use the electronvolt (eV) as a unit of energy

Energy and Momentum of Photons – Cambridge AS & A Level Physics (9702)

Learning Objectives

• Express photon energy in electronvolts (eV) and use this unit in calculations.
• Derive and apply the four exact photon relations:
  \(E = hu\), \(E = \dfrac{hc}{\lambda}\), \(p = \dfrac{E}{c}\), \(p = \dfrac{h}{\lambda}\).
• Explain how photon energy and momentum underpin the photo‑electric effect, wave‑particle duality and atomic line spectra.
• Interpret experimental evidence (radiation‑pressure, Compton scattering, photo‑electric measurements) that confirms the photon model.

Connections to Other Syllabus Topics

The photon is introduced in Topic 7.4 (Electromagnetic spectrum). Its energy in eV links directly to:

• Topic 22.2 – Photo‑electric effect (work function \(\phi\) expressed in eV).
• Topic 22.4 – Atomic line spectra (energy‑level differences \(\Delta E\) in eV).
• Topic 24 – Medical physics (radiation pressure and scattering, where photon momentum in eV c⁻¹ is convenient).

Key Concepts

• Photon: a mass‑less quantum of electromagnetic radiation that carries energy \(E\) and momentum \(p\).
• Electronvolt (eV): the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 V.
  \(1\;\text{eV}=1.602\times10^{-19}\;\text{J}\).
• Energy–frequency relation: \(E = hu\) (Planck’s constant \(h = 6.626\times10^{-34}\;\text{J s}\)).
• Energy–wavelength relation: \(E = \dfrac{hc}{\lambda}\).
• Momentum–energy relation: \(E = pc\) (derived from special relativity for a particle with rest mass \(m=0\)).
• Momentum–wavelength relation: \(p = \dfrac{h}{\lambda}\).
• Wave‑particle duality: photons exhibit interference & diffraction (wave) and photo‑electric emission & radiation pressure (particle).
• Photo‑electric effect: electrons are emitted when \(E \ge \phi\). Maximum kinetic energy \(K_{\max}=E-\phi\).
• Atomic line spectra: a transition \(\Delta E\) between quantised levels produces a photon of energy \(\Delta E\) and wavelength \(\lambda = hc/\Delta E\).

Fundamental Photon Relations (Exact)

Quantity	Formula	Units (SI)
Energy – frequency	\(E = hu\)	J (or eV)
Energy – wavelength	\(E = \dfrac{hc}{\lambda}\)	J (or eV)
Momentum – energy	\(E = pc\)	J·s m⁻¹ (or eV c⁻¹)
Momentum – wavelength	\(p = \dfrac{h}{\lambda}\)	kg·m s⁻¹

Derivation of \(E = pc\) for a Photon

Special relativity gives the general energy–momentum relation

\[ E^{2}=p^{2}c^{2}+m^{2}c^{4}. \]

For a photon the rest mass \(m=0\); therefore

\[ E^{2}=p^{2}c^{2}\;\Longrightarrow\;E = pc \quad (\text{positive root}). \]

Useful Constants

Constant	Symbol	Value	Units
Speed of light	c	2.998 × 10⁸	m s⁻¹
Planck constant	h	6.626 × 10⁻³⁴	J s
Reduced Planck constant	ħ	1.055 × 10⁻³⁴	J s
1 electronvolt	1 eV	1.602 × 10⁻¹⁹	J

Expressing Photon Energy in eV

Convert joules to electronvolts by dividing by the conversion factor:

\[ E\;(\text{eV}) = \frac{E\;(\text{J})}{1.602\times10^{-19}}. \]

Combining with \(E = hc/\lambda\) gives a handy formula when \(\lambda\) is in nanometres:

\[ \boxed{E\;(\text{eV}) = \frac{1240}{\lambda\;(\text{nm})}}. \]

When frequency is given in terahertz (THz):

\[ \boxed{E\;(\text{eV}) = 4.1357\times10^{-3}\,u\;(\text{THz})}. \]

Photon Momentum – SI and eV c⁻¹

From \(p = E/c\) the momentum can be expressed directly in terms of the photon energy:

\[ p = \frac{E}{c}. \]

If the energy is given in eV it is often convenient to keep the speed of light in the denominator, defining the unit eV c⁻¹:

\[ p\;(\text{eV}\,c^{-1}) = \frac{E\;(\text{eV})}{c}\times\frac{1.602\times10^{-19}\;\text{J}}{1\;\text{eV}}. \]

Numerically,

\[ p\;(\text{eV}\,c^{-1}) \approx 5.344\times10^{-28}\,E\;(\text{eV})\;\text{kg·m s}^{-1}. \]

Experimental Evidence (AO3)

• Radiation‑pressure measurement: A laser beam of power \(P\) exerts a force \(F = P/c\) on a suspended mirror. The force directly demonstrates the relation \(p = E/c\).
• Compton scattering: X‑rays scattered from a free electron show a wavelength shift \[ \Delta\lambda = \frac{h}{m_ec}(1-\cos\theta), \] derived from conservation of energy and momentum for photons, confirming that photons carry momentum.
• Photo‑electric effect: The linear relationship \[ eV_s = hu - \phi \] between stopping potential \(V_s\) and frequency \(u\) demonstrates quantised photon energy \(E = hu\) and the role of the work function \(\phi\) (both expressed in eV).

Worked Example 1 – Energy & Momentum of Visible Light

Problem: Find the energy (eV) and momentum (kg·m s⁻¹) of a photon with wavelength \(\lambda = 500\;\text{nm}\) (green light).

1. Energy in joules: \[ E = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{500\times10^{-9}} = 3.97\times10^{-19}\;\text{J}. \]
2. Convert to electronvolts: \[ E\;(\text{eV}) = \frac{3.97\times10^{-19}}{1.602\times10^{-19}} \approx 2.48\;\text{eV}. \]
3. Momentum: \[ p = \frac{E}{c} = \frac{3.97\times10^{-19}}{2.998\times10^{8}} = 1.33\times10^{-27}\;\text{kg·m s}^{-1}. \]
4. Momentum in eV c⁻¹ (optional): \[ p\;(\text{eV}\,c^{-1}) = 5.34\times10^{-28}\times2.48 \approx 1.33\times10^{-27}\;\text{kg·m s}^{-1}. \]

Worked Example 2 – Photo‑electric Effect

Problem: Light of wavelength 250 nm shines on a metal with work function \(\phi = 2.0\;\text{eV}\). Calculate the maximum kinetic energy of the emitted electrons and the required stopping potential.

1. Photon energy: \[ E = \frac{1240}{250}=4.96\;\text{eV}. \]
2. Maximum kinetic energy: \[ K_{\max}=E-\phi=4.96-2.0=2.96\;\text{eV}. \]
3. Stopping potential: \[ V_s = \frac{K_{\max}}{e}=2.96\;\text{V}. \]

Summary Table

Quantity	Formula	SI Units	Common eV Form
Energy	\(E = hu = \dfrac{hc}{\lambda}\)	J or eV	\(E\;(\text{eV}) = \dfrac{1240}{\lambda\;(\text{nm})}\)
Momentum	\(p = \dfrac{E}{c} = \dfrac{h}{\lambda}\)	kg·m s⁻¹	\(p\;(\text{eV}\,c^{-1}) = 5.34\times10^{-28}\,E\;(\text{eV})\)
Frequency	\(u = \dfrac{c}{\lambda}\)	Hz	\(E\;(\text{eV}) = 4.1357\times10^{-3}\,u\;(\text{THz})\)
Maximum kinetic energy (photo‑electric)	\(K_{\max}=E-\phi\)	J or eV	Use \(E\) in eV and \(\phi\) in eV.

Practice Questions

1. Calculate the energy (eV) of an X‑ray photon with wavelength \(0.10\;\text{nm}\).
2. A photon has an energy of \(3.0\;\text{eV}\). Determine its wavelength in nanometres.
3. Find the momentum (kg·m s⁻¹) of a photon whose energy is \(2.5\;\text{eV}\). Express the result also in eV c⁻¹.
4. Show algebraically that \(E = pc\) follows from the two relations \(E = hc/\lambda\) and \(p = h/\lambda\).
5. In a photo‑electric experiment, light of wavelength \(400\;\text{nm}\) ejects electrons from a metal with work function \(1.8\;\text{eV}\). Calculate the stopping potential required to halt the most energetic electrons.
6. Explain how the Compton wavelength‑shift formula provides experimental proof of photon momentum.