Wave–Particle Duality and de Broglie Wavelength | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Classify the evidence before using an equation

Diffraction is the spreading of waves. Interference occurs when waves overlap and form a pattern. These are signs of wave behaviour.

A localised impact or a one-to-one energy transfer is evidence of particle behaviour. The table compares the observations.

Swipe left or right to read every column →

Object	Particle evidence	Wave evidence
Electromagnetic radiation	Photoelectric effect: energy is transferred by individual photons.	Diffraction and interference.
Electrons	Individual impacts are detected at particular points.	A beam produces a diffraction pattern.

Key idea

Wave–particle duality means that one classical model is not sufficient for every experiment. The observed behaviour determines the useful model.

Duality does not mean that an electron changes into a water-like wave, or that a photon is a tiny classical ball. “Wave” and “particle” name the behaviours shown by different measurements.

A moving particle has a de Broglie wavelength

Definition

1 mark

What is meant by the de Broglie wavelength?

Model answer: The de Broglie wavelength is the wavelength associated with a moving particle.

This exact one-mark wording has been assessed repeatedly (9702/41/M/J/23 Q7(a)) (9702/41/M/J/25 Q8(a)).

\lambda=\dfrac{h}{p}

Symbols

$\lambda$ = de Broglie wavelength of the moving particle (m)
$h$ = Planck constant (J s)
$p$ = momentum of the particle (N s)

For a non-relativistic particle, use the momentum relation below.

p=mv

Symbols

$p$ = momentum of the particle (N s)
$m$ = mass of the particle (kg)
$v$ = speed of the particle (m s⁻¹)

Larger momentum gives a shorter de Broglie wavelength. Every moving object has a de Broglie wavelength, but diffraction is noticeable only when that wavelength is comparable with the spacing of the structure it passes through.

Change	Momentum p	de Broglie wavelength λ
Increase speed of the same particle	increases	decreases
Increase mass at the same speed	increases	decreases
Decrease momentum	decreases	increases

Worked example

Find an electron's de Broglie wavelength

An electron moves at 4.9 × 10⁷ m s⁻¹. Calculate its de Broglie wavelength (9702/41/M/J/25 Q8(b)).

Calculate the electron momentum.
$p=mv$
$p=4.46\times10^{-23}\ \mathrm{N\,s}$
Divide Planck's constant by the momentum.
$\lambda=\dfrac{h}{p}$
$\lambda=1.49\times10^{-11}\ \mathrm{m}$

Answer

\lambda=1.5\times10^{-11}\ \mathrm{m}

Your turnrelationship check

The momentum of a moving particle is doubled. What happens to its de Broglie wavelength?

Show worked answer

It halves because de Broglie wavelength is inversely proportional to momentum.

Classify the evidence before using an equation

Diffraction is the spreading of waves. Interference occurs when waves overlap and form a pattern. These are signs of wave behaviour.

A localised impact or a one-to-one energy transfer is evidence of particle behaviour. The table compares the observations.

Swipe left or right to read every column →

Object	Particle evidence	Wave evidence
Electromagnetic radiation	Photoelectric effect: energy is transferred by individual photons.	Diffraction and interference.
Electrons	Individual impacts are detected at particular points.	A beam produces a diffraction pattern.

Key idea

Wave–particle duality means that one classical model is not sufficient for every experiment. The observed behaviour determines the useful model.

Duality does not mean that an electron changes into a water-like wave, or that a photon is a tiny classical ball. “Wave” and “particle” name the behaviours shown by different measurements.

A moving particle has a de Broglie wavelength

Definition

1 mark

What is meant by the de Broglie wavelength?

Model answer: The de Broglie wavelength is the wavelength associated with a moving particle.

This exact one-mark wording has been assessed repeatedly (9702/41/M/J/23 Q7(a)) (9702/41/M/J/25 Q8(a)).

\lambda=\dfrac{h}{p}

Symbols

$\lambda$ = de Broglie wavelength of the moving particle (m)
$h$ = Planck constant (J s)
$p$ = momentum of the particle (N s)

For a non-relativistic particle, use the momentum relation below.

p=mv

Symbols

$p$ = momentum of the particle (N s)
$m$ = mass of the particle (kg)
$v$ = speed of the particle (m s⁻¹)

Change	Momentum p	de Broglie wavelength λ
Increase speed of the same particle	increases	decreases
Increase mass at the same speed	increases	decreases
Decrease momentum	decreases	increases

Worked example

Find an electron's de Broglie wavelength

An electron moves at 4.9 × 10⁷ m s⁻¹. Calculate its de Broglie wavelength (9702/41/M/J/25 Q8(b)).

Calculate the electron momentum.
$p=mv$
$p=4.46\times10^{-23}\ \mathrm{N\,s}$
Divide Planck's constant by the momentum.
$\lambda=\dfrac{h}{p}$
$\lambda=1.49\times10^{-11}\ \mathrm{m}$

Answer

\lambda=1.5\times10^{-11}\ \mathrm{m}

Your turnrelationship check

The momentum of a moving particle is doubled. What happens to its de Broglie wavelength?

Show worked answer

It halves because de Broglie wavelength is inversely proportional to momentum.