Unit 15 · Topic 15.1

15.1 Quantum Theory and Wave-Particle Duality

Some experiments around 1900 couldn't be explained by classical physics, so quantum theory was built to explain them. Light can act as a wave or as a stream of particles called photons, each with energy E = hf. Matter can act as a wave too: a particle with momentum p has a de Broglie wavelength $\lambda = \frac{h}{p}$ .

Key terms

quantum theory
photon
Planck's constant
wave-particle duality
de Broglie wavelength

Why quantum theory was needed

By 1900, physicists had excellent theories for motion, electricity and light waves. But some results didn't fit: the sharp colored lines that glowing gases give off (15.3), the spectrum of light from hot objects (15.4) and the way light knocks electrons out of metal (15.5).

Quantum theory explains these. It's needed for anything at the scale of atoms and smaller. In quantum theory, the basic pieces of nature, such as electrons and photons, act like particles in some experiments and like waves in others. This is called wave-particle duality.

Photons

In Unit 14 you saw light interfere, which only waves can do. But light can also be modeled as a stream of particles called photons. A photon has no mass and no electric charge, and its energy is proportional to its frequency:

$E = hf = \frac{hc}{\lambda}$

Here h = 6.63 × 10⁻³⁴ J·s is Planck's constant (also 4.14 × 10⁻¹⁵ eV·s). A shortcut from the equation sheet is hc = 1240 eV·nm, so a photon's energy in electron volts is 1240 divided by its wavelength in nanometers. One electron volt (eV) is 1.60 × 10⁻¹⁹ J, the energy an electron gains crossing 1 V.

Higher frequency means more energy per photon: a violet photon carries more energy than a red one, and an X-ray photon far more. Brighter light of one color means more photons, not more energetic ones.

A photon keeps going in a straight line until it meets matter it can interact with. In a vacuum all photons move at c = 3.00 × 10⁸ m/s; in a material they move at $\frac{c}{n}$ , slower in materials with a bigger index of refraction. A photon also carries momentum, $p = \frac{h}{\lambda}$ , even though it has no mass. That matters in 15.6.

Matter waves

In 1924, Louis de Broglie proposed that particles like electrons also have a wavelength: $\lambda = \frac{h}{p} = \frac{h}{mv}$ . A smaller momentum means a longer wavelength.

This was confirmed when electrons sent through crystals, and later through double slits, made interference patterns, just like light. Even when electrons are sent through one at a time, the dots they leave on a detector gradually build up the bright and dark bands.

You only notice wave behavior when the wavelength is comparable to the size of what the particle passes through. An electron's wavelength can be about the size of an atom, so atoms and crystals show its wave nature. A baseball's wavelength is absurdly tiny, so it never shows any.

Quantized values

In bound systems, like an electron held in an atom, energy and momentum can only take certain discrete values. They're quantized, like the rungs of a ladder rather than a ramp. That idea drives the rest of this unit: the Bohr model (15.2), spectra (15.3) and blackbody radiation (15.4).

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Energy of a photon
Find the energy of a photon of green light with λ = 500 nm, in electron volts and in joules.
Show the solutionHide the solution
1. Step 1: Using hc = 1240 eV·nm: $E = \dfrac{hc}{\lambda} = \dfrac{1240\text{ eV·nm}}{500\text{ nm}} = 2.48$ eV.
2. Step 2: Convert: (2.48 eV)(1.60 × 10⁻¹⁹ J/eV) = 3.97 × 10⁻¹⁹ J.
3. Step 3: Check another way: $E = \dfrac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{500 \times 10^{-9}} = 3.98 \times 10^{-19}$ J. The tiny difference is rounding.
Answer: E ≈ 2.48 eV ≈ 3.98 × 10⁻¹⁹ J
Example 2Calculator allowed
Electron versus baseball
Find the de Broglie wavelength of (a) an electron (m = 9.11 × 10⁻³¹ kg) moving at 2.0 × 10⁶ m/s and (b) a 0.145 kg baseball moving at 40 m/s. Which one could show wave behavior?
Show the solutionHide the solution
1. Step 1: (a) $\lambda = \dfrac{h}{mv} = \dfrac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(2.0 \times 10^6)} = 3.6 \times 10^{-10}$ m, about 0.36 nm.
2. Step 2: (b) $\lambda = \dfrac{6.63 \times 10^{-34}}{(0.145)(40)} = 1.1 \times 10^{-34}$ m.
3. Step 3: Atoms in a crystal are spaced about 10⁻¹⁰ m apart, close to the electron's wavelength, so electrons diffract off crystals. The baseball's wavelength is far smaller than anything it could pass through, so it acts purely as a particle.
Answer: Electron: 3.6 × 10⁻¹⁰ m (shows wave behavior). Baseball: 1.1 × 10⁻³⁴ m (doesn't).
Example 3
Factor of change (trap)
An electron's de Broglie wavelength is λ. What is its new wavelength if (a) its speed is doubled, or (b) its kinetic energy is doubled?
Show the solutionHide the solution
1. Step 1: (a) λ = h/(mv), so doubling v halves λ: the new wavelength is λ/2.
2. Step 2: (b) Kinetic energy is $K = \frac{1}{2}mv^2$ , so doubling K multiplies v by √2, not by 2.
3. Step 3: Then the momentum grows by √2, so the wavelength becomes $\dfrac{\lambda}{\sqrt{2}} \approx 0.71\lambda$ .
4. Step 4: The trap is treating doubled kinetic energy as doubled momentum.
Answer: (a) λ/2, (b) λ/√2 ≈ 0.71λ

Common mistakes

Thinking brighter light means more energetic photons. Brightness is the number of photons; each photon's energy depends only on frequency.
Saying photons have mass because they have momentum. Photons are massless but still carry momentum $p = \frac{h}{\lambda}$ .
Forgetting to convert nm to m, or eV to J, when using h in J·s. Keep units consistent, or use hc = 1240 eV·nm.
Thinking only light has a wavelength. Every moving particle has a de Broglie wavelength; it's just too small to notice for everyday objects.

On the exam

Expect quick photon-energy and de Broglie calculations, and factor-of-change questions (what happens to λ if the speed or energy changes).
You may be asked for evidence that light acts as a wave (interference, 14.8) and evidence that it acts as particles (the photoelectric effect and Compton scattering). Name the specific experiment.

Connected topics

Videos

AP Physics 2 - Unit 15 - Lesson 5 - Wave Properties of Particles
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De Broglie wavelength | Physics | Khan Academy
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Quantum Mechanics - Part 1: Crash Course Physics #43
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Check yourself

4 questions on 15.1 Quantum Theory and Wave-Particle Duality. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

What is the energy of a photon of light with a wavelength of 500 nm? (Use hc = 1240 eV·nm.)

Question 2 of 4Calculator allowed

Light has a frequency of 6.0 × 10¹⁴ Hz. What is the energy of one photon of this light?

Question 3 of 4Calculator allowed

An electron (mass 9.11 × 10⁻³¹ kg) moves at 2.0 × 10⁶ m/s. What is its de Broglie wavelength?

Question 4 of 4Calculator allowed

An electron and a proton move at the same speed. Which has the longer de Broglie wavelength?

0 of 4 answered