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For most of the nineteenth century, physicists were convinced that light was a wave. Young's double-slit experiment (1801) showed that light interferes. Fresnel, Fraunhofer and Kirchhoff built elegant mathematical theories of diffraction. Maxwell (1865) showed that light is an oscillation of the electromagnetic field travelling at the speed c = 1/√(μ₀ε₀). By 1900, the wave theory seemed complete. Optics was a solved problem.
Then, within a few short years at the beginning of the twentieth century, experiments began to show that this neat picture was incomplete. Blackbody radiation, the photoelectric effect, atomic line spectra — none of them could be explained by a pure wave theory of light, no matter how sophisticated. In 1900 Max Planck proposed, reluctantly, that electromagnetic energy is emitted and absorbed in discrete packets. In 1905 Albert Einstein went further: he proposed that light itself consists of discrete, localised packets of energy — photons — which travel through space and interact with matter one at a time.
Module 4.5 of the OCR A-Level Physics A specification (H556) — Quantum Physics — asks you to understand this revolutionary idea and its experimental consequences. This first lesson introduces the photon model: what a photon is, how much energy it carries, and the basic constants and units you will need throughout the course.
The word quantum comes from the Latin quantus, meaning "how much". In physics it refers to the smallest discrete amount of something that can exist or be exchanged. A quantum of charge is the charge on an electron, e = 1.60 × 10⁻¹⁹ C. A quantum of angular momentum in atomic physics is ℏ = h/(2π). A quantum of electromagnetic energy at frequency f is a photon, with energy E = hf.
The essential claim of quantum physics is this: some quantities that appear continuous in classical physics are, at a deep level, discrete — they come in integer multiples of some fundamental unit, the quantum. You cannot have half a photon any more than you can have half an electron.
A photon is a quantum of electromagnetic energy — a discrete, indivisible packet of light. It has the following properties:
c = 3.00 × 10⁸ m s⁻¹ in vacuum.E = hf, where f is the frequency of the associated electromagnetic wave.p = h/λ (which you will meet again in Lesson 6 when we introduce de Broglie's relation in reverse).The photon is not a tiny particle in the sense of a classical billiard ball. It does not have a well-defined position as it travels through a double slit or a diffraction grating; indeed it will interfere with itself. But when it interacts with a detector — a photocell, a photographic plate, a retinal rod — it deposits its energy at a single point, in a single event.
This dual behaviour — spreading out like a wave in propagation, arriving like a particle in detection — is the first manifestation of wave-particle duality, which we shall develop properly in Lessons 6, 7 and 8.
The central equation of the photon model is the Planck–Einstein relation:
E = hf
where:
E is the energy of the photon, in joules (J)f is the frequency of the electromagnetic wave, in hertz (Hz)h is Planck's constant, a fundamental constant of natureThe value of Planck's constant, given in the OCR data sheet, is:
h = 6.63 × 10⁻³⁴ J s
Note the units carefully: joule seconds. Energy multiplied by time. This combination — action, in the formal language of analytical mechanics — is the natural currency of quantum physics.
Because the speed of any electromagnetic wave in vacuum is c = fλ, we can rewrite the Planck relation in terms of wavelength:
E = hf = hc/λ
This second form is often more useful, because laboratory instruments typically measure wavelength directly.
Exam Tip: Make sure you know both forms of the equation —
E = hfandE = hc/λ— and know when to use each. OCR questions frequently give wavelength and expect you to usehc/λ; others give frequency and expecthf. Converting between frequency and wavelength viac = fλis a sign you are taking the long way round.
Planck's constant h = 6.63 × 10⁻³⁴ J s is extraordinarily small on human scales, and this is why quantum effects are not obvious in everyday life. Consider: a single photon of visible green light (wavelength 550 nm) has energy
E = hc/λ
= (6.63 × 10⁻³⁴)(3.00 × 10⁸)/(550 × 10⁻⁹)
= 3.6 × 10⁻¹⁹ J
A 100-watt light bulb therefore emits roughly 10²/(3.6 × 10⁻¹⁹) ≈ 3 × 10²⁰ photons every second. Each photon's energy is so small that the human eye cannot distinguish the arrival of individual quanta (except in the most sensitive dark-adapted regime, where the retina actually can respond to single photons). The discreteness of light is completely hidden by the sheer number of photons involved.
But when we zoom in on atomic and subatomic processes — where single photons interact with single electrons — the quantum nature of light becomes not merely visible but essential.
Working with joules at the atomic scale is cumbersome, because atomic energies are typically of order 10⁻¹⁸ to 10⁻¹⁹ J. For this reason, physicists use a more convenient unit: the electronvolt (eV).
Definition: One electronvolt is the kinetic energy gained by an electron of charge e = 1.60 × 10⁻¹⁹ C when accelerated through a potential difference of one volt.
W = QV
= (1.60 × 10⁻¹⁹ C)(1 V)
= 1.60 × 10⁻¹⁹ J
So the conversion factor is:
1 eV = 1.60 × 10⁻¹⁹ J
Equivalently,
1 J = 1/(1.60 × 10⁻¹⁹) eV ≈ 6.24 × 10¹⁸ eV
The electronvolt is not an SI unit, but it is universally used in atomic, molecular and particle physics because the relevant energies come out as manageable numbers:
| Physical quantity | Typical energy |
|---|---|
| Visible photon | 1.6 – 3.3 eV |
| Ionisation energy of hydrogen | 13.6 eV |
| X-ray photon | 100 eV – 100 keV |
| Nuclear binding energy per nucleon | ~8 MeV |
| Particle-physics collisions (LHC) | up to 13 TeV |
(The prefixes follow SI convention: keV = 10³ eV, MeV = 10⁶ eV, GeV = 10⁹ eV, TeV = 10¹² eV.)
Exam Tip: When an OCR question asks for an answer in eV, make sure you convert from joules using
1 eV = 1.60 × 10⁻¹⁹ J. When it asks for joules, make sure you multiply. A common careless mistake is to write "3.2 × 10⁻¹⁹ eV" when the answer should be "2.0 eV".
A photon of red light has wavelength λ = 650 nm. Calculate its energy in (a) joules and (b) electronvolts.
Solution.
(a) Using E = hc/λ:
E = (6.63 × 10⁻³⁴ J s)(3.00 × 10⁸ m s⁻¹)/(650 × 10⁻⁹ m)
= (1.989 × 10⁻²⁵ J m)/(650 × 10⁻⁹ m)
= 3.06 × 10⁻¹⁹ J
(b) Converting to electronvolts:
E = (3.06 × 10⁻¹⁹ J)/(1.60 × 10⁻¹⁹ J/eV)
= 1.91 eV
A red photon therefore carries about 1.9 eV of energy — a value worth remembering, because it is close to the work functions of many metals and close to the band gap of silicon.
A photon has energy 2.50 eV. What is its frequency?
Solution. First convert to joules:
E = (2.50 eV)(1.60 × 10⁻¹⁹ J/eV) = 4.00 × 10⁻¹⁹ J
Then use f = E/h:
f = (4.00 × 10⁻¹⁹ J)/(6.63 × 10⁻³⁴ J s)
= 6.03 × 10¹⁴ Hz
This is in the visible range — around 500 nm (green-blue), which you can verify using λ = c/f.
Because photon energy scales linearly with frequency, the different regions of the electromagnetic spectrum correspond directly to different energy ranges for their photons:
flowchart LR
Radio["Radio<br/>~10⁻⁹ eV<br/>λ ~ m to km"]
Micro["Microwave<br/>~10⁻⁵ eV<br/>λ ~ mm to cm"]
IR["Infrared<br/>~10⁻² eV<br/>λ ~ μm"]
Vis["Visible<br/>~2 eV<br/>λ ~ 400–700 nm"]
UV["Ultraviolet<br/>~10 eV<br/>λ ~ 10–400 nm"]
Xray["X-ray<br/>~10³–10⁵ eV<br/>λ ~ 0.01–10 nm"]
Gamma["Gamma<br/>>10⁵ eV<br/>λ < 0.01 nm"]
Radio --> Micro --> IR --> Vis --> UV --> Xray --> Gamma
A radio photon is so weak that it is essentially undetectable individually — radio receivers sum the effect of enormous numbers of photons acting coherently as a classical electromagnetic wave. A gamma photon, by contrast, can knock individual nuclei apart, and its interaction is always particle-like.
At this point it is reasonable to ask: why must light be quantised? Why not a classical wave? Einstein's 1905 paper argued the case from the photoelectric effect (the subject of Lesson 3). But another crucial motivation — and in fact the original motivation — was the problem of blackbody radiation.
A blackbody (an idealised object that absorbs all radiation falling on it) at temperature T emits a characteristic spectrum of electromagnetic radiation. Classical physics predicted that this spectrum should diverge at high frequencies — the so-called "ultraviolet catastrophe". Observed spectra, however, peak at a finite frequency and fall off at both ends.
Max Planck solved this puzzle in 1900 by postulating that an oscillator of frequency f can only exchange energy with the radiation field in discrete units of hf. The high-frequency modes are then statistically suppressed (because a single quantum hf is larger than the available thermal energy k_BT), and the catastrophe is averted. Planck regarded this as a mathematical trick. It was Einstein, five years later, who argued that the quantisation is a property of the radiation itself, not merely of the way matter exchanges energy with it.
Although blackbody radiation is not on the OCR specification, it is worth knowing as the historical genesis of the photon concept.
flowchart TB
Source["Source emits<br/>photons at frequency f"]
P1["Each photon<br/>E = hf"]
P2["Photons travel at<br/>c = 3 × 10⁸ m s⁻¹"]
D["Detector<br/>absorbs photons<br/>one at a time"]
Source --> P1
P1 --> P2
P2 --> D
D --> K["Each absorption<br/>is a discrete event"]
hf and hc/λ. These are equivalent, but only when paired with the correct second factor. hf uses frequency in hertz; hc/λ uses wavelength in metres.λ = 650 without converting to 650 × 10⁻⁹ m, giving an answer out by nine orders of magnitude.e the wrong way. To convert joules to eV, divide by 1.60 × 10⁻¹⁹. To convert eV to joules, multiply. Always sanity-check your answer: an atomic-scale energy should be a few eV, not 10⁻³⁸ eV.E = mc²" confuses rest mass with relativistic energy.E = hf = hc/λ.h = 6.63 × 10⁻³⁴ J s is the fundamental quantum of action.1 eV = 1.60 × 10⁻¹⁹ J.In the next lesson we shall develop the arithmetic of photon calculations in more detail, through a variety of worked examples.