Wave-particle duality and de Broglie wavelengthEdexcel A-Level Physics: Revision notes
Section 1
Particles and waves
Light shows wave behaviour in diffraction and interference. Electrons were first identified as particles, with mass, charge and momentum.
Experiments show that both can behave in both ways. This is wave-particle duality: particles such as electrons can show wave properties, such as diffraction and interference, in suitable conditions.
Section 2
Electron diffraction
In an electron diffraction tube, electrons are accelerated through a high potential difference in a vacuum and directed at a thin graphite film. A fluorescent screen shows a pattern of concentric bright rings.
The atoms of graphite act as a diffraction grating. The rings are maxima from diffraction and interference. Only waves can do this, so the experiment is evidence for the wave nature of electrons.
The pattern works because the electron wavelength is comparable to the atomic spacing (about 10⁻¹⁰ m).
Do not say that electrons "are" waves rather than particles. They show wave properties in diffraction and particle properties when they are detected or deflected by fields.
Section 3
The de Broglie wavelength
de Broglie proposed that any particle with momentum p has a wavelength
λ = h/p = h/(mv)
where h = 6.63 × 10⁻³⁴ J s is the Planck constant and p = mv for a non-relativistic particle.
- A larger momentum gives a shorter wavelength
- Electron diffraction experiments agree with the values predicted by this equation
Check the units. p must be in kg m s⁻¹ and h in J s, so λ comes out in metres.
Section 4
Worked example
An electron travels at 4.0 × 10⁶ m s⁻¹.
p = mv = 9.11 × 10⁻³¹ × 4.0 × 10⁶ = 3.6 × 10⁻²⁴ kg m s⁻¹
λ = h/p = 6.63 × 10⁻³⁴ ÷ 3.64 × 10⁻²⁴ = 1.8 × 10⁻¹⁰ m
This is about the spacing between atoms in a crystal, so a crystal diffracts the electrons.
Section 5
Electrons accelerated through a potential difference
An electron released from rest and accelerated through a potential difference V gains kinetic energy eV, so
½mv² = eV, giving v = √(2eV/m).
For V = 250 V, v = 9.4 × 10⁶ m s⁻¹, p = 8.5 × 10⁻²⁴ kg m s⁻¹ and λ = 7.8 × 10⁻¹¹ m.
A greater potential difference gives a greater momentum, a shorter wavelength and so smaller diffraction angles: the rings in the electron diffraction pattern shrink.
Section 6
Why everyday objects do not diffract
A cricket ball of mass 0.16 kg at 25 m s⁻¹ has p = 4.0 kg m s⁻¹ and λ = 6.63 × 10⁻³⁴ ÷ 4.0 = 1.7 × 10⁻³⁴ m.
This is far smaller than any gap, so no diffraction can be detected. Everything has a de Broglie wavelength, but it is only observable when the wavelength is comparable to the size of the gap or spacing, which needs a very small momentum.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Wave-particle duality and de Broglie wavelength
- Electrons travelling at a speed of 4.0 × 10⁶ m s⁻¹ are directed at a thin crystal in an evacuated tube. The spacing between neighbouring atoms in the crystal is about 2 × 10⁻¹⁰ m. The Planck constant is 6.63 × 10⁻³⁴ J s and the mass of an electron is 9.11 × 10⁻³¹ kg.Explain why a beam of these electrons would be expected to show noticeable diffraction at the crystal.2 marks
- In an electron diffraction tube, electrons are accelerated through a potential difference in a vacuum and directed at a thin film of graphite. They then strike a fluorescent screen, where a pattern of concentric bright rings is seen.Explain how the ring pattern shows that electrons have wave properties.2 marks
- Electrons are released from rest and accelerated through a potential difference of 250 V in a vacuum, then directed at a thin crystal. The electron charge is 1.60 × 10⁻¹⁹ C, the electron mass is 9.11 × 10⁻³¹ kg and the Planck constant is 6.63 × 10⁻³⁴ J s.Calculate the speed of the electrons after acceleration.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).