use the electronvolt (eV) as a unit of energy

Published by Patrick Mutisya · 14 days ago

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

Energy and Momentum of a Photon

Learning Objective

Be able to express the energy of a photon in electronvolts (eV) and use this unit in calculations of photon energy and momentum.

Key Concepts

  • A photon is a quantum of electromagnetic radiation with no rest mass.

  • The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength.

  • Even though a photon has no mass, it carries momentum.

  • The electronvolt (eV) is a convenient unit for photon energies, especially in atomic and nuclear physics.

Fundamental Relations

All of the following equations are exact for a photon:

  • Energy–frequency relation: \$E = h\nu\$

  • Energy–wavelength relation: \$E = \frac{hc}{\lambda}\$

  • Energy–momentum relation: \$E = pc\$

  • Momentum expressed in terms of wavelength: \$p = \frac{h}{\lambda}\$

Useful Constants

ConstantSymbolValueUnits
Speed of lightc2.998 × 10⁸m s⁻¹
Planck constanth6.626 × 10⁻³⁴J s
Reduced Planck constantħ1.055 × 10⁻³⁴J s
1 electronvolt1 eV1.602 × 10⁻¹⁹J

Expressing Photon Energy in eV

To convert the photon energy from joules to electronvolts, divide by the conversion factor \$1\;\text{eV}=1.602\times10^{-19}\;\text{J}\$:

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

Combining this with \$E = hc/\lambda\$ gives a convenient formula for wavelength in nanometres:

\$E\;(\text{eV}) = \frac{1240}{\lambda\;(\text{nm})}\$

Similarly, for frequency in terahertz (THz):

\$E\;(\text{eV}) = 4.1357\times10^{-3}\,\nu\;(\text{THz})\$

Photon Momentum in 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}\$

When \$E\$ is given in electronvolts, it is often convenient to keep \$c\$ in the denominator, yielding units of eV c⁻¹. Numerically:

\$p\;(\text{eV}\,c^{-1}) = \frac{E\;(\text{eV})}{c\;(=2.998\times10^{8}\,\text{m s}^{-1})} \times (1.602\times10^{-19}\,\text{J/eV})\$

For many problems it is sufficient to use the relation \$p = h/\lambda\$ with \$\lambda\$ in metres.

Worked Example

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

  1. Calculate the 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. Find the momentum:

    \$p = \frac{E}{c} = \frac{3.97\times10^{-19}}{2.998\times10^{8}} \approx 1.33\times10^{-27}\;\text{kg·m s}^{-1}\$

Summary Table

QuantityFormulaUnitsTypical Form (eV)
Energy\$E = h\nu = \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}) = \dfrac{E\;(\text{eV})}{c}\$
Frequency\$\nu = \dfrac{c}{\lambda}\$Hz\$E\;(\text{eV}) = 4.1357\times10^{-3}\,\nu\;(\text{THz})\$

Practice Questions

  1. Calculate the energy in e \cdot of an X‑ray photon with wavelength \$0.1\;\text{nm}\$.

  2. A photon has an energy of \$3.0\;\text{eV}\$. Determine its wavelength in nanometres.

  3. Find the momentum (in kg·m s⁻¹) of a photon whose energy is \$2.5\;\text{eV}\$.

  4. Show that the relation \$E = pc\$ holds for a photon by substituting \$E = hc/\lambda\$ and \$p = h/\lambda\$.

Suggested diagram: Energy–momentum diagram for a photon showing the linear relation \$E = pc\$ and the conversion between wavelength, frequency, energy, and momentum.