use the electronvolt (eV) as a unit of energy

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: $$E = hu$$
• Energy–wavelength relation: $$E = \frac{hc}{\lambda}$$
• Energy–momentum relation: $$E = pc$$
• Momentum expressed in terms of wavelength: $$p = \frac{h}{\lambda}$$

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

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):

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

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

Quantity	Formula	Units	Typical Form (eV)
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	$p\;(\text{eV}\,c^{-1}) = \dfrac{E\;(\text{eV})}{c}$
Frequency	$u = \dfrac{c}{\lambda}$	Hz	$E\;(\text{eV}) = 4.1357\times10^{-3}\,u\;(\text{THz})$

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 $p = h/\lambda$.