2.4 Diffraction and the wave nature of electrons
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AS content for Unit 2 (Waves and Electricity). Superposition and path difference are on 2.2; photons are on 2.5. No equation is needed for the angle of spreading from a single slit.
What this page covers
7 points
- What diffraction is, and Huygens' construction for a gap and for an obstacle
- How the amount of spreading depends on the gap width compared with the wavelength
- Why a diffraction grating gives sharp maxima, and deriving \(n\lambda = d\sin\theta\)
- Slit spacing from lines per mm, angles of the orders and the highest order
- Measuring the wavelength of laser light with a grating (Core Practical 6), with uncertainties
- Electron diffraction as evidence that electrons behave as waves
- The de Broglie wavelength \(\lambda = h/p\), including electrons accelerated through a p.d.
In one minute
5 points
- Diffraction is the spreading of a wave as it passes through a gap or around the edge of an obstacle, and it is greatest when the gap is about the same size as the wavelength.
- Huygens: every point on a wavefront acts as a source of secondary wavelets; the new wavefront is the surface that touches them all.
- A diffraction grating gives sharp maxima where the path difference between adjacent slits is a whole number of wavelengths: \(n\lambda = d\sin\theta\).
- \(\sin\theta\) cannot exceed 1, so the highest order is the largest whole number not greater than \(d/\lambda\).
- Electrons diffract through thin crystals, so they have a wavelength: \(\lambda = h/p\). Faster electrons have shorter wavelengths.
Contents
13 sections
- What is diffraction, and how does Huygens' construction explain it?
- What decides how much a wave spreads?
- How does a diffraction grating produce sharp maxima?
- How do you find the highest order and use the grating equation in reverse?
- How do you measure the wavelength of light with a grating?
- What shows that electrons behave as waves?
- How do you calculate the de Broglie wavelength of an electron?
- Exam tips for 2.4
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is diffraction, and how does Huygens' construction explain it?
You can hear someone talking in the next room through an open door even when you cannot see them. Sound bends round the edges of the doorway; light, apparently, does not. Both are waves, and the difference comes down to the size of the wavelength compared with the gap.
- Diffraction happens to every kind of wave (water waves, sound, light, microwaves) and is strong evidence that something is a wave.
- It changes the direction of travel and the shape of the wavefronts. It does not change the wavelength, the frequency or the speed, because the medium is the same on both sides of the gap.
- Energy is spread over a wider region, so the amplitude in any one direction beyond the gap is smaller.
Using the construction at a gap
Plane wavefronts reach the barrier, and only the part of each wavefront inside the gap carries on. Each point in the gap sends out a semicircular wavelet. In the middle of a wide gap the wavelets overlap side by side, so their envelope is still straight. At the edges there are no neighbouring wavelets on the outside, so the envelope curves round into the region behind the barrier: this is the spreading. With a gap about one wavelength wide there is effectively only one source of wavelets, so the wave leaves as semicircular wavefronts. At an obstacle the same thing happens at each edge: wavelets from the edge of the unblocked wavefront curve into the "shadow" behind it.
Changing the spacing of the wavefronts after the gap when you sketch diffraction. The wavelength is unchanged, so the spacing must be the same on both sides.
What decides how much a wave spreads?
What matters is the ratio of wavelength to gap width, \(\lambda/b\), not either one on its own.
- Gap width ≈ λ: maximum spreading; the wave fills almost the whole region beyond the gap.
- Gap width much larger than λ: little spreading; the wave carries on almost as a straight-edged beam.
- Narrowing the gap (closer to \(\lambda\)) or increasing the wavelength both increase the spreading.
- If the gap is much smaller than \(\lambda\), very little of the wave gets through at all.
Visible light has \(\lambda \approx 4 \times 10^{-7}\) m to \(7 \times 10^{-7}\) m, far smaller than doorways, so light diffraction is noticeable only with very narrow slits or very fine edges. Sound has wavelengths from centimetres to metres, so it diffracts round everyday openings.
A courtyard is enclosed by a wall with a single opening 1.2 m wide. Outside, a bird sings at 2.8 kHz and a lorry engine produces a low note of 95 Hz. The speed of sound is 340 m s⁻¹. Calculate the wavelength of each sound and explain which one is heard better by a person inside the courtyard, to one side of the opening and out of the direct line.
Wavelengths: bird: \(\lambda = 340 \div 2800 = 0.12\ \text{m}\) (0.121 m). Engine: \(\lambda = 340 \div 95 = 3.6\ \text{m}\) (3.58 m).
Compare with the gap: bird: \(\lambda/b \approx 0.1\); engine: \(\lambda/b \approx 3.0\).
The engine note is heard better. Its wavelength is larger than the gap, so it spreads strongly round behind the wall. The bird's wavelength is about a tenth of the gap, so it diffracts only a little and mostly continues straight on.
Saying "the gap must be smaller than the wavelength for diffraction". Diffraction happens at every gap; it is most noticeable when the gap is similar to the wavelength, and a gap much smaller than \(\lambda\) lets very little through.
How does a diffraction grating produce sharp maxima?
A diffraction grating is a plate with a very large number of equally spaced, parallel, very narrow slits, often specified as a number of lines per mm. Each slit is so narrow that light diffracts from it over a wide range of angles, so light from all the slits overlaps beyond the grating. In most directions the waves from the many slits arrive with a spread of phases and cancel. Only in certain directions do they all arrive in phase and superpose constructively. The result is a few sharp, bright maxima with darkness between them, much sharper and brighter than the fringes from two slits.
The diffraction grating equation
\[n\lambda = d\sin\theta\]Here \(n\) is the order of the maximum (0, 1, 2, …), \(\lambda\) the wavelength (m), \(d\) the slit spacing, centre to centre (m), and \(\theta\) the angle between the maximum and the straight-through (zero-order) direction. The equation is in the formulae list supplied in the exam, but you must be able to derive it:
- Plane waves arrive normally, so adjacent slits are in phase.
- The path difference from adjacent slits to a distant point at angle \(\theta\) is \(d\sin\theta\).
- Constructive superposition needs a path difference of a whole number of wavelengths, \(n\lambda\). Then every pair of neighbouring slits is in step, so all the slits are.
The slit spacing comes from the lines per mm: \(d = 1 \div\) (number of lines per metre), so for \(N\) lines per mm, \(d = 1/(N \times 10^{3})\) m. The zero order is straight through; the other orders come in pairs at \(\pm\theta\). Larger \(\lambda\) or smaller \(d\) gives larger angles. With white light each order except zero becomes a spectrum, violet nearest the centre and red furthest out, and higher-order spectra can overlap.
A diode laser of wavelength 635 nm shines normally onto a grating with 400 lines per mm. Calculate (a) the slit spacing, (b) the angles of the first-order and second-order maxima, and (c) the angle between the two first-order maxima.
(a) \(d = 1 \div (400 \times 10^{3}) =\) 2.50 × 10⁻⁶ m.
(b) \(n = 1\): \(\sin\theta_1 = 635 \times 10^{-9} \div 2.50 \times 10^{-6} = 0.254\), so \(\theta_1 = 14.7°\). \(n = 2\): \(\sin\theta_2 = 2 \times 0.254 = 0.508\), so \(\theta_2 = 30.5°\). 14.7° and 30.5°
(c) The first orders lie either side of the centre: \(2 \times 14.7° =\) 29.4°.
\(\theta_2\) is not \(2\theta_1\): it is \(\sin\theta\) that doubles, not \(\theta\).
How do you find the highest order and use the grating equation in reverse?
Since \(\sin\theta\) can be at most 1, \(n\lambda \le d\), so \(n \le d/\lambda\). The highest order is the largest whole number not greater than \(d/\lambda\): always round down. The total number of maxima is \(2n_\text{max} + 1\), both sides plus the central one. Used in reverse, a measured angle and a known \(d\) give the wavelength: \(\lambda = d\sin\theta/n\).
For the 400 lines per mm grating and 635 nm laser of worked example 2: (a) determine the highest order and the total number of maxima; (b) the grating is replaced by one with 800 lines per mm. How many maxima are seen now?
(a) \(d/\lambda = 2.50 \times 10^{-6} \div 635 \times 10^{-9} = 3.94\), so the highest order is \(n = 3\) (at 49.6°). Counting both sides and the centre: \(2 \times 3 + 1 =\) 7 maxima.
(b) \(d = 1.25 \times 10^{-6}\) m; \(d/\lambda = 1.97\), so the highest order is 1: 3 maxima.
Rounding 3.94 up to 4 would need \(\sin\theta = 1.02\), which is impossible.
Light from a gas discharge lamp passes normally through a grating with 450 lines per mm. For one bright line, the angle between the two second-order maxima, on opposite sides of the centre, is 58.0°. (a) Calculate the wavelength of the line. (b) Determine the highest order in which this line can be seen.
\(\theta\) is measured from the centre, so halve the angle between the pair: \(\theta_2 = 58.0° \div 2 = 29.0°\). \(d = 1 \div (450 \times 10^{3}) = 2.222 \times 10^{-6}\) m.
(a) \(\lambda = d\sin\theta/n = 2.222 \times 10^{-6} \times \sin 29.0° \div 2 =\) 5.39 × 10⁻⁷ m (539 nm).
(b) Test \(\sin\theta = n\lambda/d\): \(n = 3\): 0.727; \(n = 4\): 0.970; \(n = 5\): 1.21 > 1. Highest order: 4th.
The fourth order is at about 76° and would be faint and hard to see in practice.
Using lines per mm directly as \(d\), or forgetting the factor \(10^{3}\). Write \(d = 1/(N \times 10^{3})\) m as its own line and sense-check: grating spacings are a few micrometres.
How do you measure the wavelength of light with a grating?
This is Core Practical 6, described here in our own words.
- Set-up: mount a grating of known lines per mm in a holder, with a laser shining normally onto it. Place a white screen (or a wall) parallel to the grating, about 1–2 m away.
- Check it is normal: the maxima either side of the centre should be the same distance from the central spot; if not, rotate the grating until they are.
- Measure \(D\): the perpendicular distance from grating to screen, with a tape measure or metre rules (mm resolution).
- Measure \(2x\): the distance between the two maxima of the same order on opposite sides of the centre, with a metre rule, after recording the spot centres on paper fixed to the screen. Repeat and take the mean.
- Calculate: \(x = (2x)/2\), \(\tan\theta = x/D\), then \(\lambda = d\sin\theta/n\). Repeat for the second order and compare.
- Safety: never look into the laser beam or its reflections; use a low-power laser, keep the beam below eye level, remove shiny surfaces such as watches, and switch the laser off when not in use.
Measuring across the centre doubles the length measured, so the same reading uncertainty is a smaller percentage, and you avoid locating the exact centre of the central spot. A larger \(D\) or a higher order gives larger distances and smaller percentage uncertainties. Grating angles are often too large for the small-angle approximation, so always find \(\theta\) from \(\tan\theta = x/D\), then take the sine.
A student uses a grating labelled 300 lines per mm. The screen is 1.500 m from the grating. Three measurements of the distance between the two first-order maxima are 0.592 m, 0.586 m and 0.598 m. Determine the wavelength of the laser light, and the percentage uncertainty in the mean distance between the first-order maxima.
Mean and uncertainty (half the range): \(2x = 0.592\ \text{m}\); uncertainty \(= (0.598 - 0.586) \div 2 = \pm 0.006\ \text{m}\).
Angle: \(x = 0.296\ \text{m}\); \(\tan\theta = 0.296 \div 1.500 = 0.197\), so \(\theta = 11.2°\).
Wavelength: \(d = 1 \div (300 \times 10^{3}) = 3.33 \times 10^{-6}\ \text{m}\); \(\lambda = 3.33 \times 10^{-6} \times \sin 11.2° =\) 6.45 × 10⁻⁷ m.
Percentage uncertainty: \(0.006 \div 0.592 \times 100 =\) 1.0%. Halving \(2x\) to get \(x\) leaves the percentage uncertainty unchanged.
Writing "measure the distance between the dots" in a method. Name the instrument and what is repeated: "measure the distance between the two first-order maxima with a metre rule, repeat and take a mean".
What shows that electrons behave as waves?
In an electron diffraction tube, electrons released from a hot cathode are accelerated through a p.d. of a few kilovolts towards an anode and pass through a very thin film of graphite (carbon). On the fluorescent screen at the end you see a bright central spot surrounded by concentric bright rings.
- The regularly spaced layers of atoms in the graphite act like a grating. The atomic spacing (about \(10^{-10}\) m) is comparable with the electrons' wavelength, so the electrons diffract strongly.
- Rings arise because graphite is made of many tiny crystals at random orientations: each set of atomic layers sends electrons out at the same angle from the beam, in every direction around it.
- Why this is evidence for waves: diffraction and superposition are wave behaviours. Particles travelling in straight lines would give a single bright patch, not rings with dark regions between them.
- The test: increasing the accelerating p.d. makes the rings smaller. Faster electrons have more momentum, so (by \(\lambda = h/p\)) a shorter wavelength, and they diffract through smaller angles, exactly as a wave would.
Explaining the rings as electrons "bouncing off" or being repelled by atoms. Collisions or charge would not produce a regular pattern of bright and dark rings whose size depends on the electrons' speed. The explanation must use wavelength and superposition.
How do you calculate the de Broglie wavelength of an electron?
\[\lambda = \frac{h}{p}\]Here \(\lambda\) is the de Broglie wavelength (m), \(h\) the Planck constant, \(6.63 \times 10^{-34}\) J s, and \(p\) the momentum of the particle (kg m s⁻¹), \(p = mv\). Both \(\lambda = h/p\) and \(p = mv\) are in the formulae list. The relation applies to any moving particle, but large objects have such tiny wavelengths that their wave behaviour can never be observed.
Electrons accelerated through a potential difference
An electron accelerated from rest through a p.d. \(V\) gains kinetic energy equal to the work done on it:
\[eV = \tfrac12 mv^2\]This comes from \(V = W/Q\) and \(E_k = \tfrac12 mv^2\); derive it, as it is not given. Find \(v\), then \(p = mv\), then \(\lambda = h/p\), each on its own line. Combining the steps gives \(\lambda = h/\sqrt{2meV}\), so \(\lambda \propto 1/\sqrt{V}\): four times the p.d. halves the wavelength.
Electrons are accelerated from rest through a potential difference of 2.5 kV. \(e = 1.60 \times 10^{-19}\) C, \(m_e = 9.11 \times 10^{-31}\) kg, \(h = 6.63 \times 10^{-34}\) J s. (a) Calculate the speed, the momentum and the de Broglie wavelength of the electrons. (b) Explain why these electrons are diffracted by a crystal with atomic spacing about \(1 \times 10^{-10}\) m.
(a) \(E_k = eV = 1.60 \times 10^{-19} \times 2500 = 4.0 \times 10^{-16}\) J. Speed: \(v = \sqrt{2 \times 4.0 \times 10^{-16} \div 9.11 \times 10^{-31}} = 2.96 \times 10^{7}\ \text{m s}^{-1}\). Momentum: \(p = 9.11 \times 10^{-31} \times 2.96 \times 10^{7} = 2.70 \times 10^{-23}\ \text{kg m s}^{-1}\). Wavelength: \(\lambda = 6.63 \times 10^{-34} \div 2.70 \times 10^{-23} =\) 2.46 × 10⁻¹¹ m.
(b) The spacing is only about 4 times \(\lambda\): similar orders of magnitude, so significant diffraction.
The speed is about 10% of the speed of light, low enough for \(\tfrac12 mv^2\) to be a good approximation at this level.
A researcher wants electrons with a de Broglie wavelength of \(2.1 \times 10^{-10}\) m, equal to an atomic spacing in a crystal under study. \(m_p = 1.67 \times 10^{-27}\) kg. (a) Calculate the accelerating p.d. needed. (b) Calculate the speed a proton would need to have the same de Broglie wavelength.
Momentum: \(p = h/\lambda = 6.63 \times 10^{-34} \div 2.1 \times 10^{-10} = 3.16 \times 10^{-24}\ \text{kg m s}^{-1}\).
(a) \(v = p/m = 3.47 \times 10^{6}\ \text{m s}^{-1}\); \(E_k = \tfrac12 mv^2 = 5.47 \times 10^{-18}\) J. Then \(V = E_k/e = 5.47 \times 10^{-18} \div 1.60 \times 10^{-19} =\) 34 V (34.2 V).
(b) Same wavelength means the same momentum: \(v_p = 3.16 \times 10^{-24} \div 1.67 \times 10^{-27} =\) 1.9 × 10³ m s⁻¹ (1891 m s⁻¹).
The proton is about 1830 times more massive, so it needs a speed about 1830 times smaller for the same momentum.
Exam tips for 2.4
A full Huygens answer says: each point on a wavefront is a source of secondary wavelets; the wavelets spread out at the wave speed; the new wavefront is the envelope (common tangent) of the wavelets. Then apply it to the edges of the gap.
For "will this wave diffract noticeably?", calculate \(\lambda\) (often from \(v = f\lambda\)) and compare it with the gap width. State the comparison explicitly: "\(\lambda\) is similar to the gap, so the wave spreads a lot."
Write \(d = 1/\)(lines per metre) as the first line: 300 lines per mm is \(3.00 \times 10^{5}\) lines per m, so \(d = 3.33 \times 10^{-6}\) m. For the highest order take the whole-number part of \(d/\lambda\), even if it is 3.94; double it and add 1 for the number of maxima.
Not the same speed or the same kinetic energy. Compare particles through \(p = h/\lambda\) first, then bring in their masses.
To show electron diffraction is wave behaviour, link three ideas: a diffraction pattern (rings) is seen; diffraction is a property of waves; the wavelength from \(\lambda = h/p\) is similar to the atomic spacing, and the ring size changes with p.d. as the equation predicts.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Diffraction changes the wavelength of a wave. | The wave stays in the same medium, so its speed and frequency, and therefore \(\lambda\), are unchanged. | Diffraction changes the direction and shape of the wavefronts only. |
| The narrower the gap, the better the diffraction, always. | A gap much smaller than \(\lambda\) lets almost none of the wave through. | Spreading is greatest when the gap is similar to the wavelength. |
| \(\theta\) in \(n\lambda = d\sin\theta\) is the angle between two maxima on opposite sides. | \(\theta\) is measured from the straight-through direction. | The angle between the \(\pm n\) maxima is \(2\theta\); halve it first. |
| The second-order angle is twice the first-order angle. | \(\sin\theta\) is proportional to \(n\), not \(\theta\). | \(\sin\theta_2 = 2\sin\theta_1\); find each angle separately. |
| A grating gives as many orders as you like. | \(\sin\theta\) cannot exceed 1. | Highest order = whole-number part of \(d/\lambda\). |
| Electrons are particles, so the rings must be caused by collisions. | Collisions cannot produce a regular bright-and-dark pattern that shrinks as the p.d. rises. | The rings are a diffraction pattern: electrons have wavelength \(\lambda = h/p\). |
How this topic connects
2.2 Superposition and stationary waves
The grating equation is the condition for constructive superposition, path difference \(= n\lambda\), using the same path-difference and phase ideas as two-source interference.
2.5 Photons, the photoelectric effect and line spectra
Light diffracts like a wave but also arrives as photons; electrons are particles that also diffract. Together these make up wave–particle duality, and \(h\) links both sides. A grating is also how line spectra are measured.
Unit 1 mechanics, Unit 4 particle physics
The electron calculation joins \(V = W/Q\), \(E_k = \tfrac12 mv^2\) and \(p = mv\). To probe smaller structures such as nuclei, physicists use particles with much larger momentum, because a shorter de Broglie wavelength resolves finer detail.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(n\lambda = d\sin\theta\) | angles of maxima from a diffraction grating; d and λ in m | Yes |
| \(\lambda = h/p\) | de Broglie wavelength of a moving particle; p in kg m s⁻¹ | Yes |
| \(p = mv\) | momentum | Yes |
| \(v = f\lambda\) | wavelength of a wave, to compare with a gap | Yes |
| \(E_k = \tfrac12 mv^2\) and \(V = W/Q\) | energy of an accelerated electron | Yes |
| \(d = 1/\)(lines per metre) | slit spacing of a grating | No — learn it |
| \(n_\text{max}\) = whole-number part of \(d/\lambda\) | highest order | No — derive it from sin θ ≤ 1 |
| \(\tan\theta = x/D\) | angle from positions on a screen | No — derive it from the geometry |
| \(eV = \tfrac12 mv^2\) | electron accelerated from rest through p.d. V | No — derive it from W = QV |
Check your understanding
1. Explain why sound from a room can be heard through an open window from round the side, but the room cannot be seen from there.
Sound wavelengths (roughly 0.1 m to a few metres) are similar to the window width, so the sound diffracts strongly round the edges. Light wavelengths (about \(10^{-7}\) m) are millions of times smaller than the window, so light diffracts negligibly and travels in almost straight lines.
2. A grating with 120 lines per mm gives a first-order maximum at 4.2°. Calculate the wavelength.
\(d = 1 \div 1.20 \times 10^{5} = 8.33 \times 10^{-6}\) m; \(\lambda = 8.33 \times 10^{-6} \times \sin 4.2° =\) 6.10 × 10⁻⁷ m (610 nm).
3. Why are the maxima from a diffraction grating much sharper than the fringes from two slits?
With many slits, the waves only all arrive in phase when the path difference between neighbours is exactly \(n\lambda\). Slightly away from that angle the many waves have a spread of phases and cancel, so each maximum is narrow; more slits also let through more light, so the maxima are bright.
4. An electron has a de Broglie wavelength of 5.0 × 10⁻¹¹ m. Calculate its momentum.
\(p = h/\lambda = 6.63 \times 10^{-34} \div 5.0 \times 10^{-11} =\) 1.3 × 10⁻²³ kg m s⁻¹.
5. In an electron diffraction tube, the accelerating p.d. is increased. Explain what happens to the rings.
The electrons gain more kinetic energy, so they have greater speed and momentum. Their de Broglie wavelength \(h/p\) is shorter, so they are diffracted through smaller angles: the rings become smaller (closer to the centre).
6. Electrons are accelerated from rest through 10 kV. Calculate their de Broglie wavelength.
\(\lambda = h/\sqrt{2meV} = 6.63 \times 10^{-34} \div \sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 10\,000} =\) 1.23 × 10⁻¹¹ m, half the 2.5 kV value, as \(\lambda \propto 1/\sqrt{V}\) predicts.
Specification checklist
Numbers refer to the specification points.
- (50) I can say what diffraction is and use Huygens' construction to explain how a wave spreads through a gap or round an obstacle.
- (50) I can explain how the amount of spreading depends on the gap width compared with the wavelength.
- (51) I can derive and use \(n\lambda = d\sin\theta\), find \(d\) from lines per mm, and find the highest order.
- (52) I can describe how to find the wavelength of light with a diffraction grating, process the measurements and estimate their uncertainty.
- (53) I can explain how electron diffraction shows that electrons behave as waves.
- (54) I can use \(\lambda = h/p\), including for electrons accelerated from rest through a known p.d.