Quantum physics

All four subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.

Syllabus points

22.1 Energy and momentum of a photon

• Understand that electromagnetic radiation has a particulate nature.
• Understand that a photon is a quantum of electromagnetic energy.
• Recall and use E = hf.
• Use the electronvolt (eV) as a unit of energy.
• Understand that a photon has momentum and that the momentum is given by p = E/c.

22.2 Photoelectric effect

• Understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation.
• Understand and use the terms threshold frequency and threshold wavelength.
• Explain photoelectric emission in terms of photon energy and work function energy.
• Recall and use hf = Φ + ½mv²_max.
• Explain why the maximum kinetic energy of photoelectrons is independent of intensity, whereas the photoelectric current is proportional to intensity.

22.3 Wave-particle duality

• Understand that the photoelectric effect provides evidence for a particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence for a wave nature.
• Describe and interpret qualitatively the evidence provided by electron diffraction for the wave nature of particles.
• Understand the de Broglie wavelength as the wavelength associated with a moving particle.
• Recall and use λ = h/p.

22.4 Energy levels in atoms and line spectra

• Understand that there are discrete electron energy levels in isolated atoms (for example atomic hydrogen).
• Understand the appearance and formation of emission and absorption line spectra.
• Recall and use hf = E₁ - E₂.

The photon

Electromagnetic radiation has a particulate nature as well as a wave nature. A photon is a quantum of electromagnetic energy, meaning the smallest indivisible packet in which that energy can be carried.

E = hf

where h is the Planck constant, 6.63 × 10⁻³⁴ J s. Since c = fλ, this is also

E = hc / λ

so short wavelength means high photon energy. An ultraviolet photon carries far more energy than an infrared photon, which is the reason ultraviolet damages skin and infrared merely warms it.

Worked example. Find the energy of a photon of wavelength 500 nm.

E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / (500 × 10⁻⁹) = 3.978 × 10⁻¹⁹

so about 3.98 × 10⁻¹⁹ J.

The electronvolt

Energies of this size are awkward in joules, so they are quoted in electronvolts. One electronvolt is the energy transferred when an electron moves through a potential difference of one volt:

1 eV = 1.60 × 10⁻¹⁹ J

To convert from joules to electronvolts, divide by 1.60 × 10⁻¹⁹; to go the other way, multiply. Continuing the example above:

3.978 × 10⁻¹⁹ / (1.60 × 10⁻¹⁹) = 2.486

so the photon carries about 2.49 eV, which is a much easier number to hold in mind for visible light.

Momentum of a photon

A photon has momentum even though it has no mass:

p = E / c

Combining with E = hf and c = fλ gives p = h/λ, which is the same relationship that reappears below as the de Broglie equation. That is not a coincidence; it is the point.

Worked example. The momentum of the 500 nm photon above.

p = 3.978 × 10⁻¹⁹ / (3.00 × 10⁸) = 1.326 × 10⁻²⁷

so 1.33 × 10⁻²⁷ N s. Tiny, but real: it is why a solar sail works and why radiation pressure matters inside stars.

The photoelectric effect

Shine electromagnetic radiation on a clean metal surface and electrons may be emitted. Those emitted electrons are called photoelectrons.

The observations are the reason the topic exists, because none of them can be explained by a wave model:

The photon explanation

One photon interacts with one electron and transfers all of its energy in a single event. There is no accumulation and no sharing.

To escape the surface, an electron must be supplied with at least the work function energy Φ, which is the minimum energy needed to remove an electron from the surface of the metal. If the photon carries less than Φ, the electron cannot escape and the energy is dissipated as heat, no matter how many such photons arrive.

Any surplus becomes kinetic energy:

hf = Φ + ½mv²_max

The maximum kinetic energy belongs to electrons emitted from the very surface. Electrons from deeper down lose energy on the way out, so they emerge with less, which is why the equation gives a maximum rather than a fixed value.

Threshold frequency and threshold wavelength

At the threshold frequency f₀, the photon energy is exactly enough to release an electron with no kinetic energy left over:

hf₀ = Φ

The corresponding threshold wavelength is λ₀ = hc/Φ. Note that the threshold is a maximum wavelength, not a minimum, because longer wavelength means lower energy. Getting that inequality the wrong way round is a common and expensive error.

Worked example. Sodium has a work function of 2.3 eV. Find its threshold frequency.

Convert to joules first:

Φ = 2.3 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹

f₀ = 3.68 × 10⁻¹⁹ / (6.63 × 10⁻³⁴) = 5.55 × 10¹⁴

so f₀ is about 5.55 × 10¹⁴ Hz. The corresponding wavelength:

λ₀ = 3.00 × 10⁸ / (5.55 × 10¹⁴) = 5.405 × 10⁻⁷

that is 540 nm, in the green. Light of longer wavelength than that, so yellow, orange or red, causes no emission from sodium at all.

Worked example, maximum kinetic energy. Light of wavelength 400 nm falls on that sodium surface.

E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / (400 × 10⁻⁹) = 4.9725 × 10⁻¹⁹

Maximum kinetic energy:

4.9725 × 10⁻¹⁹ - 3.68 × 10⁻¹⁹ = 1.2925 × 10⁻¹⁹

so about 1.29 × 10⁻¹⁹ J, or 0.81 eV.

Intensity versus frequency

This distinction carries several marks whenever it appears, and it follows directly from one photon meeting one electron.

• Intensity is the number of photons arriving per second. More photons means more electrons released per second, so a larger photocurrent, but each photon still carries the same energy, so the maximum kinetic energy is unchanged.
• Frequency sets the energy of each photon. A higher frequency means more energy per photon, so more is left over after the work function is paid, so a greater maximum kinetic energy. It does not change how many photons arrive, so it does not by itself change the current.

A graph of maximum kinetic energy against frequency is therefore a straight line of gradient h, crossing the frequency axis at f₀ and the energy axis at -Φ. Being asked to find the Planck constant from the gradient of such a graph is a standard question.

Wave-particle duality

Neither model alone accounts for everything:

• Wave evidence for light: interference and diffraction, including Young's double slit from topic 8. These require a wave that can superpose with itself.
• Particle evidence for light: the photoelectric effect, which requires energy delivered in discrete quanta.

So light behaves as a wave in some experiments and as a stream of particles in others, and no single classical picture covers both. de Broglie proposed that the reverse also holds, and that particles have a wavelength:

λ = h / p = h / (mv)

The evidence is electron diffraction. Fire a beam of electrons through a thin polycrystalline film of graphite and they produce concentric rings on a fluorescent screen, exactly like the diffraction pattern from a wave passing through many randomly oriented crystals. Particles behaving as particles could not do that.

Two observations from the experiment confirm the equation:

• Increasing the accelerating voltage increases the electrons' speed and therefore their momentum, so λ decreases, and the rings become smaller. That relationship, faster electrons giving tighter rings, is the direct evidence for λ = h/p.
• The spacing of the rings depends on the spacing of the atomic layers in the graphite, which is what a diffraction pattern should depend on.

Diffraction is only noticeable when the wavelength is comparable with the size of the gap. The atomic spacing in a crystal is around 10⁻¹⁰ m, which is why electrons accelerated through a few kilovolts, whose de Broglie wavelength is of that order, diffract at a crystal while a cricket ball does not diffract through a doorway.

Worked example. Find the de Broglie wavelength of an electron accelerated through 1000 V.

Kinetic energy gained:

E = 1000 × 1.60 × 10⁻¹⁹ = 1.6 × 10⁻¹⁶

From E = p²/(2m):

p² = 2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁶ = 2.915 × 10⁻⁴⁶

so p = 1.707 × 10⁻²³ N s, and

λ = 6.63 × 10⁻³⁴ / (1.707 × 10⁻²³) = 3.884 × 10⁻¹¹

that is about 3.9 × 10⁻¹¹ m, comparable with atomic spacings, so diffraction is observable.

Worked example, why we never see this for large objects. A 0.16 kg cricket ball at 30 m s⁻¹.

p = 0.16 × 30 = 4.8

λ = 6.63 × 10⁻³⁴ / 4.8 = 1.381 × 10⁻³⁴

A wavelength of 10⁻³⁴ m is unimaginably smaller than any aperture, so no diffraction is ever observed. The wave nature is not absent; it is unmeasurable.

Energy levels and line spectra

Electrons in an isolated atom can only occupy certain discrete energy levels. They cannot have energies in between.

By convention the levels are negative, with zero taken as the energy of a free electron at infinity, exactly as with gravitational potential in topic 13. The lowest level, the ground state, is the most negative. An electron given enough energy to reach zero has been ionised, and the ionisation energy is the energy needed to take an electron from the ground state to infinity.

When an electron falls from a higher level E₁ to a lower level E₂, the difference is emitted as a single photon:

hf = E₁ - E₂

Because the levels are discrete, only certain differences exist, so only certain frequencies are emitted, which is why the spectrum consists of lines rather than a continuous band. That is the entire argument, and it is what a line spectrum is evidence for.

Emission and absorption spectra

• An emission line spectrum appears as bright coloured lines on a dark background. It is produced by a hot low-pressure gas: electrons are excited to higher levels by collisions or heating, then fall back, emitting photons of the characteristic frequencies.
• An absorption line spectrum appears as dark lines on a continuous coloured background. It is produced when white light passes through a cool gas: photons whose energies exactly match a level difference are absorbed, exciting electrons to higher levels. The absorbed photons are re-emitted almost immediately, but in all directions rather than along the original beam, so the line looks dark by comparison.

The dark lines of an absorption spectrum fall at exactly the same wavelengths as the bright lines of the emission spectrum of the same element. Since every element has its own set of levels, the pattern is a fingerprint, which is how the composition of stars is determined from their light.

Worked example. An electron falls from a level at -3.4 eV to one at -13.6 eV.

Energy difference:

-3.4 - (-13.6) = 10.2

Convert to joules:

10.2 × 1.60 × 10⁻¹⁹ = 1.632 × 10⁻¹⁸

λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / (1.632 × 10⁻¹⁸) = 1.219 × 10⁻⁷

that is 122 nm, in the ultraviolet. This is the first line of the Lyman series of hydrogen, and it is invisible to the eye, which is worth noticing: not every transition in hydrogen produces visible light.

Common mistakes

• Saying a photon has mass. It has energy and momentum but no mass.
• Using E = hf with the wavelength in nanometres rather than metres.
• Multiplying by 1.60 × 10⁻¹⁹ when converting joules to electronvolts. Divide.
• Saying brighter light gives photoelectrons more kinetic energy. It gives more of them.
• Saying that if the light is bright enough, emission will happen below the threshold frequency. It never does.
• Saying several photons can combine to release one electron. One photon, one electron.
• Treating the threshold wavelength as a minimum. It is a maximum.
• Forgetting that the work function must be converted from eV to J before use with h in SI units.
• Using the speed rather than the momentum in λ = h/p.
• Saying electron diffraction shows electrons are waves rather than that they have a wave nature as well as a particle nature.
• Saying energy levels are positive, or forgetting that the ground state is the most negative.
• Explaining an absorption spectrum by saying the photons are simply destroyed. They are re-emitted in all directions.
• Writing hf = E₁ + E₂ instead of the difference.