DoAssignment.ca

F3.2 · Describe evidence for the wave model of matter

Learn to describe evidence for the wave model of matter through clear examples and targeted practice.

Ontario Grade 12 Physics

Revolutions in Modern Physics: Quantum Mechanics and Special Relativity

How electron diffraction supports the idea that matter can show wave behaviour

In everyday life, an electron behaves like a tiny particle: it has mass and can transfer energy and momentum. Yet experiments also show patterns associated with waves. The wave model of matter describes this wave-like behaviour. It does not mean that a particle is a water wave or that it spreads out like a visible ripple. In this lesson, the key evidence is diffraction: electrons striking a crystal produce a pattern that is explained by wave interference. A relationship between momentum and wavelength helps connect the observations to the model.

What you will learn

1. From familiar waves to matter waves

A wave transfers energy and can produce effects such as reflection, refraction, diffraction, and interference. Diffraction is the spreading or bending of a wave when it passes through an opening or around an obstacle. Interference occurs when waves overlap. Their effects can reinforce one another or partly cancel. A pattern of stronger and weaker intensity can result.
Matter is made of particles, such as electrons and protons. In the particle model, a particle has a definite mass and momentum. Momentum is a vector: it has both magnitude and direction. Mass and speed are scalars: they have magnitude but no direction. The wave model of matter adds a way to describe wave-like behaviour of particles. It does not remove their particle properties.
The key Grade 12 relationship is the de Broglie relationship. It assigns a wavelength to a particle with momentum. For a particle moving much more slowly than light, momentum can be calculated from mass and velocity. Use SI units: momentum in kilogram metres per second, wavelength in metres, and Planck’s constant in joule seconds.
For a moving particle, choose the particle as the system. Describe its motion in a chosen reference frame, such as the laboratory frame. A reference frame is the viewpoint from which position and motion are described. Choose the particle’s direction of travel as positive. The sign gives the momentum direction; the wavelength is a positive scalar.
λ=hp=hmv\lambda=\frac{h}{p}=\frac{h}{mv}

2. Electron diffraction: the central evidence

A crystal is a solid whose atoms are arranged in a regular repeating pattern. The gaps between neighbouring atomic layers are very small. When a beam of electrons is directed at a crystal, the electrons scatter from the atoms. Under suitable conditions, the scattered electrons form distinct intensity maxima rather than a uniform spread.
This pattern is evidence for wave behaviour. Waves scattered from regularly spaced parts of the crystal can reinforce in some directions and cancel in others. The result is a diffraction pattern. The pattern’s structure depends on the electron wavelength and the crystal spacing. A particle-only picture in which electrons simply bounce from isolated atoms does not account for the characteristic wave-like pattern as well.
In the Davisson–Germer experiment, electrons scattered from a nickel crystal produced a pattern with strong intensity at particular angles. The observed result agreed with the wavelength predicted for the electrons using their momentum. This agreement is important: the wave model did not merely label the pattern; it connected a measured pattern with a predicted wavelength.
This is experimental evidence, not a proposed classroom procedure. The observation is the scattered-electron intensity pattern. Interpreting the pattern as diffraction, and associating a wavelength with the electrons, is the model used to explain that observation.
2dsin⁡θ=nλ2d\sin\theta=n\lambda

3. Reading the evidence and using the model

The crystal relationship shown above describes constructive reinforcement for a simple set of crystal planes. Here, dd is the spacing between planes, θ\theta is the angle measured from the plane, and nn is a positive whole-number order. The relationship is useful for connecting a diffraction angle to a wavelength when the crystal spacing is known. It is not needed for every calculation in this lesson.
The de Broglie relationship makes a testable comparison. If the momentum of a particle is larger, its predicted wavelength is smaller. A smaller wavelength can produce diffraction at different angles for the same crystal spacing. Thus, particle momentum, predicted wavelength, and observed diffraction pattern can be compared.
The model does not say that every object produces an easily visible diffraction pattern. For objects with large momentum, the predicted wavelength is extremely small. The wave effect is then difficult to detect with ordinary openings or crystal spacings. Electrons have sufficiently small mass that their wavelengths can be relevant to atomic-scale structures.
When evaluating evidence, keep the observation and explanation distinct. A detector records where electrons arrive and how often. The pattern is measured evidence. The wave model explains its regular maxima and their dependence on momentum and crystal spacing. The model is supported because its predictions fit the observed pattern.
p=mvp=mv

Worked example

1. Predicting an electron’s wavelength

An electron moves at 2.0×106 m/s2.0\times10^6\ \mathrm{m/s}. Calculate its de Broglie wavelength. Use electron mass 9.11×10−31 kg9.11\times10^{-31}\ \mathrm{kg} and Planck’s constant 6.63×10−34 J s6.63\times10^{-34}\ \mathrm{J\,s}.
  1. Set the system and direction
    The system is the electron, described in the laboratory frame. Choose its direction of travel as positive. The speed is given, so the momentum magnitude is the mass times the speed.
    p=mvp=mv
  2. Calculate momentum
    Substitute the given SI values. The momentum is positive because the electron moves in the chosen positive direction.
    p=(9.11×10−31 kg)(2.0×106 m/s)=1.822×10−24 kg m/sp=(9.11\times10^{-31}\ \mathrm{kg})(2.0\times10^6\ \mathrm{m/s})=1.822\times10^{-24}\ \mathrm{kg\,m/s}
  3. Calculate wavelength
    Divide Planck’s constant by the momentum magnitude. The units reduce to metres because a joule is a kilogram metre squared per second squared.
    λ=6.63×10−34 J s1.822×10−24 kg m/s=3.64×10−10 m\lambda=\frac{6.63\times10^{-34}\ \mathrm{J\,s}}{1.822\times10^{-24}\ \mathrm{kg\,m/s}}=3.64\times10^{-10}\ \mathrm{m}
Answer: To two significant figures, the electron’s wavelength is 3.6×10−10 m3.6\times10^{-10}\ \mathrm{m}.
Check: The units are metres, as required for wavelength. The wavelength is near the scale of atomic spacing, so diffraction by a crystal is plausible. Wavelength has no direction.

Worked example

2. Comparing wavelengths at the same speed

An electron and a proton travel at the same speed, 1.5×106 m/s1.5\times10^6\ \mathrm{m/s}. Compare their de Broglie wavelengths. Use proton mass 1.67×10−27 kg1.67\times10^{-27}\ \mathrm{kg} and electron mass 9.11×10−31 kg9.11\times10^{-31}\ \mathrm{kg}.
  1. Define the comparison
    Treat each particle separately in the same laboratory frame. Choose each particle’s direction of travel as positive. Since their speeds are equal, the heavier particle has greater momentum.
    λ=hmv\lambda=\frac{h}{mv}
  2. Find each wavelength
    Use the same value of Planck’s constant for both particles. Keep the mass and speed in SI units.
    λe=6.63×10−34(9.11×10−31)(1.5×106)=4.9×10−10 m,λp=6.63×10−34(1.67×10−27)(1.5×106)=2.6×10−13 m\lambda_e=\frac{6.63\times10^{-34}}{(9.11\times10^{-31})(1.5\times10^6)}=4.9\times10^{-10}\ \mathrm{m},\quad \lambda_p=\frac{6.63\times10^{-34}}{(1.67\times10^{-27})(1.5\times10^6)}=2.6\times10^{-13}\ \mathrm{m}
  3. Compare the results
    The proton’s wavelength is much smaller because its mass, and therefore its momentum at this speed, is much greater. Both wavelengths are positive scalars.
    λeλp≈1.8×103\frac{\lambda_e}{\lambda_p}\approx1.8\times10^3
Answer: The electron’s wavelength is about 4.9×10−10 m4.9\times10^{-10}\ \mathrm{m}, while the proton’s is about 2.6×10−13 m2.6\times10^{-13}\ \mathrm{m}. The electron’s wavelength is approximately 1.8×1031.8\times10^3 times larger.
Check: Both results have units of metres. The comparison is reasonable: at equal speed, the proton is much more massive, so its wavelength is much shorter.

Worked example

3. Interpreting a diffraction observation

A beam of electrons produces strong scattered intensity at particular angles after striking a crystal. Explain what this observation supports and how momentum matters, without treating the pattern as proof that electrons are ordinary waves.
  1. Identify the measured evidence
    The measured evidence is a non-uniform pattern of electron arrivals, with strong intensity at particular angles. Do not confuse this observation with the explanation used to interpret it.
  2. Apply the wave model
    The crystal has regularly spaced atoms. Electron waves scattered from this arrangement can reinforce in some directions and partly cancel in others. This accounts for the intensity maxima as diffraction.
    2dsin⁡θ=nλ2d\sin\theta=n\lambda
  3. Connect the model to momentum
    The model predicts the electron wavelength from its momentum. Agreement between the wavelength-based prediction and the observed pattern supports the wave model of matter.
    λ=hp\lambda=\frac{h}{p}
Answer: The angle-dependent intensity maxima are evidence of electron diffraction. Their agreement with a wavelength predicted from electron momentum supports the wave model of matter.
Check: This conclusion is limited to the evidence: electrons show wave-like behaviour in diffraction. It does not claim that electrons are ordinary visible waves.

Common mistakes and how to avoid them

Saying that an electron diffraction pattern proves electrons are exactly like water waves.
Correction: The pattern supports wave-like behaviour. Electrons also have particle properties, so the evidence supports a model rather than a claim that electrons are ordinary material waves.
Using speed as momentum in the de Broglie relationship.
Correction: Momentum depends on both mass and velocity. For the low-speed relationship used here, calculate momentum with mass times velocity before finding wavelength.
Calling the model’s explanation a measured result.
Correction: The measured result is the electron-arrival pattern. Diffraction and interference are the model-based explanation of that evidence.
Giving wavelength a positive or negative direction.
Correction: Wavelength is a scalar and is reported as a positive length. Momentum is a vector and its direction depends on the chosen frame and positive direction.

Lesson summary

Check your understanding

Question 1

A particle’s momentum increases while Planck’s constant stays the same. What happens to its de Broglie wavelength?
  1. It increases.
  2. It decreases.
  3. It stays the same.
  4. It becomes a vector.
Show answer and explanation
It decreases.
The de Broglie relationship makes wavelength inversely proportional to momentum. A larger momentum gives a smaller wavelength.

Question 2

Which observation is evidence for wave-like behaviour of electrons?
  1. Electrons have mass.
  2. Electrons form a diffraction pattern after scattering from a crystal.
  3. A crystal has a repeating arrangement of atoms.
  4. Momentum has a direction.
Show answer and explanation
Electrons form a diffraction pattern after scattering from a crystal.
A crystal-scattering pattern with intensity maxima is a diffraction observation, which is explained by wave behaviour.

Question 3

In an electron diffraction experiment, what is measured evidence and what is the model-based explanation?
  1. The wavelength is measured directly, and the arrival pattern is the explanation.
  2. The arrival pattern is measured, and diffraction explains its structure.
  3. The crystal spacing is the explanation, and electron arrivals are not evidence.
  4. The momentum is the pattern, and interference is a detector reading.
Show answer and explanation
The arrival pattern is measured, and diffraction explains its structure.
The detector records the distribution of electron arrivals. The wave model uses diffraction and interference to explain the pattern.

Key terms

Diffraction
The spreading or bending of waves when they pass an opening or encounter an obstacle or structure.
Interference
The reinforcing or partly cancelling effects that occur when waves overlap.
Momentum
A vector quantity related to an object’s mass and velocity.
De Broglie wavelength
The wavelength associated with a particle’s momentum in the wave model of matter.
Reference frame
The viewpoint used to describe position and motion.
Evidence
An observation or measurement used to assess whether a model explains physical behaviour.

Continue through SPH4U

View the complete SPH4U Ontario Grade 12 Physics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation F3.2. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question