Grade 11

Advanced

8 min

Lesson 11 of 12

Wave-Particle Duality and de Broglie

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Exploring the revolutionary idea that matter, not just light, exhibits wave-like properties.

What if you were told that you, your phone, and every atom in the universe are not just solid objects, but also ripples in space? In the quantum world, the line between a 'thing' and a 'wave' completely vanishes.

Learning Objectives

1.Explain the concept of wave-particle duality for both light and matter
2.Calculate the de Broglie wavelength of moving particles using $\lambda = \frac{h}{mv}$
3.Discuss the evidence for electron diffraction and its implications for quantum mechanics

The Radical Symmetry of Nature

For centuries, physicists debated whether light was a wave or a particle. By the early 1900s, Einstein proved light behaves like a particle (the photon) during the photoelectric effect, while Maxwell proved it behaves like a wave. In 1924, Louis de Broglie proposed a shocking symmetry: if light waves can act like particles, then matter particles must act like waves. This concept, known as wave-particle duality, suggests that every moving object has an associated wavelength. However, because Planck’s constant ( $h$ ) is so incredibly small ( $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$ ), these wave properties are only noticeable for tiny masses like electrons. For a human or a car, the wavelength is so small it is effectively invisible.

Quick Check

According to de Broglie, what two properties must an object have to exhibit a wavelength?

Look at the variables required for momentum.

Answer

The object must have mass and be in motion (momentum).

Calculate the de Broglie wavelength of an electron ( $m = 9.11 \times 10^{-31} \text{ kg}$ ) moving at $1.0 \times 10^6 \text{ m/s}$ .

1. Identify the formula: $\lambda = \frac{h}{mv}$ 2. Plug in the constants: $\lambda = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(1.0 \times 10^6)}$ 3. Calculate: $\lambda \approx 7.27 \times 10^{-10} \text{ m}$ 4. This wavelength is roughly the size of an atom!

The Proof: Electron Diffraction

How do we know matter waves are real? Waves have a unique signature: diffraction. When waves pass through a gap similar to their wavelength, they spread out and interfere. In 1927, Davisson and Germer fired a beam of electrons at a nickel crystal. Instead of bouncing off like tiny marbles, the electrons created an interference pattern—the unmistakable fingerprint of a wave. This proved that electrons aren't just 'points' of mass; they are 'wave packets' that can overlap and cancel each other out. This discovery laid the foundation for the Electron Microscope, which uses the tiny wavelengths of electrons to see objects 1,000 times smaller than what light microscopes can resolve.

Quick Check

Why don't we see a baseball diffract when it passes through a doorway?

Think about the inverse relationship between mass and wavelength.

Answer

The baseball's mass is so large that its de Broglie wavelength is billions of times smaller than the width of the doorway.

Calculate the wavelength of a $0.15 \text{ kg}$ baseball thrown at $40 \text{ m/s}$ .

1. Formula: $\lambda = \frac{h}{mv}$ 2. Calculation: $\lambda = \frac{6.626 \times 10^{-34}}{(0.15)(40)}$ 3. Result: $\lambda \approx 1.1 \times 10^{-34} \text{ m}$ 4. Conclusion: This is $10^{19}$ times smaller than a proton. It is physically impossible to measure a wave this small, which is why classical physics works for large objects.

Advanced Application: Kinetic Energy and Waves

In many laboratory settings, we don't know the velocity of a particle directly, but we do know its Kinetic Energy (

K

). Since

K = \frac{1}{2}mv^2

, we can rewrite momentum (

p = mv

) as

p = \sqrt{2mK}

. Substituting this into de Broglie’s equation gives us a powerful tool:

\lambda = \frac{h}{\sqrt{2mK}}

This relationship shows that as you increase the energy of a particle (by accelerating it through a voltage), its wavelength gets shorter. This is exactly how we tune electron microscopes to achieve higher resolutions.

An electron is accelerated from rest through a potential difference of $100 \text{ V}$ , giving it a kinetic energy of $1.6 \times 10^{-17} \text{ J}$ . Find its wavelength.

1. Use the energy-wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$ 2. Substitute values: $\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2(9.11 \times 10^{-31})(1.6 \times 10^{-17})}}$ 3. Simplify the denominator: $\sqrt{2.915 \times 10^{-47}} \approx 5.4 \times 10^{-24}$ 4. Final Result: $\lambda \approx 1.23 \times 10^{-10} \text{ m}$ or $0.123 \text{ nm}$ .

Key Takeaways

1Wave-particle duality states that all matter has both corpuscular (particle) and undulating (wave) properties.
2The de Broglie wavelength is inversely proportional to momentum: $\lambda = \frac{h}{p}$ .
3Electron diffraction experiments provided the first experimental evidence that matter behaves as a wave.
4Macroscopic objects have wavelengths too small to detect because of their large mass relative to Planck's constant.

Quick Check

Multiple Choice

If the velocity of an electron is doubled, what happens to its de Broglie wavelength?

Multiple Choice

Which phenomenon best demonstrates the wave nature of electrons?

True/False

A stationary proton has a de Broglie wavelength of zero.

What's Next?

Review Tomorrow

In 24 hours, try to write down the de Broglie equation from memory and explain why we don't see your pet cat 'diffract' through a cat-door.

Practice Activity

Research how 'Electron Microscopes' use de Broglie's principle to see viruses that are invisible to normal light.

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Wave-Particle Duality and de Broglie

Learning Objectives

The Radical Symmetry of Nature

The Proof: Electron Diffraction

Advanced Application: Kinetic Energy and Waves

Key Takeaways

Quick Check

What's Next?

Wave-Particle Duality and de Broglie

Learning Objectives

The Radical Symmetry of Nature

The Proof: Electron Diffraction

Advanced Application: Kinetic Energy and Waves

Key Takeaways

Quick Check

What's Next?