Quantum Theory and Wave-Particle Duality: Study Guide (Physics 2 15.1)

Why a new theory was needed

By the end of the nineteenth century, classical physics (Newton’s laws, plus the wave theory of light and the electromagnetism you met in Units 10 to 14) explained a huge range of observations. But a few experiments with atoms and light gave results it could not explain at all:

• Atomic spectra. A hot gas of one element gives out light at only a few sharp wavelengths, not a smooth spread of colours. You will study this in Topic 15.3.
• Blackbody radiation. The classical model predicted that a hot object should give out an unlimited amount of energy at short wavelengths. It does not. This is Topic 15.4.
• The photoelectric effect. Light can knock electrons out of a metal, but whether it does depends on the light’s frequency, not on how bright it is. This is Topic 15.5.

Quantum theory was built to explain these results. It is needed whenever you describe matter and energy at the scale of atoms and smaller. Its central surprise is that the basic “things” of nature, such as photons and electrons, show both particle-like and wave-like behaviour. Which one you see depends on the experiment.

At everyday scales, quantum theory gives the same predictions as classical physics. So nothing you learned earlier is wrong. It is the large-scale limit of a deeper theory.

Light as photons

In Unit 14 you modelled light as an electromagnetic wave. That model explains interference, diffraction and polarization very well. But when light gives energy to matter, the energy arrives in separate packets. Each packet is a photon.

A photon:

• has no mass and no electric charge;
• carries energy proportional to its frequency:

E = hf = hc/λ

Here h is Planck’s constant, h = 6.63 × 10⁻³⁴ J·s, f is the frequency of the light and λ is its wavelength in a vacuum. Doubling the frequency doubles the energy of each photon. Halving the wavelength also doubles it.

Photon energies are tiny in joules, so physicists often use the electron-volt (eV). One electron-volt is the energy an electron gains when it moves through a potential difference of 1 V:

1 eV = 1.60 × 10⁻¹⁹ J

In these units h = 4.14 × 10⁻¹⁵ eV·s. A red photon of wavelength 700 nm carries about 1.78 eV; a violet photon of 400 nm carries about 3.11 eV. The ratio of the energies, 1.75, is just the inverse ratio of the wavelengths (700/400).

Brightness and photon energy are different things. A brighter beam of the same colour has more photons per second, not more energetic photons. If a beam has power P and every photon has energy E, the number of photons arriving per second is

N/t = P/E

How photons travel

• A photon moves in a straight line unless it interacts with matter (for example, it is absorbed, reflected or scattered).
• In a vacuum, every photon travels at c = 3.00 × 10⁸ m/s, whatever its energy.
• In a medium, photons travel more slowly. The speed is inversely proportional to the index of refraction: v = c/n. A larger n means a slower photon.

When light enters glass, its frequency does not change, because the source sets the frequency. So the energy of each photon, E = hf, does not change either. Since v = fλ, a lower speed means a shorter wavelength inside the medium: λ_medium = λ/n. Use the vacuum wavelength in E = hc/λ.

Evidence for both models

A two-slit experiment shows both kinds of behaviour at once. Send light at two narrow slits, but make it so dim that only one photon is in the apparatus at a time. Use a detector screen that records each arrival.

• Each photon is detected at one point, as a single dot. That is particle-like.
• Where any one photon lands cannot be predicted. But after thousands of photons, the dots build up the bright and dark fringes of a two-slit interference pattern. That is wave-like.

The wave model tells you the probability of a photon arriving at each point: high at the bright fringes, close to zero at the dark ones. The particle model tells you that energy is delivered in whole photons. You need both.

Matter waves

In 1924 Louis de Broglie suggested that the link works the other way round too: if waves of light behave like particles, then particles of matter should behave like waves. He proposed that any particle with momentum p has a wavelength, now called the de Broglie wavelength:

λ = h/p = h/(mv)

The wavelength increases as the momentum decreases. A slow, light particle has a long wavelength. A fast, heavy object has an extremely short one.

The evidence came in 1927, when Davisson and Germer fired electrons at a nickel crystal and found that they were scattered strongly only in certain directions. The rows of atoms acted like a diffraction grating, exactly as for X-rays. Since then, interference experiments like the one in Figure 1 have been done with electrons, neutrons and even large molecules. The electrons arrive one at a time as dots and still build up an interference pattern. So electrons, like photons, show both particle-like and wave-like behaviour.

When do matter waves matter?

Wave effects such as diffraction are noticeable only when the wavelength is comparable to the size of the gap, obstacle or system the wave meets. (You saw the same rule for light and sound in Unit 14.) So:

• For an electron in an atom, λ is about the size of the atom. Quantum theory is essential.
• For a thrown ball, λ is about 10⁻³⁴ m, far smaller than an atomic nucleus. Wave effects can never be seen, and classical mechanics works perfectly.

Particles with kinetic energy from a potential difference

Electrons in experiments are often accelerated from rest through a potential difference ΔV. The electric field does work qΔV on each one, so its kinetic energy becomes K = qΔV. Since K = p²/(2m), the momentum is p = √(2mK), and

λ = h/√(2mK) = h/√(2mqΔV)

This shows the functional dependence: λ ∝ 1/√ΔV for a given particle, and λ ∝ 1/√m for particles with the same charge and the same ΔV.

Bound systems and quantized energy

Think back to a string fixed at both ends (Topic 14.6). Only standing waves that fit a whole number of half-wavelengths into the string are possible, so only certain frequencies are allowed.

An electron held in an atom is a bound system: it cannot escape without extra energy. Its matter wave is confined, a little like the wave on the string. As a result, quantum theory predicts that the energy and momentum of a bound system can have only certain discrete, or quantized, values. The electron cannot have any energy in between. This is the idea behind atomic energy levels, which the next topic develops with the Bohr model.

A free particle, by contrast, is not confined, so its kinetic energy can take any value.

Worked example 1: photons from a laser

Question. A green laser pointer emits light of vacuum wavelength 532 nm with a power of 5.0 mW. (a) Find the energy of one photon in joules and in electron-volts. (b) How many photons leave the laser each second? (c) The beam enters a glass block with n = 1.50. Find the speed, wavelength and energy of the photons in the glass.

1. (a) Frequency: f = c/λ = (3.00 × 10⁸ m/s) ÷ (532 × 10⁻⁹ m) = 5.64 × 10¹⁴ Hz.
2. Energy per photon: E = hf = (6.63 × 10⁻³⁴ J·s)(5.639 × 10¹⁴ Hz) = 3.74 × 10⁻¹⁹ J.
3. In electron-volts: E = 3.739 × 10⁻¹⁹ J ÷ 1.60 × 10⁻¹⁹ J/eV = 2.34 eV.
4. (b) Photons per second: N/t = P/E = (5.0 × 10⁻³ J/s) ÷ (3.739 × 10⁻¹⁹ J) = 1.34 × 10¹⁶ photons per second.
5. (c) Speed in glass: v = c/n = (3.00 × 10⁸ m/s) ÷ 1.50 = 2.00 × 10⁸ m/s.
6. Wavelength in glass: λ_glass = λ/n = 532 nm ÷ 1.50 = 355 nm.
7. Energy: the frequency is unchanged, so E = hf is still 2.34 eV.

Answer. (a) 3.74 × 10⁻¹⁹ J = 2.34 eV. (b) 1.3 × 10¹⁶ photons per second (2 significant figures, as for the power). (c) 2.00 × 10⁸ m/s, 355 nm, 2.34 eV.

Interpretation. Even a weak laser sends out more than ten thousand million million photons each second. That is why light looks smooth and continuous in everyday life. Notice that 355 nm is in the ultraviolet range for light in a vacuum, but the photons in the glass are still green photons with 2.34 eV each. Colour and photon energy go with frequency, not with the wavelength inside a medium.

Worked example 2: an electron wave and a ball

Question. An electron is accelerated from rest through a potential difference of 150 V. (a) Find its de Broglie wavelength. (b) Compare it with the de Broglie wavelength of a 0.15 kg ball thrown at 30 m/s. (c) By what factor does the electron’s wavelength change if the potential difference is increased to 600 V?

1. (a) Kinetic energy: K = eΔV = (1.60 × 10⁻¹⁹ C)(150 V) = 2.40 × 10⁻¹⁷ J.
2. Momentum: p = √(2mK) = √(2 × 9.11 × 10⁻³¹ kg × 2.40 × 10⁻¹⁷ J) = 6.61 × 10⁻²⁴ kg·m/s.
3. Wavelength: λ = h/p = (6.63 × 10⁻³⁴ J·s) ÷ (6.613 × 10⁻²⁴ kg·m/s) = 1.00 × 10⁻¹⁰ m.
4. Check on speed: v = p/m = 7.26 × 10⁶ m/s, about 2.4% of c, so the non-relativistic formula K = ½mv² is fine.
5. (b) Ball: p = mv = (0.15 kg)(30 m/s) = 4.5 kg·m/s, so λ = (6.63 × 10⁻³⁴ J·s) ÷ (4.5 kg·m/s) = 1.47 × 10⁻³⁴ m.
6. Ratio: λ_electron/λ_ball = 1.00 × 10⁻¹⁰ ÷ 1.47 × 10⁻³⁴ = 6.8 × 10²³.
7. (c) λ ∝ 1/√ΔV. The potential difference is multiplied by 4, so λ is multiplied by 1/√4 = 1/2. The new wavelength is 5.01 × 10⁻¹¹ m.

Answer. (a) 1.00 × 10⁻¹⁰ m. (b) About 1.5 × 10⁻³⁴ m, roughly 10²⁴ times smaller. (c) The wavelength halves.

Interpretation. 1.0 × 10⁻¹⁰ m is about the spacing between atoms in a solid. So a beam of 150 eV electrons is diffracted by a crystal, just as X-rays are. This is the kind of effect Davisson and Germer saw (their electrons had a lower energy, but the idea is the same). The ball’s wavelength is so small that there is no gap or obstacle on which it could ever show diffraction.

Common misconceptions

• “Brighter light has more energetic photons.” For one colour, every photon has the same energy E = hf. A brighter beam just has more photons per second.
• “Photons have mass because they carry energy.” A photon has zero mass and zero charge. It still carries energy (and, as you will see in Topic 15.6, momentum).
• “Light slows down in glass, so its photons lose energy.” The frequency stays the same in the medium, so the photon energy stays the same. The wavelength gets shorter instead. When the light leaves the glass, it travels at c again.
• Using λ in a medium in E = hc/λ. That formula needs the vacuum wavelength. It is safer to work with frequency.
• “An electron is really a particle, and the wave is just how it moves.” The electron is not travelling along a wavy path. The wave describes where it is likely to be found. Each detection is a single point; the pattern appears only from many detections.
• “Bigger objects have longer de Broglie wavelengths.” It is the other way round: λ = h/(mv), so a larger momentum means a shorter wavelength.
• “Quantum theory replaces classical physics everywhere.” For everyday objects, quantum theory and classical physics give the same answers. You only need quantum theory when an object’s de Broglie wavelength is similar to the size of the gap, obstacle or region it is in.
• Forgetting to convert eV to J. In λ = h/√(2mK) or E = hf with h in J·s, the energy must be in joules.

Where this leads

Next, the Bohr model of atomic structure uses the electron’s de Broglie wavelength to explain why an atom has only certain allowed orbits and energies. Photon energy E = hf returns in emission and absorption spectra, blackbody radiation and the photoelectric effect. Test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate.