.
(2)
(d)
The grating is replaced by one with 1200 lines per mm. Explain whether a second-order maximum can now be observed.
(2)
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Answer
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Notes
Part (a)
11 lines enclose 10 spacings: d = 25.0 ÷ 10 = 2.50 μm = 2.50 × 10−6 m
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Part (b)
1 ÷ 2.50 × 10−6 = 4.00 × 105 lines per m = 400 lines per mm
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Part (c)
sin θ = 2 × 532 × 10−9 ÷ 2.50 × 10−6 = 0.426
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θ = 25.2° (allow error carried forward)
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Part (d)
d = 1 ÷ 1.20 × 106 = 8.33 × 10−7 m, so sin θ = 2 × 532 × 10−9 ÷ 8.33 × 10−7 = 1.28
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sin θ cannot be greater than 1, so there is no second-order maximum
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Exam tip: If nλ/d comes out greater than 1, that order does not exist. A grating with more lines per mm spreads the maxima further apart but gives fewer of them.
To measure the wavelength of light from a laser, a student uses a grating with 300 lines per mm. She measures the distance x on a screen from the central maximum to one first-order maximum, for different grating-to-screen distances D. Figure 1 shows her results and the best-fit line.
(a)
Explain why x is expected to be directly proportional to D.
(1)
(b)
Use the graph to determine tan θ.
(2)
(c)
Calculate the wavelength of the laser light.
(2)
(d)
The best-fit line does not pass through the origin. Suggest an error in the measurement of D that would explain this, and state whether it affects the wavelength found from the gradient.
(1)
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Answer
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Notes
Part (a)
x = D tan θ, and θ is fixed because d, λ and n do not change, so x ∝ D
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Part (b)
gradient from a large triangle, e.g. (31.0 − 11.0) cm ÷ (1.50 − 0.50) m = 20 cm m−1
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tan θ = 0.20 m ÷ 1.00 m = 0.20 (allow 0.19–0.21)
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Part (c)
θ = tan−1 0.20 = 11.3° and d = 1 ÷ 3.00 × 105 = 3.33 × 10−6 m
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λ = d sin θ = 3.33 × 10−6 × 0.196 = 6.5 × 10−7 m (allow error carried forward)
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Part (d)
extending the line, x = 0 at D = −0.05 m, so every D was measured about 5 cm too short (e.g. from a point in front of the grating); this systematic error shifts the line but does not change the gradient, so λ is not affected
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Exam tip: A constant error in one variable shifts a straight-line graph without changing its gradient. Using the gradient removes the effect of such an error.
Figure 1 shows how the kinetic energy gained by a particle that starts from rest depends on the accelerating potential difference V, for an electron and for an α-particle. me = 9.11 × 10−31 kg, mα = 6.64 × 10−27 kg, h = 6.63 × 10−34 J s, 1 eV = 1.60 × 10−19 J.
(a)
Explain why the line for the α-particle has twice the gradient of the line for the electron.
(1)
(b)
Use the graph to find, in joules, the kinetic energy of an electron accelerated through 2.5 kV.
(1)
(c)
Calculate the de Broglie wavelength of this electron.
(2)
(d)
Calculate the de Broglie wavelength of an α-particle accelerated from rest through the same potential difference.
(2)
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Answer
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Notes
Part (a)
kinetic energy gained = QV, and the α-particle has charge +2e, twice the size of the electron's charge
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Part (b)
2.5 keV = 2500 × 1.60 × 10−19 = 4.0 × 10−16 J
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Part (c)
using Ek = ½mv2: p = √(2mEk) = √(2 × 9.11 × 10−31 × 4.0 × 10−16) = 2.70 × 10−23 kg m s−1
✓ 1
λ = h/p = 2.5 × 10−11 m
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Part (d)
from the graph Ek = 5.0 keV = 8.0 × 10−16 J, so p = √(2 × 6.64 × 10−27 × 8.0 × 10−16) = 3.26 × 10−21 kg m s−1
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λ = h/p = 2.0 × 10−13 m
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Common error: Using the electron's kinetic energy for the α-particle. Through the same p.d. the α-particle gains twice the energy, and its mass is far larger.
Electrons accelerated from rest through a potential difference V have a de Broglie wavelength λ. Relativistic effects are negligible. The potential difference is increased to 4V. What is the new de Broglie wavelength?
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Answer
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Notes
eV = p2/2m, so p = √(2meV) ∝ √V, and λ = h/p ∝ 1/√V. Multiplying V by 4 multiplies p by 2, so the wavelength becomes λ/2.
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Answer B
Answer: B
Exam tip: Write the wavelength in terms of V before using ratios: λ = h/√(2meV), so λ ∝ V−1/2.
Electrons are to be accelerated from rest so that their de Broglie wavelength is 1.5 × 10−10 m. e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s. What accelerating potential difference is needed?
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Answer
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Notes
p = h/λ = 4.42 × 10−24 kg m s−1. Kinetic energy = p2/2m = 1.07 × 10−17 J. V = Ek/e = 67 V.
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Answer C
Answer: C
Exam tip: Work in steps: momentum from λ = h/p, speed from p = mv, kinetic energy ½mv2, then V = Ek/e.
A student is given a diffraction grating marked with its number of lines per mm, a laser, a large screen, a metre rule and a tape measure.
(a)
Describe the method, including the measurements, how the data are processed and one safety precaution.
(6)
(b)
With a different grating of 350 lines per mm, the student finds the first-order maximum at 13.2° to the straight-through direction. Calculate the wavelength of the laser light.
(2)
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Answer
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Notes
Part (a)
shine the laser normally onto the grating; calculate d = 1 ÷ (lines per metre)
✓ 1
place the screen parallel to the grating and measure the grating–screen distance D with the tape measure
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measure the distance between the two maxima of the same order on opposite sides of the centre (2x) with the metre rule, and halve it
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repeat the measurements / measure for several orders and take means
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find θ from tan θ = x/D, then λ = d sin θ / n
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safety: never look into the beam or its reflections / remove reflective surfaces / keep the beam below eye level
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Part (b)
d = 1 ÷ 3.50 × 105 = 2.86 × 10−6 m
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λ = d sin θ = 2.86 × 10−6 × sin 13.2° = 6.52 × 10−7 m
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Exam tip: Name the instrument for each length, say what is repeated, and give the equations used to turn the readings into λ. A safety point must be specific to lasers.
A student uses an electron diffraction tube to test the de Broglie equation. For each accelerating potential difference V she measures the radius r of the inner ring on the screen: V = 2.0 kV: r = 16.9 mm; V = 3.0 kV: r = 13.7 mm; V = 4.0 kV: r = 11.9 mm; V = 5.0 kV: r = 10.6 mm. For these small angles, r is proportional to the de Broglie wavelength λ of the electrons.
(a)
An electron of mass m and charge e is accelerated from rest through a p.d. V. Show that its de Broglie wavelength is λ = h/√(2meV).
(2)
(b)
Use the data to show that r√V is approximately constant.
(2)
(c)
Explain how this result supports the idea that electrons behave as waves with λ = h/p.
(1)
(d)
Predict the radius of the inner ring at 6.0 kV.
(1)
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Answer
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Notes
Part (a)
eV = ½mv2, so v = √(2eV/m)
✓ 1
p = mv = √(2meV), and λ = h/p = h/√(2meV)
✓ 1
Part (b)
values of r√V (with V in kV): 23.9, 23.7, 23.8, 23.7 mm kV1/2
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these agree to within about 1%, so r√V is approximately constant
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Part (c)
r ∝ 1/√V matches λ ∝ 1/√V from part (a), so the ring size depends on the p.d. exactly as the de Broglie wavelength does
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Part (d)
r = 23.8 ÷ √6.0 = 9.7 mm
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Model sketch
Figure: r against 1/√V; the points lie close to a straight line through the origin, so r√V is constant
Exam tip: To test a proportionality such as r ∝ 1/√V, calculate r√V for every row and comment on how closely the values agree.
Slow-moving neutrons are used to study crystal structures. A beam of neutrons is to have a de Broglie wavelength of 1.8 × 10−10 m. mn = 1.67 × 10−27 kg, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s, 1 eV = 1.60 × 10−19 J.
(a)
Calculate the speed of the neutrons.
(2)
(b)
Calculate the kinetic energy of each neutron in eV.
(2)
(c)
An electron has the same kinetic energy as these neutrons. Explain whether its de Broglie wavelength is longer or shorter, and calculate it.
Figure 1 shows two electron diffraction patterns, X and Y, produced by the same graphite film in the same tube. Pattern X was obtained with an accelerating potential difference of 3.0 kV. The radius of each ring is proportional to the de Broglie wavelength of the electrons. Relativistic effects are negligible. What accelerating potential difference was used for pattern Y?
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Answer
Mark
Notes
Inner ring radii: X = 16 mm, Y = 8 mm, so λ is halved for Y and the momentum p = h/λ is doubled. With eV = ½mv2, p = √(2meV) ∝ √V, so doubling p needs 4 times the p.d.: 4 × 3.0 = 12 kV.
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Answer D
Answer: D
Exam tip: Smaller rings mean a shorter wavelength and so a larger momentum. Since λ ∝ 1/√V, halving λ needs four times the p.d.
Figure 1 shows how the de Broglie wavelength λ depends on 1/v, where v is the speed, for two particles X and Y. Relativistic effects are negligible. What is the value of (mass of X) ÷ (mass of Y)?
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Answer
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Notes
λ = h/(mv), so the gradient of λ against 1/v is h/m. Gradient for X = 6.0 ÷ 2.0 = 3.0; for Y = 2.0 ÷ 4.0 = 0.50 (same units). Mass is inversely proportional to the gradient, so mX/mY = 0.50 ÷ 3.0 = 0.17.
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Answer A
Answer: A
Exam tip: Write the equation in the form y = kx: λ = (h/m) × (1/v). The steeper line belongs to the lighter particle.
A student uses one laser and six different diffraction gratings. For each grating she notes the number of lines per mm, N, and measures the angle θ of the first-order maximum. Figure 1 shows a graph of sin θ against N with the best-fit line.
(a)
Show that sin θ is expected to be directly proportional to N, and state what the gradient represents.
(2)
(b)
Use the graph to determine the wavelength of the laser light.
(3)
(c)
Give one advantage of finding λ from the gradient of this graph rather than from one grating.
(1)
(d)
Explain why this laser gives no first-order maximum with a grating of 1600 lines per mm.
(1)
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Answer
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Notes
Part (a)
d = 1/(lines per metre) = 1/(1000N), so sin θ = λ/d = 1000λN
✓ 1
λ is constant, so sin θ ∝ N; the gradient is 1000λ (λ if N is in lines per m)
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Part (b)
gradient from a large triangle, e.g. 0.320 ÷ 500 = 6.4 × 10−4 (per line per mm)
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lines per mm converted to lines per m: λ = 6.4 × 10−4 ÷ 1000
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λ = 6.4 × 10−7 m (allow 6.3–6.5 × 10−7 m)
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Part (c)
it uses all six results, so random errors are averaged out / an anomalous result would show up away from the line
✓ 1
Part (d)
sin θ = λN = 6.4 × 10−7 × 1.6 × 106 = 1.02, which is greater than 1, so no angle is possible
✓ 1
Common error: Forgetting that N is in lines per mm. Multiply by 1000 to get lines per metre, or the wavelength comes out 1000 times too large.
Electrons are accelerated from rest through different potential differences V, and their de Broglie wavelength λ is found from the ring pattern they produce with a graphite film. Theory gives λ = h/√(2meeV). Figure 1 shows a graph of λ against 1/√V with the best-fit line. e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg.
(a)
Explain why the graph is expected to be a straight line through the origin, and write down an expression for its gradient.
(2)
(b)
Use the graph to determine the gradient.
(2)
(c)
Calculate the value of the Planck constant given by this experiment.
(2)
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Answer
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Notes
Part (a)
λ = [h/√(2mee)] × (1/√V), which has the form y = kx
✓ 1
h, me and e are constants, so a straight line through the origin with gradient h/√(2mee)
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Part (b)
large triangle on the line, e.g. 30.0 × 10−12 m ÷ 2.50 × 10−2 V−1/2
Exam tip: Take the gradient from two well-separated points on the best-fit line, not from data points, and include the powers of ten from both axis labels.
An electron starts from rest at the cathode of an electron gun and is accelerated uniformly to the anode, then continues at constant speed. Figure 1 is its velocity–time graph. e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s.
(a)
State the speed of the electron as it leaves the anode.
(1)
(b)
Use the graph to determine the distance between the cathode and the anode.
(2)
(c)
Calculate the accelerating potential difference between the cathode and the anode.
(2)
(d)
Calculate the de Broglie wavelength of the electron as it leaves the anode.
(1)
(e)
Determine the de Broglie wavelength of the electron at time t = 2.0 ns.
(1)
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Answer
Mark
Notes
Part (a)
6.0 × 106 m s−1
✓ 1
Part (b)
distance = area under the graph from 0 to 4.0 ns
✓ 1
= ½ × 4.0 × 10−9 × 6.0 × 106 = 0.012 m (12 mm)
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Part (c)
Ek = ½mv2 = ½ × 9.11 × 10−31 × (6.0 × 106)2 = 1.64 × 10−17 J
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