understand that electromagnetic radiation has a particulate nature

Electromagnetic Radiation – Photon (Particulate) Nature

Learning Objectives

• Explain why light can be described as a stream of photons.
• Use the quantitative relations for photon energy and momentum.
• Apply these ideas to the photo‑electric effect, Compton scattering and radiation pressure.
• Connect the photon model to other Cambridge quantum‑physics topics (wave‑particle duality, de Broglie wavelength, atomic transitions).

Key Definitions (Cambridge terminology)

Fundamental Constants

Photon Energy

Planck’s relation links the energy of a photon to its frequency (or wavelength):

\[

E = h\nu = \frac{hc}{\lambda}

\]

Useful exam shortcut (λ in nm, E in eV)

\[

E\;[\text{eV}] \approx \frac{1240}{\lambda\;[\text{nm}]}

\]

Photon Momentum

For a mass‑less particle the relativistic relation \(E^{2}=p^{2}c^{2}\) reduces to

\[

E = pc \;\;\Longrightarrow\;\; p = \frac{E}{c}

\]

Substituting the energy expression gives the compact formula

\[

p = \frac{h}{\lambda}

\]

Thus momentum is inversely proportional to wavelength.

Derivation without the Lorentz factor

A photon has zero rest mass, so the classical definition \(p = mv\) cannot be used. The only combination of the known quantities \(E\) (J) and \(c\) (m s⁻¹) that yields the dimensions of momentum (kg m s⁻¹) is \(p = E/c\). Inserting \(E = hc/\lambda\) immediately gives \(p = h/\lambda\).

Experimental Evidence for the Particulate Description

1. Photo‑electric effect

   • Electrons are emitted only if the incident light has a frequency \(\nu \ge \nu_{0}\) (threshold frequency), regardless of intensity.
   • Maximum kinetic energy of the emitted electrons:

     \[

     K_{\max}=h\nu-\phi

     \]

   • Increasing the light intensity raises the number of photons, therefore more electrons are emitted, but it does not increase \(K_{\max}\).

2. Compton scattering

   • When X‑rays scatter from loosely bound electrons the wavelength changes by

     \[

     \Delta\lambda = \frac{h}{m_{e}c}\,(1-\cos\theta)

     \]

   • The shift follows directly from conservation of photon momentum before and after the collision.

3. Radiation pressure

   • For a beam of intensity \(I\) (W m⁻²) that is completely absorbed:

     \[

     P = \frac{I}{c}

     \]

   • If the beam is perfectly reflected the pressure doubles:

     \[

     P = \frac{2I}{c}

     \]

   • These formulas are obtained by counting the momentum transferred per photon (\(p = h/\lambda\)) and the photon flux \(\frac{I}{E}\).

Links to Other Cambridge Quantum‑Physics Topics (Syllabus 22.3–22.4)

• Wave‑particle duality – Light shows wave behaviour (interference, diffraction) and particle behaviour (photo‑electric, Compton). The photon model supplies the particle side.
• de Broglie wavelength – Matter particles have \(\lambda = h/p\). The same expression for photons highlights the symmetry between light and matter.
• Atomic transitions – An electron moving between energy levels emits or absorbs a photon with \(\Delta E = h\nu\). This underpins spectroscopy and connects directly to the photon‑energy formula.

Worked Example (Exam‑style)

Problem: A monochromatic beam has wavelength \(\lambda = 500\ \text{nm}\). Determine (a) the photon energy in electron‑volts, (b) the photon momentum in kg·m s⁻¹, and (c) the radiation pressure on a perfectly absorbing surface if the beam intensity is \(I = 2.0\ \text{W m}^{-2}\).

Solution:

1. Energy (eV)

   \[

   E\;[\text{eV}] \approx \frac{1240}{\lambda\;[\text{nm}]} = \frac{1240}{500}=2.48\ \text{eV}

   \]

   (If required in joules: \(E = 2.48\ \text{eV}\times1.602\times10^{-19}=3.97\times10^{-19}\ \text{J}\).)

2. Momentum

   \[

   p = \frac{h}{\lambda}= \frac{6.626\times10^{-34}}{5.00\times10^{-7}}

   = 1.33\times10^{-27}\ \text{kg·m s}^{-1}

   \]

   (Alternatively \(p = E/c = 3.97\times10^{-19}/3.00\times10^{8}=1.32\times10^{-27}\ \text{kg·m s}^{-1}\).)

3. Radiation pressure (absorption)

   \[

   P = \frac{I}{c}= \frac{2.0}{3.00\times10^{8}}

   = 6.7\times10^{-9}\ \text{N m}^{-2}

   \]

   (For a perfectly reflecting surface the answer would be twice this value.)

Quick Reference Table – Photon Properties for Common Wavelengths

Conversion used: \(p = E/c\) and \(E\) (J) = \(E\) (eV) × \(1.602\times10^{-19}\).

Key Formulas to Remember (Cambridge 9702)

• Photon energy: \(E = h\nu = \dfrac{hc}{\lambda}\) or \(E\;[\text{eV}] \approx \dfrac{1240}{\lambda\;[\text{nm}]}\)
• Photon momentum: \(p = \dfrac{E}{c} = \dfrac{h}{\lambda}\)
• Photo‑electric kinetic energy: \(K_{\max}=h\nu-\phi\)
• Compton wavelength shift: \(\Delta\lambda = \dfrac{h}{m_{e}c}\,(1-\cos\theta)\)
• Radiation pressure (absorption): \(P = \dfrac{I}{c}\) ; (reflection): \(P = \dfrac{2I}{c}\)
• de Broglie wavelength (matter waves): \(\lambda = \dfrac{h}{p}\)

Summary

Electromagnetic radiation can be regarded as a stream of photons. Each photon carries a quantised energy \(E = h\nu\) and a corresponding momentum \(p = h/\lambda = E/c\). This particle picture explains the photo‑electric effect, Compton scattering and radiation pressure, and it fits naturally into the broader quantum‑physics framework of wave‑particle duality, de Broglie matter waves, and atomic transitions. Mastery of the formulas above enables students to tackle the full range of Cambridge A‑Level questions on photon energy and momentum.