understand that a photon is a quantum of electromagnetic energy

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

Learning Objectives (AO1‑AO3)

• Explain why light exhibits both wave‑ and particle‑like behaviour (AO1).
• State and use the fundamental relations for photon energy, momentum and frequency (AO1).
• Derive the photon‑momentum expression from relativistic principles (AO2).
• Apply these relations to solve quantitative problems in the photo‑electric effect, radiation pressure and Compton scattering (AO2‑AO3).
• Interpret experimental data that determine photon energy (AO3).

Syllabus Mapping (Section 22)

Syllabus Item	Content Covered in These Notes
22.1 – Photon energy & momentum (recall & use equations)	Sections 1–3, derivation of \(p = h/\lambda\), constants table.
22.2 – Photo‑electric effect (experiment, threshold, stopping potential)	Section 4, worked example, experimental note.
22.3 – Wave‑particle duality & de Broglie wavelength	Section 5, de Broglie hypothesis and electron‑diffraction example.
22.4 – Atomic energy‑level transitions & line spectra (apply \(hf = \Delta E\))	Section 6, Balmer‑series derivation and example.

1. Photon as a Quantum of Electromagnetic Energy

• Historical background – Planck (1900) proposed that electromagnetic radiation is emitted/absorbed in discrete packets of energy \(E = hu\). Einstein (1905) identified these packets as “photons”.
• Key properties of a photon:
  • Zero rest mass: \(m_{0}=0\).
  • Travels at the speed of light in vacuum: \(c = 2.998\times10^{8}\ \text{m s}^{-1}\).
  • Energy proportional to frequency: \(E = hu\).
  • Momentum despite having no mass: \(p = h/\lambda\).

2. Fundamental Relations

Quantity	Relation	Notes
Energy	\(E = hu\)	Planck’s quantisation; \(u\) = frequency (Hz)
Energy (wavelength form)	\(E = \dfrac{hc}{\lambda}\)	Using \(c = \lambdau\)
Momentum	\(p = \dfrac{h}{\lambda} = \dfrac{E}{c}\)	From relativistic energy–momentum relation for \(m_{0}=0\)
Frequency–wavelength link	\(c = \lambdau\)	Speed of light in vacuum

Derivation of Photon Momentum

1. Wave relation: \(c = \lambdau\) → \(u = c/\lambda\).
2. Planck’s law: \(E = hu\) → \(E = hc/\lambda\).
3. Relativistic energy–momentum relation: \[ E^{2} = (pc)^{2} + (m_{0}c^{2})^{2}. \] With \(m_{0}=0\) this reduces to \(E = pc\).
4. Substituting \(E = hc/\lambda\) gives \(p = E/c = h/\lambda\).

3. Key Constants

Symbol	Quantity	Value
\(h\)	Planck constant	\(6.626\times10^{-34}\ \text{J·s}\)
\(c\)	Speed of light in vacuum	\(2.998\times10^{8}\ \text{m·s}^{-1}\)
\(\lambda\)	Wavelength of a photon	Variable (m)
\(u\)	Frequency of a photon	Variable (Hz)
\(e\)	Elementary charge	\(1.602\times10^{-19}\ \text{C}\)

4. Photo‑electric Effect (Syllabus 22.2)

• Experiment – Monochromatic light of frequency \(u\) strikes a clean metal surface, causing electrons to be emitted.
• Einstein’s photo‑electric equation: \[ hf = \phi + \tfrac12 m_{e}v_{\max}^{2}, \] where \(\phi\) is the work function.
• Threshold frequency: \(u_{0} = \phi/h\). No electrons are emitted if \(u < u_{0}\).
• Stopping potential \(V_{s}\) satisfies \(eV_{s}= \tfrac12 m_{e}v_{\max}^{2}\), giving the practical form \[ hf = \phi + eV_{s}. \]

Worked Example – Stopping Potential

Problem: Light of wavelength \(250\ \text{nm}\) falls on a sodium surface (\(\phi = 2.28\ \text{eV}\)). Find the stopping potential.

1. Frequency: \(\displaystyle u = \frac{c}{\lambda}= \frac{2.998\times10^{8}}{2.50\times10^{-7}} = 1.20\times10^{15}\ \text{Hz}\).
2. Photon energy: \(\displaystyle E = hu = (6.626\times10^{-34})(1.20\times10^{15}) = 7.95\times10^{-19}\ \text{J}=4.97\ \text{eV}\).
3. Stopping potential: \[ V_{s}= \frac{hf-\phi}{e}= \frac{4.97\ \text{eV}-2.28\ \text{eV}}{1\ \text{eV/e}} = 2.69\ \text{V}. \]

5. Wave‑Particle Duality & de Broglie Wavelength (Syllabus 22.3)

• de Broglie hypothesis – any particle of momentum \(p\) has an associated wavelength \[ \lambda = \frac{h}{p}. \]
• For photons, \(p = h/\lambda\) so the de Broglie relation reduces to the photon wavelength expression already given.

Example – Electron Diffraction

Electrons accelerated through a potential \(V = 150\ \text{V}\) have kinetic energy \(eV\). Their momentum is

\[ p = \sqrt{2m_{e}eV}, \]

so the de Broglie wavelength is

\[ \lambda = \frac{h}{\sqrt{2m_{e}eV}} \approx 0.10\ \text{nm}, \]

comparable with inter‑atomic spacings, allowing diffraction.

6. Atomic Energy‑Level Transitions & Line Spectra (Syllabus 22.4)

• When an atom drops from an initial level \(i\) to a lower level \(f\), a photon is emitted with energy \[ hf = E_{i} - E_{f}. \]
• For hydrogen‑like atoms the Rydberg formula gives the wavenumber of any spectral line: \[ \frac{1}{\lambda}=R_{Z}\!\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right), \] where \(R_{Z}=R_{\!H}Z^{2}\) and \(R_{\!H}=1.097\times10^{7}\ \text{m}^{-1}\).
• Balmer series (visible lines) corresponds to \(n_{f}=2\) and \(n_{i}=3,4,5,\dots\). Substituting \(n_{f}=2\) gives \[ \frac{1}{\lambda}=R_{\!H}\!\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right),\qquad n=3,4,5,\dots \]

Worked Example – H‑α Line (\(n=3\rightarrow n=2\))

\[ \frac{1}{\lambda}=1.097\times10^{7}\!\left(\frac{1}{4}-\frac{1}{9}\right) =1.097\times10^{7}\times\frac{5}{36}=1.52\times10^{6}\ \text{m}^{-1}, \] \[ \lambda = \frac{1}{1.52\times10^{6}} = 6.58\times10^{-7}\ \text{m}=658\ \text{nm}. \]

7. Applications of Photon Momentum

• Radiation pressure – For a beam of intensity \(I\):
  • Perfect absorber: \(P = \dfrac{I}{c}\).
  • Perfect reflector: \(P = \dfrac{2I}{c}\).
• Compton scattering – Change in wavelength when a photon scatters from a free electron: \[ \Delta\lambda = \frac{h}{m_{e}c}\,(1-\cos\theta), \] demonstrating transfer of photon momentum to the electron.

8. Experimental Determination of Photon Energy (AO3)

A typical laboratory set‑up:

1. A monochromator selects a narrow wavelength band (\(\lambda\pm\Delta\lambda\)).
2. The beam strikes a calibrated silicon photodiode; the photocurrent \(I\) is proportional to the photon flux.
3. Using the measured electrical power \(P_{\text{elec}} = I V_{\text{bias}}\) and the relation \(E = hc/\lambda\), the energy per photon is obtained.
4. Uncertainty contributors:
   • Wavelength calibration: \(\pm0.5\ \text{nm}\).
   • Diode responsivity: \(\pm2\%\).
   • Stray light and alignment errors.
   Combined standard uncertainty is typically \(\approx 3\%\).

9. Worked Example – Photon Rate from Measured Power

A 650 nm laser delivers an average power of \(5.0\ \text{mW}\). Find the number of photons emitted per second.

1. Photon energy: \[ E = \frac{hc}{\lambda}= \frac{(6.626\times10^{-34})(2.998\times10^{8})}{6.50\times10^{-7}} = 3.05\times10^{-19}\ \text{J}. \]
2. Photon rate: \[ N = \frac{P}{E}= \frac{5.0\times10^{-3}}{3.05\times10^{-19}} \approx 1.64\times10^{16}\ \text{s}^{-1}. \]

10. Suggested Diagram

11. Summary

A photon is a mass‑less quantum of electromagnetic radiation. Its energy and momentum are directly linked to its frequency and wavelength:

\[ E = hu = \frac{hc}{\lambda},\qquad p = \frac{h}{\lambda} = \frac{E}{c}. \]

These simple relations bridge the wave description (\(\lambda,u\)) and the particle description (energy, momentum) and underpin the photo‑electric effect, radiation pressure, Compton scattering, atomic line spectra and the broader concept of wave‑particle duality.

12. Further Reading / Links to the Syllabus

• Section 22.1 – Energy & momentum of a photon.
• Section 22.2 – Photo‑electric effect (experimental set‑up, threshold, stopping potential).
• Section 22.3 – Wave‑particle duality & de Broglie wavelength.
• Section 22.4 – Energy levels & line spectra (Bohr model, hydrogen series, Rydberg formula).
• Section 23 – Quantum physics – applications to semiconductors and lasers.
• Paper 5 – Practical skills: measuring photon energy and radiation pressure.