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7.1 - Quantum Theory and Wave-Particle Duality

How Quantum Theory Got Started

Quantum theory was developed to explain things about matter and energy that classical mechanics just couldn’t handle.
Some of these phenomena include: - Atomic spectra: the discrete emission and absorption lines. - Blackbody radiation: how intensity is spread out across wavelength. - The photoelectric effect: light knocking electrons out of metals.
Classical physics couldn’t explain any of these, and that pushed people toward a whole new theory.
Here’s what classical physics predicted for each one, and what actually happens: - Atomic spectra: an electron circling a nucleus is accelerating, so classical physics says it should give off light nonstop and spiral into the nucleus. Atoms would glow at every wavelength and then collapse. Instead, atoms are stable and only give off a few sharp wavelengths (Topic 7.3). - Blackbody radiation: classical physics predicts that a hot object’s intensity keeps climbing at shorter and shorter wavelengths (the “ultraviolet catastrophe”). Real objects show a peak and then a drop-off (Topic 7.4). - The photoelectric effect: classical physics predicts that bright enough light of any color should knock electrons loose, and that brighter light should make faster electrons. Instead, there’s a cutoff frequency, and brightness only changes how many electrons come out (Topic 7.5).

Why We Need Quantum Theory

Quantum theory is what you need to describe how matter behaves at atomic and subatomic scales.
At everyday sizes, classical mechanics works fine. But at atomic scales (around 10−1010^{-10} m and smaller), quantum effects take over and classical predictions fall apart.

Wave-Particle Duality

In quantum theory, fundamental particles can act both like particles and like waves.
This is one of the strangest and most mind-bending ideas in physics: - Sometimes matter or light acts like a particle (localized and discrete). - Sometimes it acts like a wave (interference and diffraction). - Which behavior you see depends on how you observe or measure it.
Neither picture on its own is the full story. You need both to fully describe a quantum object.

Photons: Light as Particles

Light can be modeled both as a wave and as discrete particles called photons.

Photon Properties

A photon is a massless, electrically neutral particle whose energy is proportional to its frequency.
E=hfE = hf
where: - EE = energy of the photon (J) - hh = Planck’s constant = 6.63×10−346.63 \times 10^{-34} J·s - ff = frequency of the light (Hz)
Key properties: - Mass: m=0m = 0 (massless). - Charge: q=0q = 0 (electrically neutral). - Energy: E=hfE = hf (proportional to frequency). - Speed: cc in vacuum.

How Photons Move

Photons travel in straight lines unless they run into matter.
Those interactions include: - Reflection. - Refraction. - Absorption. - Scattering.
Between interactions, photons go in straight paths. In everyday gravitational fields, they don’t noticeably curve the way massive particles do.

Photon Speed

The speed of a photon depends on the medium it’s moving through.

Speed in Free Space

The speed of every photon in free space equals the speed of light:
c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s}
This is a universal constant that doesn’t care about the photon’s energy or frequency.

Speed in a Medium

In general, a photon’s speed through a medium is inversely proportional to that medium’s index of refraction:
v=cnv = \frac{c}{n}
where: - vv = speed of light in the medium - cc = speed of light in vacuum - nn = index of refraction (always ≥1\geq 1)
A higher index means slower light.

Connecting the Wave and Particle Pictures for Light

Light shows both wave and particle behavior, and the two models tie together.
Wave properties: wavelength (λ\lambda) and frequency (ff).
c=fλc = f\lambda
or
λ=cf\lambda = \frac{c}{f}
Particle properties: energy (EE).
E=hfE = hf
Putting them together:
E=hf=hcλE = hf = \frac{hc}{\lambda}
  • Higher frequency means shorter wavelength means higher-energy photons.
  • Lower frequency means longer wavelength means lower-energy photons.

Electron Volts

Joules are huge compared to the energy of one photon or one electron, so modern physics problems often use the electron volt (eV) instead. One electron volt is the energy an electron gains when it moves through a potential difference of 1 volt. Using ΔU=qΔV\Delta U = q\Delta V:
1 eV=(1.60×10−19 C)(1 V)=1.60×10−19 J1 \text{ eV} = (1.60 \times 10^{-19} \text{ C})(1 \text{ V}) = 1.60 \times 10^{-19} \text{ J}
The AP Physics 2 equation sheet lists Planck’s constant both ways, plus a shortcut for hchc: - h=6.63×10−34h = 6.63 \times 10^{-34} J·s =4.14×10−15= 4.14 \times 10^{-15} eV·s - hc=1.99×10−25hc = 1.99 \times 10^{-25} J·m =1240= 1240 eV·nm
With hc=1240hc = 1240 eV·nm, a photon’s energy in eV is just 1240 divided by its wavelength in nanometers, which saves a lot of powers of ten.
Example: a green laser pointer emits light with a wavelength of 532 nm. Find the energy of one photon in electron volts and in joules.
E=hcλ=1240 eV⋅nm532 nm=2.33 eVE = \frac{hc}{\lambda} = \frac{1240 \text{ eV} \cdot \text{nm}}{532 \text{ nm}} = 2.33 \text{ eV}E=(2.33 eV)(1.60×10−19 J/eV)=3.73×10−19 JE = (2.33 \text{ eV})(1.60 \times 10^{-19} \text{ J/eV}) = 3.73 \times 10^{-19} \text{ J}
Answer: each photon carries 2.33 eV, or 3.73×10−193.73 \times 10^{-19} J.

Matter Waves: Particles as Waves

Particles can show wave properties too, as versions of Young’s double-slit experiment have shown.
When electrons, neutrons, or other particles pass through a double slit, they build up an interference pattern, which is unmistakably wave-like behavior.
Here’s the strange part. If you send electrons through the slits one at a time, each electron lands on the screen as a single dot, the way a particle would. But after thousands of electrons, the dots pile up into bright and dark fringes, the same pattern light makes. The electron’s wave passes through both slits and interferes, but each electron is detected at one spot like a particle.

de Broglie Wavelength

A wave model of matter is captured by the de Broglie wavelength, which gets longer as a particle’s momentum gets smaller.
λ=hp\lambda = \frac{h}{p}
where: - λ\lambda = de Broglie wavelength (m) - hh = Planck’s constant = 6.63×10−346.63 \times 10^{-34} J·s - pp = momentum = mvmv (kg·m/s)
The key relationships: - Bigger momentum means shorter wavelength (less wave-like behavior). - Smaller momentum means longer wavelength (more wave-like behavior).
Example: electrons moving at 2.00×1052.00 \times 10^5 m/s pass through two slits 1.00×10−61.00 \times 10^{-6} m apart. A detector screen sits 1.50 m away. How far from the center is the first bright fringe?
First find the electron’s de Broglie wavelength:
λ=hmv=6.63×10−34(9.11×10−31)(2.00×105)=3.64×10−9 m\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(2.00 \times 10^{5})} = 3.64 \times 10^{-9} \text{ m}
Then use the double-slit equation from Unit 6, d(ymaxL)≈mλd\left(\frac{y_{\text{max}}}{L}\right) \approx m\lambda, with m=1m = 1:
y=λLd=(3.64×10−9)(1.50)1.00×10−6=5.46×10−3 my = \frac{\lambda L}{d} = \frac{(3.64 \times 10^{-9})(1.50)}{1.00 \times 10^{-6}} = 5.46 \times 10^{-3} \text{ m}
Answer: the first bright fringe is 5.46 mm from the center. Particles with no wave behavior would just make two stripes behind the slits, so the fringes are direct evidence that electrons act like waves.

When Quantum Effects Actually Matter

Quantum theory is necessary when the de Broglie wavelength is about the same size as the system itself.
Examples: - An electron in an atom: λ∼\lambda \sim the size of the atom (~10−1010^{-10} m), so quantum effects are essential. - A baseball: λ∼10−34\lambda \sim 10^{-34} m, way smaller than the ball, so quantum effects don’t matter. - Electrons in materials: λ\lambda is close to the lattice spacing, so quantum effects are important.

Quantization

Energy and momentum come in discrete, or quantized, values for bound systems described by quantum theory.
Quantization means certain quantities can only take on specific, discrete values instead of any value you want.
Examples: - Electron energy levels in atoms (only specific energies are allowed). - Angular momentum in atoms (quantized). - Vibrational and rotational states in molecules.
Free particles (unbound) can have a continuous range of energies. Bound particles (like electrons in atoms) are stuck with discrete energy levels.
Why being bound causes quantization: a bound particle’s matter wave is trapped in a small region, so it has to form a standing wave, just like a wave on a string fixed at both ends (Unit 6). Only certain wavelengths fit, so only certain momenta (p=h/λp = h/\lambda) and energies are allowed. A free particle isn’t trapped, so any wavelength works and its energy can take any value. Topic 7.2 shows this for the electron in a hydrogen atom.

Practice Problems

1. Find the energy of a red-light photon with wavelength 650 nm.
2. An electron has 100 eV of kinetic energy. Find its de Broglie wavelength. (Use 1 eV=1.6×10−191 \text{ eV} = 1.6 \times 10^{-19} J and me=9.11×10−31m_e = 9.11 \times 10^{-31} kg.)
3. Which has more energy, a blue-light photon (λ = 450 nm) or a red-light photon (λ = 650 nm)?
4. A photon has energy 3.0×10−193.0 \times 10^{-19} J. What are its frequency and wavelength?
5. Compare the de Broglie wavelengths of: - (a) An electron moving at 10610^6 m/s. - (b) A baseball (mass 0.15 kg) moving at 40 m/s.
6. Explain why we don’t see wave behavior in everyday objects like cars or people.
7. Light enters glass with index of refraction 1.5. By what factor does its speed drop? Does its frequency change? Does its wavelength change?
8. An X-ray photon has a wavelength of 0.10 nm. Find its energy in joules and in electron volts.

Solutions

1. First find the frequency using c=fλc = f\lambda:
f=cλ=3.00×108650×10−9=4.62×1014 Hzf = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{650 \times 10^{-9}} = 4.62 \times 10^{14} \text{ Hz}
Then find the energy:
E=hf=(6.63×10−34)(4.62×1014)=3.06×10−19 JE = hf = (6.63 \times 10^{-34})(4.62 \times 10^{14}) = 3.06 \times 10^{-19} \text{ J}
Or go straight there: E=hcλ=(6.63×10−34)(3.00×108)650×10−9=3.06×10−19E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{650 \times 10^{-9}} = 3.06 \times 10^{-19} J.
In eV: E=3.06×10−191.6×10−19=1.91E = \frac{3.06 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.91 eV.
Answer: the energy is 3.06×10−193.06 \times 10^{-19} J, or 1.91 eV.
2. First get the momentum from the kinetic energy:
KE=12mv2=p22mKE = \frac{1}{2}mv^2 = \frac{p^2}{2m}p=2m⋅KE=2(9.11×10−31)(100×1.6×10−19)p = \sqrt{2m \cdot KE} = \sqrt{2(9.11 \times 10^{-31})(100 \times 1.6 \times 10^{-19})}p=2.915×10−47=5.40×10−24 kg\cdotpm/sp = \sqrt{2.915 \times 10^{-47}} = 5.40 \times 10^{-24} \text{ kg·m/s}
Then use the de Broglie equation:
λ=hp=6.63×10−345.40×10−24=1.23×10−10 m=0.123 nm\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.40 \times 10^{-24}} = 1.23 \times 10^{-10} \text{ m} = 0.123 \text{ nm}
Answer: the de Broglie wavelength is 1.23×10−101.23 \times 10^{-10} m, or 0.123 nm, which is right around atomic sizes.
3. Using E=hcλE = \frac{hc}{\lambda}, the energy is inversely proportional to wavelength. Blue light has the shorter wavelength (450 nm < 650 nm), so it has the higher energy.
Working it out: - Blue: E=hc450×10−9=4.42×10−19E = \frac{hc}{450 \times 10^{-9}} = 4.42 \times 10^{-19} J. - Red: E=hc650×10−9=3.06×10−19E = \frac{hc}{650 \times 10^{-9}} = 3.06 \times 10^{-19} J.
Answer: blue-light photons have more energy, about 1.4 times as much.
4. Find the frequency:
f=Eh=3.0×10−196.63×10−34=4.52×1014 Hzf = \frac{E}{h} = \frac{3.0 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.52 \times 10^{14} \text{ Hz}
Find the wavelength:
λ=cf=3.00×1084.52×1014=6.64×10−7 m=664 nm\lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{4.52 \times 10^{14}} = 6.64 \times 10^{-7} \text{ m} = 664 \text{ nm}
Answer: the frequency is 4.52×10144.52 \times 10^{14} Hz and the wavelength is 664 nm (red-orange light).
5.
  1. Electron:
      2. p=mv=(9.11×10−31)(106)=9.11×10−25 kg\cdotpm/sp = mv = (9.11 \times 10^{-31})(10^6) = 9.11 \times 10^{-25} \text{ kg·m/s}
λ=hp=6.63×10−349.11×10−25=7.28×10−10 m=0.728 nm\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-25}} = 7.28 \times 10^{-10} \text{ m} = 0.728 \text{ nm}
  1. Baseball:
      2. p=mv=(0.15)(40)=6.0 kg\cdotpm/sp = mv = (0.15)(40) = 6.0 \text{ kg·m/s}
λ=hp=6.63×10−346.0=1.1×10−34 m\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.0} = 1.1 \times 10^{-34} \text{ m}
Answer: - Electron: 7.28×10−107.28 \times 10^{-10} m, right around atomic sizes, so wave effects can show up. - Baseball: 1.1×10−341.1 \times 10^{-34} m, absurdly small, so wave effects are completely negligible.
The baseball’s wavelength is roughly 102410^{24} times smaller.
6. The de Broglie wavelength λ=h/p\lambda = h/p is inversely proportional to momentum.
Everyday objects have a lot of mass and usually decent speeds, which gives them enormous momentum compared to subatomic particles. That makes their wavelengths incredibly tiny, many orders of magnitude below atomic scales.
For wave effects to show up, the wavelength has to be close to the size of the system or the openings and obstacles the object runs into. Since everyday objects have wavelengths around 10−3010^{-30} m or smaller, and real structures are macroscopic (bigger than 10−610^{-6} m), the ratio λ/L\lambda/L is so small that wave effects are undetectable.
Answer: everyday objects have huge momentum, so their de Broglie wavelengths are far too small (under 10−3010^{-30} m) next to their size or the structures they interact with. Wave effects need λ to be comparable to the system size.
7. Speed drop:
v=cn=c1.5=23cv = \frac{c}{n} = \frac{c}{1.5} = \frac{2}{3}c
The speed drops by a factor of 1.5 (down to 67% of the original).
Frequency: does not change. Frequency is set by the source and stays the same across boundaries.
Wavelength: using v=fλv = f\lambda with ff held constant:
λglass=vf=c/nf=1n⋅cf=λairn\lambda_{\text{glass}} = \frac{v}{f} = \frac{c/n}{f} = \frac{1}{n} \cdot \frac{c}{f} = \frac{\lambda_{\text{air}}}{n}
The wavelength drops by a factor of 1.5 (to 67% of the original).
Answer: the speed drops by a factor of 1.5, the frequency is unchanged, and the wavelength drops by a factor of 1.5.
8. Find the energy:
E=hcλ=(6.63×10−34)(3.00×108)0.10×10−9E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{0.10 \times 10^{-9}}E=1.989×10−251.0×10−10=1.99×10−15 JE = \frac{1.989 \times 10^{-25}}{1.0 \times 10^{-10}} = 1.99 \times 10^{-15} \text{ J}
Convert to eV:
E=1.99×10−151.6×10−19=1.24×104 eV=12.4 keVE = \frac{1.99 \times 10^{-15}}{1.6 \times 10^{-19}} = 1.24 \times 10^{4} \text{ eV} = 12.4 \text{ keV}
Answer: the energy is 1.99×10−151.99 \times 10^{-15} J, or 12.4 keV.

Common Mistakes

  • Mixing up which wavelength equation to use. For photons, λ=c/f\lambda = c/f. For matter, λ=h/p\lambda = h/p. Don’t swap them.
  • Forgetting how small h is. Planck’s constant, h=6.63×10−34h = 6.63 \times 10^{-34} J·s, is tiny, which is exactly why quantum effects don’t show up at everyday scales.
  • Thinking photons have mass. Photons are massless. They carry energy and momentum, but zero rest mass.
  • Using the wrong momentum for photons. For photons, p=E/c=h/λp = E/c = h/\lambda, not p=mvp = mv (since m=0m = 0).
  • Picking wave or particle exclusively. Quantum objects show both wave and particle properties depending on the experiment. Neither model alone is complete.
  • Getting frequency wrong in a medium. When light enters a medium, the frequency stays constant, while the wavelength and speed both drop by a factor of nn.
  • Wrong units for Planck’s constant. h=6.63×10−34h = 6.63 \times 10^{-34} J·s. Watch your units: energy in joules, frequency in Hz (1/s).
  • Forgetting to convert wavelength units. Wavelengths often come in nm. Convert to meters: 1 nm =10−9= 10^{-9} m.
  • Thinking higher frequency means lower energy. It’s the opposite. E=hfE = hf, so higher frequency means higher energy.
  • Confusing the de Broglie wavelength with a photon’s wavelength. Both use λ\lambda, but you calculate them differently. Photon: λ=c/f\lambda = c/f. Matter: λ=h/p=h/(mv)\lambda = h/p = h/(mv).
  • Not knowing when quantum effects matter. They’re significant when the de Broglie wavelength is comparable to the system size. For macroscopic objects, λ is vanishingly small.
  • Thinking quantization applies to everything. Only bound systems have quantized energy levels. Free particles have a continuous range of energies.
  • Getting the wavelength-energy relationship backwards. E∝1/λE \propto 1/\lambda (inverse), not directly proportional. Shorter wavelength means higher energy.
  • Forgetting photon momentum. Even though they’re massless, photons carry momentum: p=h/λ=E/cp = h/\lambda = E/c.
  • Assuming classical and quantum give the same answers. At atomic scales, classical mechanics fails and you need quantum mechanics.
  • Not using eV when it helps. Atomic and molecular energies are often cleaner in electron volts than in joules. 1 eV =1.6×10−19= 1.6 \times 10^{-19} J.
  • Treating wave-particle duality as just a teaching device. It’s a real, fundamental property of nature. Experiments confirm quantum objects genuinely show both sides.

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