recall and use E = hf

Energy and Momentum of a Photon (Cambridge IGCSE/A‑Level 9702)

Learning Objective (AO1 + AO2)

Recall the fundamental photon relations and use them to calculate photon energy, momentum and related quantities in a range of contexts.

• \(E = hf\)
• \(E = \dfrac{hc}{\lambda}\)
• \(p = \dfrac{E}{c} = \dfrac{h}{\lambda} = \dfrac{hf}{c}\)

Key Concepts

• Photons are quanta of electromagnetic radiation. They have zero rest mass but carry energy and momentum.
• Energy–frequency relation: \[

  E = hf \qquad\text{with } h = 6.626\times10^{-34}\ \text{J·s}

  \]

• Using the wave relation \(c = \lambda f\) the same energy can be written in terms of wavelength: \[

  E = \frac{hc}{\lambda}

  \]

• For a mass‑less particle \(E = pc\); therefore photon momentum is \[

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

  \]

• Photon momentum, although tiny, produces measurable effects such as radiation pressure, the photo‑electric effect, and Compton scattering.

Useful Constants

Typical Photon Energies

Derivation of Photon Momentum

Starting from the energy–frequency relation and the wave‑speed definition:

\[

E = hf,\qquad c = \lambda f

\]

Eliminate \(f\) to obtain the wavelength form of the energy:

\[

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

\]

For a particle with zero rest mass the relativistic relation \(E = pc\) holds, giving

\[

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

\]

Photon‑Related Phenomena (AO2)

Photo‑electric Effect

• Maximum kinetic energy of ejected electrons: \[

  K_{\max}=hf-\phi

  \]

• Threshold frequency: \[

  f_{\text{thr}}=\frac{\phi}{h}

  \]

• Both equations are used to determine the work function \(\phi\) or the kinetic energy for a given frequency.

Compton Scattering

The change in photon wavelength after scattering from a free electron is

\[

\Delta\lambda = \frac{h}{m_ec}(1-\cos\theta)

\]

derived from conservation of energy (\(hf\)) and momentum (\(h/\lambda\)).

Radiation Pressure

• Absorbing surface: \(F = \dfrac{P}{c}\)
• Perfectly reflecting surface: \(F = \dfrac{2P}{c}\) (momentum reversal doubles the transfer).

de Broglie Wavelength (Matter Waves)

Any particle of momentum \(p\) has an associated wavelength

\[

\lambda_{\text{dB}} = \frac{h}{p}

\]

For a non‑relativistic particle with kinetic energy \(K\), \(p = \sqrt{2mK}\) and

\[

\lambda_{\text{dB}} = \frac{h}{\sqrt{2mK}}

\]

Example: an electron with \(K = 100\ \text{eV}\) has

\[

\lambda_{\text{dB}} = \frac{6.626\times10^{-34}}{\sqrt{2(9.109\times10^{-31})(100\times1.602\times10^{-19})}}

\approx 1.23\times10^{-10}\ \text{m} \;(0.123\ \text{nm})

\]

Energy Levels & Line Spectra

• When an electron jumps between two atomic energy levels \(E1\) and \(E2\), a photon is emitted or absorbed with energy \[

  hf = |E1-E2|

  \]

• Balmer series (visible hydrogen lines) – e.g. the \(H_\alpha\) line at \(\lambda = 656.3\ \text{nm}\) corresponds to \[

  \Delta E = \frac{hc}{\lambda}= \frac{(6.626\times10^{-34})(2.998\times10^{8})}{656.3\times10^{-9}}

  \approx 3.03\times10^{-19}\ \text{J}=1.89\ \text{eV}

  \]

Worked Example Problems (AO2)

1. Energy of a green photon – \(\lambda = 550\ \text{nm}\)

   \[

   E = \frac{hc}{\lambda}= \frac{(6.626\times10^{-34})(2.998\times10^{8})}{550\times10^{-9}}

   = 3.61\times10^{-19}\ \text{J}=2.25\ \text{eV}

   \]

2. Momentum of an X‑ray photon – \(\lambda = 0.1\ \text{nm}\)

   \[

   p = \frac{h}{\lambda}= \frac{6.626\times10^{-34}}{0.1\times10^{-9}}

   = 6.63\times10^{-24}\ \text{kg·m·s}^{-1}

   \]

3. Radiation pressure on an absorbing surface – laser power \(P = 5\ \text{W}\)

   \[

   F = \frac{P}{c}= \frac{5}{2.998\times10^{8}}

   = 1.67\times10^{-8}\ \text{N}

   \]

4. Radiation pressure on a reflecting surface – same 5 W laser

   \[

   F = \frac{2P}{c}= 3.34\times10^{-8}\ \text{N}

   \]

5. Photo‑electric threshold frequency – \(\phi = 2.0\ \text{eV}\)

   \[

   f_{\text{thr}} = \frac{\phi}{h}= \frac{2.0\times1.602\times10^{-19}}{6.626\times10^{-34}}

   = 4.8\times10^{14}\ \text{Hz}

   \]

6. Maximum kinetic energy for a given frequency – \(\lambda = 250\ \text{nm}\) (so \(f = c/\lambda\)) and \(\phi = 1.5\ \text{eV}\)

   \[

   f = \frac{c}{\lambda}= \frac{2.998\times10^{8}}{250\times10^{-9}}=1.20\times10^{15}\ \text{Hz}

   \]

   \[

   K_{\max}=hf-\phi = (6.626\times10^{-34})(1.20\times10^{15})-1.5\,\text{eV}

   = 4.96\times10^{-19}\ \text{J}-1.5\,\text{eV}=2.6\ \text{eV}

   \]

7. Compton wavelength shift – incident \(\lambda = 0.071\ \text{nm}\), \(\theta = 60^{\circ}\)

   \[

   \Delta\lambda = \frac{h}{m_ec}(1-\cos60^{\circ})

   = 2.43\times10^{-12}\,(1-0.5)=1.22\times10^{-12}\ \text{m}

   \]

   \[

   \lambda' = 0.071\ \text{nm}+0.00122\ \text{nm}=0.0722\ \text{nm}

   \]

8. de Broglie wavelength of a 100 eV electron (see de Broglie section) – \(\lambda_{\text{dB}} = 0.123\ \text{nm}\).
9. Energy difference for the H\(_\alpha\) line – \(\lambda = 656.3\ \text{nm}\) gives \(\Delta E = 1.89\ \text{eV}\).

Common Misconceptions

• Photons have zero rest mass, yet they carry momentum because \(p = h/\lambda\) does not involve mass.
• The relation \(E = hf\) is universal for all electromagnetic radiation, not just visible light.
• When converting between wavelength and frequency, keep units consistent (nm → m, THz → Hz).
• Radiation pressure on a mirror is twice that on an absorbing surface; the factor of 2 is often omitted.
• De Broglie wavelength applies to *matter* particles, not to photons (which already have \(\lambda = h/p\)).
• Photon energy equals the *difference* between atomic energy levels, not the absolute value of a single level.

Suggested Classroom Activities

1. Calculate photon energies for the seven colours of the visible spectrum and plot \(E\) versus \(\lambda\). Discuss the linear relationship between \(E\) and \(1/\lambda\).
2. Demonstrate radiation pressure using a light‑mill or a small solar‑sail model; compare measured forces with the predictions \(F = P/c\) (absorbing) and \(F = 2P/c\) (reflecting).
3. Measure the wavelength of a He‑Ne laser with a diffraction grating, then compute its momentum and the corresponding radiation pressure on a suspended mirror.
4. Perform a photo‑electric experiment (metal cathode + variable‑frequency LED) to verify the threshold frequency and the equation \(K_{\max}=hf-\phi\).
5. Use a computer simulation of Compton scattering to visualise the wavelength shift and confirm energy‑momentum conservation.
6. Calculate de Broglie wavelengths for electrons, neutrons and larger particles (e.g., C\(_{60}\) molecules) and discuss why wave behaviour becomes unobservable for macroscopic masses.
7. Analyse a spectral line (e.g., the Balmer series) by converting its wavelength to the energy difference between atomic levels.