Photons and the Planck Constant Revision Notes | A Level AQA Physics
A photon is a discrete packet of electromagnetic radiation with energy E = hf, where h is the Planck constant. Free AQA notes and 13 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Photons and the Planck Constant is part of Particles in A Level AQA Physics.
6 min
Reading time
15
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A Level AQA
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What this lesson covers
A photon is a discrete packet of electromagnetic radiation with energy E = hf, where h is the Planck constant. Free AQA notes and 13 questions answered.
Describe the photon model of electromagnetic radiation and explain why it replaced a purely wave-based description
State the relationship between photon energy, frequency, and wavelength using $E = hf = \frac{hc}{\lambda}$
Calculate photon energy given frequency or wavelength, converting between eV and joules where required
Explain the physical significance of the Planck constant and identify its correct SI unit
Key ideas to keep in view
Use this lesson to stay inside Particles while connecting the detail back to the wider Particles and radiation topic.
Photon: A discrete quantum of electromagnetic radiation that carries a fixed packet of energy proportional to its frequency. Used when describing the particle-like behaviour of light.
Planck constant (h): The fundamental physical constant of proportionality linking photon energy to frequency. Value: $h = 6.63 \times 10^{-34}$ J s.
Photon energy: The energy carried by a single photon, calculated using $E = hf$ or $E = \frac{hc}{\lambda}$. Measured in joules (J) or electronvolts (eV).
Electronvolt (eV): A unit of energy equal to $1.6 \times 10^{-19}$ J, convenient for expressing the small energies of individual photons and atomic transitions.
Revision notes
Read the core explanations from Photons and the Planck Constant before testing yourself from memory.
1. Why Light Needs a New Model
Picture a dimmer switch controlling a light bulb. Classical wave theory says you can dial the energy of light up or down continuously — like turning a tap. But experiments at the atomic scale told a completely different story: energy arrives in fixed, indivisible packets. That discovery changed physics permanently.
By the early twentieth century, phenomena such as the photoelectric effect could not be explained by treating light as a continuous wave. Max Planck proposed that electromagnetic radiation is emitted and absorbed in discrete chunks of energy. Einstein extended this idea, naming each chunk a photon.
The photon model does not replace the wave model — both are needed. Light shows wave behaviour in diffraction and interference, and particle behaviour when it interacts with matter. At A Level, you need to be comfortable switching between these two descriptions depending on the context of the question.
2. The Planck Constant and the Photon Energy Equation
The energy of a photon is not random — it is set precisely by the frequency of the radiation. Higher frequency means a more energetic photon. The constant of proportionality that links them is the Planck constant, given the symbol $h$.
Because the speed of light relates frequency and wavelength by $c = f\lambda$, you can substitute $f = \frac{c}{\lambda}$ directly into $E = hf$ to get the wavelength form of the equation. This is the version to reach for when a question gives you wavelength instead of frequency.
Notice the inverse relationship in $E = \frac{hc}{\lambda}$: as wavelength increases, energy decreases. This is why gamma rays (very short wavelength) are far more energetic than radio waves (very long wavelength). The Planck constant $h = 6.63 \times 10^{-34}$ J s is extraordinarily small, which is why individual photon energies are tiny on a human scale — but at the atomic level, they are exactly the right size to drive electronic transitions.
3. Working with Photon Energies: Units and Conversions
Photon energies calculated using $E = hf$ come out in joules, which are extremely small numbers — often of the order $10^{-19}$ J. To make these numbers more manageable, physicists use the electronvolt (eV) as an alternative unit.
The conversion is worth memorising as a reflex: joules to eV means dividing by $1.6 \times 10^{-19}$, and eV to joules means multiplying by $1.6 \times 10^{-19}$. Examiners regularly give photon energies in eV and ask for a frequency or wavelength — so you need to convert to joules before substituting into $E = hf$.
4. Worked Calculations
Theory is only useful when you can apply it under exam conditions. The two worked examples below cover the most common question types: one using frequency, one using wavelength. Work through each step before revealing the next — that active engagement is what builds exam fluency.
Worked example: Calculating Photon Energy from Frequency
A source emits electromagnetic radiation of frequency $5.0 \times 10^{14}$ Hz. Calculate the energy of one photon. Give your answer in joules and in electronvolts. [3 marks]
$E = hf$ — Frequency is given directly, so the $E = hf$ form is the correct choice. [M1] for selecting and writing the formula.
$E = 6.63 \times 10^{-34} \times 5.0 \times 10^{14}$ — Both $h$ and $f$ are now substituted with their correct SI values. [M1] for correct substitution.
$E \approx 3.3 \times 10^{-19}$ J — The data was given to 2 significant figures, so the answer is rounded to 2 significant figures. [A1] for correct answer with units.
$E = \frac{3.3 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{3.3}{1.6} \approx 2.1$ eV — Dividing joules by $1.6 \times 10^{-19}$ converts to eV. This photon sits in the visible light range, which is consistent with a frequency of $5.0 \times 10^{14}$ Hz.
Answer: $E \approx 3.3 \times 10^{-19}$ J or approximately $2.1$ eV
Worked example: Calculating Photon Energy from Wavelength
A laser emits light of wavelength $4.50 \times 10^{-7}$ m. Calculate the energy of one photon emitted by the laser. [3 marks]
$E = \frac{hc}{\lambda}$ — Wavelength is given, not frequency, so this form avoids a separate step to find $f$. [M1] for correct equation.
$\lambda = 4.50 \times 10^{-7}$ m = 450 nm, which is violet light. Visible photons typically have energies of $1.8$–$3.1$ eV. Converting: $\frac{4.42 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.76$ eV. — The answer sits comfortably within the expected range for visible light, confirming the calculation is physically reasonable.
Returning to the big picture: the photon model tells us that electromagnetic radiation is quantised. A beam of light is not a smooth, continuous flow of energy — it is a stream of individual photons, each carrying a precise energy $E = hf$. The intensity of a beam depends on how many photons arrive per second, not on the energy of each individual photon.
This distinction matters enormously for exam questions. A dim source of ultraviolet light and a bright source of infrared light can deliver the same total power, but the UV photons are individually far more energetic. When a UV photon strikes a metal surface, it can eject an electron in a way that no number of infrared photons can — because no single infrared photon carries enough energy to overcome the work function. That is the photoelectric effect, and it is direct evidence for the photon model.
The chart above reinforces the inverse relationship between wavelength and photon energy. Radio photons are so low in energy that their quantum nature is almost never relevant in everyday contexts. Gamma photons, by contrast, carry enough energy to ionise atoms and damage biological tissue — a direct consequence of their extremely high frequency.
6. Required Practical Context: Measuring the Planck Constant
The Planck constant is not just a number to memorise — it can be measured experimentally using LEDs of known wavelength. When an LED just begins to emit light, the voltage across it (the threshold voltage $V_0$) provides the minimum energy needed to produce a photon. At threshold, the electrical energy supplied equals the photon energy: $eV_0 = hf$, where $e$ is the elementary charge.
7. Exam Technique and Common Mistakes
The photon energy equations are among the most reliably tested calculations in the Particles and Radiation section. Marks are lost in predictable ways, and knowing those patterns in advance is a genuine advantage.
For extended-response questions asking you to explain the photon model, the mark scheme rewards a clear chain of reasoning: state that electromagnetic radiation consists of discrete quanta called photons, state that each photon carries energy $E = hf$, and link this to the physical consequence being discussed (such as why only high-frequency radiation can eject electrons in the photoelectric effect).
Every spec point in this lesson — the photon model, the Planck constant, and the equation $E = hf = \frac{hc}{\lambda}$ — has now appeared in both teaching content and retrieval or exam-application contexts. The core message to carry into the exam: photon energy is fixed by frequency alone, the Planck constant is the bridge between the quantum and classical worlds, and unit consistency is the single most reliable way to protect your marks.
The photon-energy relationships required by AQA are E = hf and, because f = c/λ in a vacuum, E = hc/λ. These equations show that higher-frequency electromagnetic radiation carries more energy per photon, while longer wavelength radiation carries less.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
A photon is a discrete quantum, or packet, of electromagnetic radiation. Each photon carries a fixed amount of energy set entirely by its frequency, has zero rest mass and travels at the speed of light in a vacuum.
Do photons have mass?
Photons have zero rest mass and travel at the speed of light in a vacuum. Each photon is a discrete packet of electromagnetic radiation carrying a fixed amount of energy determined entirely by its frequency.
How do you calculate photon energy from wavelength?
Use E = hc/λ. For a laser of wavelength 4.50 × 10⁻⁷ m, E = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) ÷ 4.50 × 10⁻⁷ = 4.42 × 10⁻¹⁹ J, which is about 2.76 eV, within the range for visible light.
How do you calculate photon energy from frequency?
Use E = hf with h = 6.63 × 10⁻³⁴ J s. For radiation of frequency 5.0 × 10¹⁴ Hz, E = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ ≈ 3.3 × 10⁻¹⁹ J. Dividing by 1.6 × 10⁻¹⁹ converts this to about 2.1 eV.
Why does higher frequency light have more energy per photon?
Photon energy is fixed by frequency alone through E = hf, so higher frequency means a more energetic photon. Since E = hc/λ, energy falls as wavelength increases, which is why gamma rays are far more energetic than radio waves.
How can you measure the Planck constant using LEDs?
Use LEDs of known wavelength and find the threshold voltage V₀ at which each LED just begins to emit light. At threshold, the electrical energy supplied equals the photon energy, so eV₀ = hf, where e is the elementary charge.
What is Planck's constant in A Level Physics?
The Planck constant, h, is 6.63 × 10⁻³⁴ J s. It is the constant of proportionality linking the energy of a photon to its frequency in E = hf, and its tiny value is why individual photon energies are so small on a human scale.
What is the unit of the Planck constant?
The Planck constant has the SI unit J s. From E = hf, h = E/f, and dividing joules by Hz, which is s⁻¹, gives joule seconds.
How do you give a photon energy in electronvolts?
A photon energy in joules is converted to eV by dividing by 1.6 × 10⁻¹⁹. Radiation of frequency 5.0 × 10¹⁴ Hz has photons of about 3.3 × 10⁻¹⁹ J, which is about 2.1 eV.
Does the intensity of light change the energy of each photon?
Intensity does not change the energy of each photon. The intensity of a beam depends on how many photons arrive per second, while the energy of each photon is fixed by its frequency through E = hf.
Why can UV light eject electrons from a metal but infrared cannot?
A single UV photon carries enough energy to overcome the work function of a metal and eject an electron. No single infrared photon carries enough energy, so no number of infrared photons can eject an electron in the same way.
What is the energy of a photon of visible light?
Visible photons typically have energies of 1.8–3.1 eV, of the order 10⁻¹⁹ J. Violet light of wavelength 450 nm, for example, has photons of about 4.42 × 10⁻¹⁹ J, or about 2.76 eV.
Who came up with the idea of the photon?
Max Planck proposed that electromagnetic radiation is emitted and absorbed in discrete chunks of energy, and Einstein extended this idea, naming each chunk a photon. This was needed because the photoelectric effect could not be explained by a continuous wave.