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

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: $E = hu = \frac{hc}{\lambda}$.
• 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 $E = hu$ gives the alternative form:

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

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 Type	Wavelength / Frequency	Energy $E$ (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.

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.