22.1 Energy and momentum of a photon
- Syllabus
- 9702–2028–2029
- Topic
- 22.1
- Level
- A2
Learning objectives
22.1.1Electromagnetic radiation has a particulate nature• understand that electromagnetic radiation has a particulate nature22.1.2A photon is a quantum of electromagnetic energy• understand that a photon is a quantum of electromagnetic energy22.1.3E = hf• recall and use E = hf22.1.4The electronvolt (eV) as a unit of energy• use the electronvolt (eV) as a unit of energy22.1.5A photon has momentum and that the momentum is given by p = E / c• understand that a photon has momentum and that the momentum is given by p = E / c
Electromagnetic radiation behaves as both a wave and a stream of particles
Electromagnetic radiation shows wave behaviour such as interference and diffraction and particle behaviour such as quantised photon interactions.
Use the model that explains the observation; neither classical wave nor classical particle language alone covers every experiment.
A diffraction pattern supports wave behaviour, while the photoelectric effect requires discrete energy transfers.
Wave–particle duality is not a claim that light alternates between two physical substances during travel.
A photon is a quantum packet of electromagnetic energy
A photon is a discrete packet of electromagnetic energy with energy proportional to frequency and zero rest mass.
Photons travel at c in vacuum and are absorbed or emitted as whole quanta in interactions.
A higher-frequency ultraviolet photon carries more energy than a visible red photon.
A photon is not a tiny classical wave crest, and increasing intensity increases photon number at fixed frequency rather than each photon’s energy.
Calculate photon energy, wavelength and photon rate with E=hf
one−photonenergyE=hf=hc/λ,withh=6.63×10−34Jsandc=3.00×108ms−1
For λ=630 nm, E=hc/λ=(6.63×10⁻³⁴)(3.00×10⁸)/(630×10⁻⁹)=3.16×10⁻¹⁹ J=1.97 eV.
For photon energy 74 keV, first convert E=74×10³×1.60×10⁻¹⁹ J, then λ=hc/E=1.68×10⁻¹¹ m.
beampowerP=(numberofphotonspersecond)×Ephotonphotonrate=P/Ephoton
A 1.0×10⁻² W beam of 1.97 eV photons emits (1.0×10⁻²)/(1.97×1.60×10⁻¹⁹)=3.2×10¹⁶ photons s⁻¹.
E=hf is energy per photon. Higher f means larger E; larger λ means smaller E. Convert nm to m and eV to J before using SI constants.
An electronvolt is the energy gained by one elementary charge through one volt
One electronvolt is 1 eV=1.60×10⁻¹⁹ J, the energy transfer when a charge of magnitude e moves through 1 V.
Use eV for particle-scale energies and convert to joules when applying SI equations or comparing macroscopic work.
A 5.0 eV electron has energy about 8.0×10⁻¹⁹ J.
An electronvolt is an energy unit, not a voltage or an electron’s mass.
Photon momentum p=E/c transfers force when light is absorbed or reflected
photonmomentummagnitudep=E/c=hf/c=h/λ
A photon has zero rest mass but non-zero momentum directed along its propagation. Shorter wavelength or higher energy means greater momentum.
For E=3.11×10⁻¹⁹ J, p=E/c=(3.11×10⁻¹⁹)/(3.00×10⁸)=1.04×10⁻²⁷ kg m s⁻¹ (or N s).
| Normal interaction at a stationary surface | Photon momentum change magnitude | Momentum delivered to surface |
|---|---|---|
| photon absorbed | p | p |
| photon reflected straight back | 2p | 2p |
averageforce=momentumtransferredperunittimeForcompleteabsorptionofbeampowerP:F=P/c;forperfectreflection:F=2P/c.
Do not use p=mv for a photon. Reflection transfers twice the normal momentum of absorption because the photon momentum reverses, rather than merely falling to zero.