AP Physics 2 (2025)
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15.1 Quantum Theory and Wave-Particle Duality

Verified for the 2025 AP Physics 2 (2025) examCitation:

Quantum theory explains phenomena classical mechanics couldn't, like atomic spectra and the photoelectric effect. It describes matter at atomic scales, where particles exhibit both particle-like and wave-like behavior, a concept known as wave-particle duality.

Properties of Wave-Particle Duality

Quantum Theory Fundamentals

Quantum theory emerged in the early 20th century to explain observations that classical physics simply could not account for. These phenomena included atomic spectra (specific patterns of light emitted by atoms), blackbody radiation (the emission of electromagnetic radiation from heated objects), and the photoelectric effect (electrons ejected from metals when struck by light).

  • Quantum theory provides a framework for understanding matter and energy at the atomic and subatomic scales
  • At these tiny scales, the rules of classical physics break down and new principles emerge
  • One of the most revolutionary concepts is that fundamental particles don't behave exclusively as either particles or waves, but exhibit characteristics of both
  • This dual nature applies to electrons, protons, neutrons, and other subatomic particles

Light as Wave and Particle

Light had been traditionally understood as a wave following Maxwell's electromagnetic theory. However, quantum theory revealed its particle nature as well.

  • Light consists of discrete packets of energy called photons
  • Photons have several unique properties:
    • They are massless particles
    • They carry no electric charge
    • Their energy is directly proportional to their frequency according to: E=hfE=hf where hh is Planck's constant (6.63×10346.63 \times 10^{-34} J·s) and ff is frequency
  • The wavelength of light relates to its frequency through: λ=cf\lambda=\frac{c}{f} where cc is the speed of light
  • Photons travel in straight lines unless they interact with matter
  • When interacting with matter, photons can undergo:
    • Reflection (bouncing off surfaces)
    • Refraction (bending when passing between different media)
    • Diffraction (bending around obstacles)

Photon Speed in Media

The speed at which photons travel depends on the medium through which they're moving.

  • In a vacuum (free space), all photons travel at the universal speed limit: c=3.00×108 m/sc=3.00 \times 10^{8} \mathrm{~m/s}
  • When photons travel through a medium like water or glass, they slow down
  • The speed of photons in a medium is related to the index of refraction (nn) of that medium: v=cnv = \frac{c}{n}
  • Materials with higher indices of refraction (like diamond, n2.4n ≈ 2.4) slow photons more than materials with lower indices (like air, n1.0003n ≈ 1.0003)
  • This slowing effect is what causes light to bend (refract) when passing between different media

Wave Properties of Particles

One of the most profound insights of quantum theory is that particles like electrons also exhibit wave-like behavior. This was first proposed by Louis de Broglie in 1924.

  • Particles demonstrate wave properties that can be observed in experiments similar to Young's double-slit experiment
  • When electrons pass through two narrow slits, they create an interference pattern on a detector screen—behavior previously associated only with waves
  • The wavelength of a particle is given by the de Broglie relation: λ=hp\lambda=\frac{h}{p} where hh is Planck's constant and pp is the particle's momentum
  • Important implications of the de Broglie wavelength:
    • Smaller particles with less momentum have longer wavelengths
    • Quantum effects become significant when a particle's de Broglie wavelength is comparable to the size of the system it's in
    • For everyday objects like baseballs, the de Broglie wavelength is so incredibly small that quantum effects are unnoticeable

Quantization in Bound Systems

In quantum systems where particles are confined or bound (like electrons in atoms), energy and momentum can only take on specific, discrete values—they are quantized.

  • Electrons in atoms can only exist in specific energy levels, not at arbitrary energies between these levels
  • When an electron transitions between energy levels, it must absorb or emit a photon with energy exactly equal to the energy difference between levels
  • The quantization of energy explains why atoms emit or absorb light only at specific frequencies (creating spectral lines)
  • Other examples of quantization include:
    • Vibrational energy levels in molecules
    • Energy levels of electrons in solids (which leads to band structure)
    • The "particle in a box" model, which shows that a confined particle can only have certain energy values

Practice Problem 1: Photon Energy

A radio station broadcasts at a frequency of 99.5 MHz. Calculate the energy of a single photon emitted by this radio station.

Solution

To find the energy of a photon, we use the equation E=hfE = hf where:

  • hh is Planck's constant = 6.63×10346.63 \times 10^{-34} J·s
  • ff is the frequency = 99.5 MHz = 99.5×10699.5 \times 10^6 Hz

Substituting these values: E=(6.63×1034 J\cdotps)×(99.5×106 Hz)E = (6.63 \times 10^{-34} \text{ J·s}) \times (99.5 \times 10^6 \text{ Hz}) E=6.60×1026E = 6.60 \times 10^{-26} J

This energy is extremely small, which is why radio waves don't have enough energy to cause ionization or other quantum effects that higher-energy photons (like X-rays) can cause.

Practice Problem 2: De Broglie Wavelength

An electron is accelerated from rest through a potential difference of 100 V. Calculate the de Broglie wavelength of this electron.

Solution

To find the de Broglie wavelength, we need to:

  1. Calculate the kinetic energy gained by the electron
  2. Find the momentum of the electron
  3. Apply the de Broglie relation

Step 1: The energy gained equals the charge of the electron times the potential difference: E=qV=(1.60×1019 C)×(100 V)=1.60×1017E = qV = (1.60 \times 10^{-19} \text{ C}) \times (100 \text{ V}) = 1.60 \times 10^{-17} J

Step 2: This energy is converted to kinetic energy, so: KE=12mv2=1.60×1017KE = \frac{1}{2}mv^2 = 1.60 \times 10^{-17} J

The momentum p=mvp = mv can be found by: p=2mKEp = \sqrt{2mKE} p=2×(9.11×1031 kg)×(1.60×1017 J)p = \sqrt{2 \times (9.11 \times 10^{-31} \text{ kg}) \times (1.60 \times 10^{-17} \text{ J})} p=5.41×1024p = 5.41 \times 10^{-24} kg}\cdot\text{m/s

Step 3: Apply the de Broglie relation: λ=hp=6.63×1034 J\cdotps5.41×1024 kgm/s=1.23×1010\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \text{ J·s}}{5.41 \times 10^{-24} \text{ kg}\cdot\text{m/s}} = 1.23 \times 10^{-10} m

This wavelength (0.123 nm) is comparable to the size of atoms, which is why electron microscopes can resolve atomic structures.