Wave-Particle Duality

Section: Quantum Physics | Syllabus: Cambridge A Level Physics (9702)

Wave-Particle Duality

Evidence for Dual Nature

Electromagnetic radiation and matter exhibit both wave-like and particle-like properties.

Wave evidence: Interference and diffraction (e.g., Young's double slit).
Particle evidence: Photoelectric effect, Compton scattering.

Electron Diffraction

In 1927, electron diffraction provided evidence that particles (electrons) also behave as waves. When a beam of electrons passes through a thin film of graphite (polycrystalline), a diffraction pattern of concentric rings is observed on a screen.

Explanation

The gap between atomic planes in graphite ( 10^-10 m) is comparable to the de Broglie wavelength of the electrons, causing diffraction. This proves electrons have wave properties.

de Broglie Wavelength

Louis de Broglie proposed that any moving particle has an associated wavelength.

λ = (h/p)

where p is momentum (mv).
λ = (h/mv) Why don't we see diffraction of cricket balls?

For macroscopic objects, the mass m is large, so momentum mv is large. This makes the wavelength λ = h/mv extremely small (much smaller than any aperture), so no diffraction is observable.

Worked Examples

Worked Example: Wavelength of an Electron

Question: An electron travels at 2.0 × 10^7 m s^-1. Calculate its de Broglie wavelength. (m_e = 9.11 × 10^-31 kg)

Solution

1. Calculate momentum: p = mv = (9.11 × 10^-31)(2.0 × 10^7) = 1.82 × 10^-23 kg m s^-1

2. Calculate wavelength: λ = 6.63 × 10^-341.82 × 10^-23 = 3.6 × 10^-11 m

Worked Example: de Broglie Wavelength of Accelerated Electrons

Question: Calculate the de Broglie wavelength of electrons that have been accelerated across a potential difference of 3.0 kV. Give your answer in pm.

Solution

1. Calculate kinetic energy gained: E_k = eV = (1.60 × 10^-19)(3000) = 4.80 × 10^-16 J

2. Find velocity using E_k = (1/2)mv^2:

v = √(2E_k/m) = √2 × 4.80 × 10^-169.11 × 10^-31 = 3.25 × 10^7 m s^-1

3. Calculate momentum: p = mv = (9.11 × 10^-31)(3.25 × 10^7) = 2.96 × 10^-23 kg m s^-1

4. Apply de Broglie equation: λ = (h/p) = 6.63 × 10^-342.96 × 10^-23 = 2.24 × 10^-11 m

λ = 22.4 pm

Worked Example: Why Macroscopic Objects Don't Show Wave Properties

Question: A tennis ball has mass 67 g and a velocity of 25 m s^-1. Give two reasons why it would be meaningless to consider the wave properties of the ball.

Solution

1. Calculate the de Broglie wavelength:

λ = (h/mv) = 6.63 × 10^-34(0.067)(25) = 4.0 × 10^-34 m

Reason 1: This wavelength (10^-34 m) is immeasurably small - far smaller than any aperture, atomic spacing, or measuring device could detect. No diffraction would be observable.

Reason 2: The wavelength is approximately 10^-33 times smaller than the ball itself, so wave effects would be completely negligible compared to the ball's dimensions.

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Wave-Particle Duality