Specification: AQA A-level Physics (7408) 3.2.2.4 | Paper: 1 | Data and formulae booklet: given in the exam
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Wave–particle duality: all matter and radiation show both wave and particle properties. Light behaves as a wave in diffraction and interference, and as particles (photons) in the photoelectric effect; particles such as electrons can also behave as waves.
de Broglie wavelength: the wavelength associated with a moving particle, λ = h ÷ mv, where mv is the particle’s momentum. The greater the momentum, the shorter the wavelength.
Electron diffraction: a beam of electrons passing through a thin layer of polycrystalline graphite produces a pattern of concentric rings. Diffraction is a wave property, so this is the evidence that electrons behave as waves. It is significant only when λ is comparable to the atomic spacing (about 10−10 m).
The equation
de Broglie wavelength = Planck constant ÷ momentum λ = h ÷ mv (= h ÷ p)
| Quantity | Symbol | Unit |
|---|---|---|
| de Broglie wavelength | λ | metres (m) |
| Planck constant | h | joule seconds (J s), 6.63 × 10−34 J s |
| mass of the particle | m | kilograms (kg) |
| speed of the particle | v | metres per second (m s⁻¹) |
| momentum | p = mv | kg m s⁻¹ |
Rearranged: v = h ÷ mλ | m = h ÷ λv | mv = h ÷ λ | For a charge accelerated from rest through a pd V: ½mv² = eV, so mv = √(2meV) and λ = h ÷ √(2meV). Convert nm to m (× 10−9) and keV to J (× 1.60 × 10−16) first.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) An electron travels at 2.0 × 106 m s⁻¹. Calculate its de Broglie wavelength, to 2 significant figures. (h = 6.63 × 10−34 J s, mₑ = 9.11 × 10−31 kg)
Show solution
F: λ = h ÷ mv
I: λ = 6.63 × 10−34 ÷ (9.11 × 10−31 × 2.0 × 106) = 6.63 × 10−34 ÷ 1.822 × 10−24
F: λ is already the subject
A: 3.638… × 10−10 m = 3.6 × 10−10 m (2 s.f.)
Q2 (Core) A particle has a momentum of 4.5 × 10−24 kg m s⁻¹. Calculate its de Broglie wavelength, to 3 significant figures. (h = 6.63 × 10−34 J s)
Show solution
F: λ = h ÷ p
I: λ = 6.63 × 10−34 ÷ 4.5 × 10−24
F: λ is already the subject
A: 1.473… × 10−10 m = 1.47 × 10−10 m (3 s.f.)
Q3 (Core) An electron has a de Broglie wavelength of 0.15 nm. Calculate its speed, to 2 significant figures. (h = 6.63 × 10−34 J s, mₑ = 9.11 × 10−31 kg)
Show solution
Convert first: 0.15 nm = 0.15 × 10−9 m = 1.5 × 10−10 m
F: λ = h ÷ mv
I: 1.5 × 10−10 = 6.63 × 10−34 ÷ (9.11 × 10−31 × v)
F: v = h ÷ mλ = 6.63 × 10−34 ÷ (9.11 × 10−31 × 1.5 × 10−10) = 6.63 × 10−34 ÷ 1.3665 × 10−40
A: 4.851… × 106 m s⁻¹ = 4.9 × 106 m s⁻¹ (2 s.f.)
Q4 (Core) A 58 g tennis ball is served at 180 km h⁻¹. Calculate its de Broglie wavelength, to 2 significant figures, and explain why the ball is never seen to diffract. (h = 6.63 × 10−34 J s)
Show solution
Convert first: 58 g = 0.058 kg; 180 km h⁻¹ = 180 ÷ 3.6 = 50 m s⁻¹
F: λ = h ÷ mv
I: λ = 6.63 × 10−34 ÷ (0.058 × 50) = 6.63 × 10−34 ÷ 2.9
F: λ is already the subject
A: 2.286… × 10−34 m = 2.3 × 10−34 m (2 s.f.). This is far smaller than any gap or object (even a nucleus is about 10−15 m), so diffraction is far too small to observe.
Watch out: using the mass in grams or the speed in km h⁻¹. Both give a wavelength that is wrong by a factor of 1000 or 3.6.
Q5 (Exam) Neutron diffraction is used to study crystal structure. A beam of neutrons has a de Broglie wavelength of 1.8 × 10−10 m, similar to the spacing of atoms in a crystal. Calculate (a) the speed of the neutrons, to 2 significant figures, and (b) their kinetic energy in eV, to 2 significant figures. (h = 6.63 × 10−34 J s, mₙ = 1.675 × 10−27 kg, 1 eV = 1.60 × 10−19 J)
Show solution
Step 1 – (a) speed
F: λ = h ÷ mv
I: 1.8 × 10−10 = 6.63 × 10−34 ÷ (1.675 × 10−27 × v)
F: v = 6.63 × 10−34 ÷ (1.675 × 10−27 × 1.8 × 10−10) = 6.63 × 10−34 ÷ 3.015 × 10−37
A (a): 2199.0… m s⁻¹ = 2.2 × 103 m s⁻¹ (2 s.f.)
Step 2 – (b) kinetic energy
Eₖ = ½mv² = 0.5 × 1.675 × 10−27 × 2199.0…² = 4.049… × 10−21 J
A (b): 4.049… × 10−21 ÷ 1.60 × 10−19 = 0.0253… eV = 0.025 eV (2 s.f.)
Watch out: keep the unrounded speed for Step 2. Squaring a rounded 2200 m s⁻¹ makes the error twice as large.
Q6 (Exam) An electron and a proton both travel at 3.0 × 106 m s⁻¹. Calculate the de Broglie wavelength of each, to 2 significant figures, and the ratio λₑ ÷ λₚ, to 3 significant figures. (h = 6.63 × 10−34 J s, mₑ = 9.11 × 10−31 kg, mₚ = 1.673 × 10−27 kg)
Show solution
Step 1 – electron
F: λ = h ÷ mv
I: λₑ = 6.63 × 10−34 ÷ (9.11 × 10−31 × 3.0 × 106) = 6.63 × 10−34 ÷ 2.733 × 10−24
A: 2.425… × 10−10 m = 2.4 × 10−10 m (2 s.f.)
Step 2 – proton
λₚ = 6.63 × 10−34 ÷ (1.673 × 10−27 × 3.0 × 106) = 6.63 × 10−34 ÷ 5.019 × 10−21
A: 1.320… × 10−13 m = 1.3 × 10−13 m (2 s.f.)
Step 3 – ratio
at the same speed λ ∝ 1 ÷ m, so λₑ ÷ λₚ = mₚ ÷ mₑ = 1.673 × 10−27 ÷ 9.11 × 10−31 = 1836.4…
A: 1840 (3 s.f.). Only the electron’s wavelength is comparable to atomic spacing (~10−10 m), so only the electron beam would diffract noticeably through graphite.
Q7 (Exam) In an electron diffraction tube, electrons travelling at 8.0 × 106 m s⁻¹ pass through a thin graphite target and produce rings on a fluorescent screen. (a) Calculate the de Broglie wavelength of the electrons, to 2 significant figures. (b) Explain why the rings show that electrons behave as waves. (c) State and explain what happens to the rings when the electron speed is increased. (h = 6.63 × 10−34 J s, mₑ = 9.11 × 10−31 kg)
Show solution
F: λ = h ÷ mv
I: λ = 6.63 × 10−34 ÷ (9.11 × 10−31 × 8.0 × 106) = 6.63 × 10−34 ÷ 7.288 × 10−24
F: λ is already the subject
A (a): 9.097… × 10−11 m = 9.1 × 10−11 m (2 s.f.), comparable to the spacing between carbon atoms
A (b): the rings are a diffraction pattern. Diffraction is a property of waves only, so the electrons must be behaving as waves as they pass through the gaps between the atoms.
A (c): higher speed means greater momentum, so λ = h ÷ mv is smaller. The electrons diffract less, so the rings get smaller (closer together).
Watch out: saying the rings are caused by electrons hitting the atoms like particles, or that faster electrons give bigger rings. Smaller λ always means less spreading.
Q8 (Exam) An electron is accelerated from rest through a potential difference of 150 V. Calculate (a) its speed, to 3 significant figures, and (b) its de Broglie wavelength, to 3 significant figures. (h = 6.63 × 10−34 J s, mₑ = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Step 1 – kinetic energy gained
Eₖ = eV = 1.60 × 10−19 × 150 = 2.4 × 10−17 J
Step 2 – (a) speed
½mv² = eV, so v = √(2eV ÷ m) = √(2 × 2.4 × 10−17 ÷ 9.11 × 10−31) = √(5.268… × 1013)
A (a): 7.258… × 106 m s⁻¹ = 7.26 × 106 m s⁻¹ (3 s.f.)
Step 3 – (b) wavelength
F: λ = h ÷ mv
I: λ = 6.63 × 10−34 ÷ (9.11 × 10−31 × 7.258… × 106) = 6.63 × 10−34 ÷ 6.612… × 10−24
A (b): 1.002… × 10−10 m = 1.00 × 10−10 m (3 s.f.), just right for diffraction by atoms
Watch out: eV here means e × V (charge × potential difference), not the unit electronvolt. The electron gains 150 eV of kinetic energy, which is 2.4 × 10−17 J.
Q9 (Stretch) An electron and a proton each have a kinetic energy of 2.0 keV. (a) Show that λ = h ÷ √(2mEₖ). (b) Calculate the de Broglie wavelength of each, to 3 significant figures. (c) Calculate the ratio λₑ ÷ λₚ, to 3 significant figures, and compare it with the ratio at equal speeds. (h = 6.63 × 10−34 J s, mₑ = 9.11 × 10−31 kg, mₚ = 1.673 × 10−27 kg, 1 eV = 1.60 × 10−19 J)
Show solution
Step 1 – (a) derivation
Eₖ = ½mv² = (mv)² ÷ 2m, so p = mv = √(2mEₖ) and λ = h ÷ p = h ÷ √(2mEₖ)
Convert first: 2.0 keV = 2.0 × 103 × 1.60 × 10−19 = 3.2 × 10−16 J
Step 2 – (b) electron
I: λₑ = 6.63 × 10−34 ÷ √(2 × 9.11 × 10−31 × 3.2 × 10−16) = 6.63 × 10−34 ÷ 2.414… × 10−23
A: 2.745… × 10−11 m = 2.75 × 10−11 m (3 s.f.)
Step 3 – (b) proton
I: λₚ = 6.63 × 10−34 ÷ √(2 × 1.673 × 10−27 × 3.2 × 10−16) = 6.63 × 10−34 ÷ 1.034… × 10−21
A: 6.407… × 10−13 m = 6.41 × 10−13 m (3 s.f.)
Step 4 – (c) ratio
at equal kinetic energy λ ∝ 1 ÷ √m, so λₑ ÷ λₚ = √(mₚ ÷ mₑ) = √1836.4… = 42.85…
A (c): 42.9 (3 s.f.), much smaller than the ratio of 1840 at equal speeds, because the electron must move much faster than the proton to have the same kinetic energy
Watch out: at equal kinetic energy the ratio is √(mₚ ÷ mₑ), not mₚ ÷ mₑ. Read whether the question fixes the speed or the energy.
Q10 (Stretch) Electrons accelerated from rest through a potential difference V are used in a diffraction experiment. (a) Show that λ = h ÷ √(2meV). (b) Calculate λ when V = 4.0 kV, to 3 significant figures. (c) The spacing between rows of carbon atoms in graphite is 2.13 × 10−10 m. Calculate the pd that gives electrons a de Broglie wavelength equal to this spacing, to 3 significant figures. Ignore relativistic effects. (h = 6.63 × 10−34 J s, mₑ = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Step 1 – (a) derivation
½mv² = eV, so m²v² = 2meV and mv = √(2meV); then λ = h ÷ mv = h ÷ √(2meV)
Step 2 – (b) wavelength at 4.0 kV
Convert first: 4.0 kV = 4000 V
I: λ = 6.63 × 10−34 ÷ √(2 × 9.11 × 10−31 × 1.60 × 10−19 × 4000) = 6.63 × 10−34 ÷ √(1.16608 × 10−45) = 6.63 × 10−34 ÷ 3.414… × 10−23
A (b): 1.941… × 10−11 m = 1.94 × 10−11 m (3 s.f.)
Step 3 – (c) required pd
F: squaring, λ² = h² ÷ 2meV, so V = h² ÷ (2meλ²)
I: V = (6.63 × 10−34)² ÷ (2 × 9.11 × 10−31 × 1.60 × 10−19 × (2.13 × 10−10)²) = 4.39569 × 10−67 ÷ 1.322… × 10−68
A (c): 33.23… V = 33.2 V (3 s.f.)
Watch out: λ ∝ 1 ÷ √V, not 1 ÷ V. Making the pd four times bigger only halves the wavelength.
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