understand that a photon has momentum and that the momentum is given by p = E / c

Published by Patrick Mutisya · 14 days ago

Energy and Momentum of a Photon – A-Level Physics 9702

Energy and Momentum of a Photon

Learning Objective

Students will understand that a photon, despite having no rest mass, carries momentum and that its momentum is given by the relation \$p = \frac{E}{c}\$ where \$E\$ is the photon’s energy and \$c\$ is the speed of light in vacuum.

Key Concepts

  • Photon as a quantum of electromagnetic radiation.

  • Energy of a photon: \$E = h\nu = \frac{hc}{\lambda}\$.

  • Momentum of a photon: \$p = \frac{E}{c} = \frac{h}{\lambda}\$.

  • Radiation pressure and its applications (e.g., solar sails, photon pressure on surfaces).

Derivation of the Momentum Formula

Starting from the relativistic energy–momentum relation for a particle with rest mass \$m_0\$:

\$E^2 = (pc)^2 + (m_0 c^2)^2\$

For a photon \$m_0 = 0\$, so the equation reduces to:

\$E = pc \quad\Longrightarrow\quad p = \frac{E}{c}\$

Substituting \$E = h\nu\$ gives the alternative form:

\$p = \frac{h\nu}{c} = \frac{h}{\lambda}\$

Numerical Example

  1. Find the momentum of a photon of wavelength \$500\ \text{nm}\$ (green light).

Given: \$h = 6.626\times10^{-34}\ \text{J·s}\$, \$c = 3.00\times10^8\ \text{m·s}^{-1}\$, \$\lambda = 500\ \text{nm}=5.00\times10^{-7}\ \text{m}\$.

\$p = \frac{h}{\lambda} = \frac{6.626\times10^{-34}}{5.00\times10^{-7}} = 1.33\times10^{-27}\ \text{kg·m·s}^{-1}\$

The momentum is extremely small, illustrating why macroscopic objects are not noticeably affected by individual photons.

Photon Momentum in Different Contexts

Radiation TypeWavelength / FrequencyEnergy \$E\$ (J)Momentum \$p\$ (kg·m·s\$^{-1}\$)
Visible light (green)\$\lambda = 5.00\times10^{-7}\ \text{m}\$\$E = \dfrac{hc}{\lambda}=3.98\times10^{-19}\$\$p = \dfrac{h}{\lambda}=1.33\times10^{-27}\$
Ultraviolet (UV)\$\lambda = 2.00\times10^{-8}\ \text{m}\$\$E = 9.94\times10^{-18}\$\$p = 3.31\times10^{-26}\$
X‑ray\$\lambda = 1.00\times10^{-10}\ \text{m}\$\$E = 1.99\times10^{-15}\$\$p = 6.63\times10^{-24}\$
Gamma ray\$\lambda = 1.00\times10^{-12}\ \text{m}\$\$E = 1.99\times10^{-13}\$\$p = 6.63\times10^{-22}\$

Applications in A‑Level Exams

  • Calculating radiation pressure: \$P = \dfrac{I}{c}\$ for perfectly absorbing surface, \$P = \dfrac{2I}{c}\$ for perfectly reflecting surface, where \$I\$ is the intensity.

  • Photon momentum in the photoelectric effect – linking energy and momentum to explain electron ejection.

  • Solar sail thrust: \$F = \dfrac{P A}{c}\$ where \$A\$ is the sail area.

Common Misconceptions

  • “Photons have no momentum because they have no mass.” – Momentum is a property of energy and motion, not solely of mass.

  • Confusing \$p = \dfrac{E}{c}\$ with \$p = mv\$ – the latter applies only to massive particles with a defined rest mass.

  • Assuming radiation pressure is negligible – in space, cumulative photon pressure can produce measurable forces over long periods.

Suggested Classroom Activities

  1. Demonstration: Use a laser pointer to exert a measurable force on a lightweight mirror suspended by a fine thread.

  2. Problem‑solving: Calculate the thrust on a solar sail of given area at 1 AU from the Sun.

  3. Discussion: Compare photon momentum with that of macroscopic particles (e.g., a baseball) to highlight scale differences.

Suggested diagram: A photon striking a surface and transferring momentum, illustrating both absorption and reflection cases.

Summary

Even though photons are massless, they carry energy and therefore momentum, given by \$p = E/c = h/\lambda\$. This principle underlies many phenomena from radiation pressure to the operation of solar sails, and it is a core topic in the Cambridge A‑Level Physics syllabus.