What is the energy of a photon and how is it calculated?

The energy of a photon is calculated using the formula E = hf, where E is the energy, h is Planck's constant (6.626 x 10^-34 Js), and f is the frequency of the photon. This relationship highlights the direct proportionality between the energy of a photon and its frequency, a fundamental concept in Quantum Physics covered in the Cambridge International A Levels curriculum.

How is the momentum of a photon determined?

The momentum of a photon is given by the formula p = E/c, where p is the momentum, E is the energy of the photon, and c is the speed of light in a vacuum (approximately 3 x 10^8 m/s). Alternatively, it can also be expressed as p = h/λ, where λ is the wavelength of the photon. This concept is crucial in understanding the wave-particle duality in Quantum Physics.

Why do photons have momentum if they are massless?

Photons are massless particles, yet they possess momentum due to their energy. According to the theory of relativity, momentum can be defined in terms of energy and speed of light, allowing massless particles like photons to have momentum. This is an important concept in Quantum Physics and is often explored in the Cambridge International A Levels Physics syllabus.

How does the wavelength of a photon relate to its energy and momentum?

The wavelength of a photon is inversely proportional to both its energy and momentum. As the wavelength decreases, the energy and momentum increase. This relationship is expressed through the equations E = hc/λ and p = h/λ, where h is Planck's constant and c is the speed of light. Understanding this relationship is key in Quantum Physics studies, including those at the Cambridge International A Levels.

What role does Planck's constant play in the energy and momentum of a photon?

Planck's constant (h) is a fundamental constant in Quantum Physics that relates the energy of a photon to its frequency (E = hf) and its momentum to its wavelength (p = h/λ). It serves as a bridge between the wave and particle nature of light, making it a pivotal element in the study of photon energy and momentum, as covered in the Cambridge International A Levels Physics curriculum.

Why is "Saying photon has rest mass" a common mistake in Energy and momentum of a photon?

Why it happens: Confusion with massive particles. How to avoid it: Photon REST mass = 0. Has energy and momentum but no rest mass. Source: 9702 Examiner Reports 2022-2024

Quantum PhysicsCambridge International A Levels Physics (9702)

Energy and momentum of a photon

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Photon Energy/Momentum Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

$E = hf$ , $p = h/\lambda$ .

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

22.1.1 — Compute photon energy and momentum.

Photon properties

Quantum of light.

Photon. Quantum (smallest packet) of EM radiation.

Energy. $E = hf = hc/\lambda$ .

Momentum. $p = h/\lambda = E/c$ .

Rest mass = 0. But carries energy and momentum.

Units. Energy in J (sometimes eV; 1 eV = $1.6 \times 10^{-19}$ J).

Cambridge tip. Use $hc/\lambda$ when wavelength given; $hf$ when frequency.

$E = hf$ .
$p = h/\lambda$ .
Zero rest mass.

See the full worked example for energy and momentum of a photon →

How it’s examined

Paper 4 typically 5 marks.

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/42 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Energy and momentum of a photon

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1Photon energy (5 marks)
Extended• photon
Question
Find energy of a photon with $\lambda = 500$ nm. $h = 6.63 \times 10^{-34}$ J·s. (5 marks)
Step-by-step solution
1. Step 1
  $E = hc/\lambda$ .
  $E = \dfrac{(6.63 \times 10^{-34})(3 \times 10^8)}{500 \times 10^{-9}} \approx 3.98 \times 10^{-19} \text{ J}$
Answer
$E \approx 3.98 \times 10^{-19}$ J.
2Photon momentum (5 marks)
Extended• photon
Question
Find momentum of photon at 500 nm. (5 marks)
Step-by-step solution
1. Step 1
  $p = h/\lambda$ .
  $p = \dfrac{6.63 \times 10^{-34}}{500 \times 10^{-9}} \approx 1.33 \times 10^{-27} \text{ kg m/s}$
Answer
$p \approx 1.33 \times 10^{-27}$ kg·m/s.

Key Formulae — Energy and momentum of a photon

The formulae you need to memorise for energy and momentum of a photon on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Photon energy
$E = hf = \dfrac{hc}{\lambda}$
When to use
Energy quantum of light. $h = 6.63 \times 10^{-34}$ J·s.
Photon momentum
$p = \dfrac{h}{\lambda} = \dfrac{E}{c}$
When to use
Photons have momentum despite zero rest mass.

Key Definitions and Keywords — Energy and momentum of a photon

Definitions to memorise and the exact keywords mark schemes credit for energy and momentum of a photon answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Photon
Examiner keyword
Quantum of electromagnetic radiation. Has energy $hf$ and momentum $h/\lambda$ .

Common Mistakes and Misconceptions — Energy and momentum of a photon

The traps other students keep falling into on energy and momentum of a photon questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Saying photon has rest mass
9702 Examiner Reports 2022-2024
Why it happens
Confusion with massive particles.
How to avoid it
Photon REST mass = 0. Has energy and momentum but no rest mass.

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