understand that electromagnetic radiation has a particulate nature

Cambridge A‑Level Physics 9702 – Energy and Momentum of a Photon

Energy and Momentum of a Photon

Learning Objective

Understand that electromagnetic radiation possesses a particulate (photon) nature and be able to calculate the energy and momentum associated with a photon.

Key Concepts

  • A photon is a quantum of electromagnetic radiation.
  • Energy of a photon is directly proportional to its frequency.
  • Even though photons are mass‑less, they carry momentum.
  • The particulate description explains phenomena such as the photoelectric effect, Compton scattering and radiation pressure.

Energy of a Photon

The energy $E$ of a photon is given by Planck’s relation:

$$E = hu = \frac{hc}{\lambda}$$

where

  • $h = 6.626\times10^{-34}\ \text{J·s}$ is Planck’s constant,
  • $u$ is the frequency of the radiation,
  • $c = 3.00\times10^{8}\ \text{m·s}^{-1}$ is the speed of light in vacuum,
  • $\lambda$ is the wavelength.

Momentum of a Photon

From the relativistic energy‑momentum relation $E^{2}=p^{2}c^{2}+m_{0}^{2}c^{4}$ and noting that the rest mass $m_{0}=0$ for a photon, we obtain:

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

Thus the momentum $p$ is inversely proportional to the wavelength.

Derivation from Relativistic Principles (Optional)

Starting with $E = \gamma m_{0}c^{2}$ and $p = \gamma m_{0}v$, for $m_{0}=0$ the Lorentz factor $\gamma$ becomes infinite, but the ratio $E/p$ remains finite and equals $c$. This leads directly to $p = E/c$ for a mass‑less particle.

Experimental Evidence for the Particulate Nature

  1. Photoelectric Effect – Electrons are emitted from a metal only if the incident light has a frequency above a threshold, regardless of intensity. The kinetic energy of the emitted electrons follows $K_{\max}=hu-\phi$, where $\phi$ is the work function.
  2. Compton Scattering – X‑rays scattered from electrons show a wavelength shift $\Delta\lambda = \frac{h}{m_{e}c}(1-\cos\theta)$, consistent with photon momentum transfer.
  3. Radiation Pressure – Light exerts a measurable pressure $P = \frac{I}{c}$ (or $2I/c$ for perfect reflection) on surfaces, explained by momentum transfer from photons.
Suggested diagram: Schematic of the photoelectric effect showing incident photons, metal surface, and emitted electrons.

Sample Calculations

Table 1 illustrates the energy and momentum of photons at selected wavelengths.

Wavelength $\lambda$ (nm) Frequency $u$ (THz) Energy $E$ (eV) Momentum $p$ (kg·m s⁻¹)
400 750 3.10 5.27 × 10⁻²⁸
600 500 2.07 3.51 × 10⁻²⁸
800 375 1.55 2.62 × 10⁻²⁸
1000 300 1.24 2.07 × 10⁻²⁸

Conversion factors used: $1\ \text{eV}=1.602\times10^{-19}\ \text{J}$, $p = E/c$.

Key Formulas to Remember

  • $E = hu = \dfrac{hc}{\lambda}$
  • $p = \dfrac{E}{c} = \dfrac{h}{\lambda}$
  • Photoelectric kinetic energy: $K_{\max}=hu-\phi$
  • Compton wavelength shift: $\Delta\lambda = \dfrac{h}{m_{e}c}(1-\cos\theta)$
  • Radiation pressure (absorption): $P = \dfrac{I}{c}$

Summary

Electromagnetic radiation can be described as a stream of photons, each carrying a quantised amount of energy $E = hu$ and a corresponding momentum $p = h/\lambda$. This particulate description successfully explains phenomena that wave theory alone cannot, confirming the dual nature of light.