recall and use E = hf

Energy and Momentum of a Photon

Learning Objective

Recall and use the relation $E = hf$ to calculate the energy of a photon, and apply the associated momentum formula $p = \dfrac{E}{c} = \dfrac{h}{\lambda}$ in a variety of contexts.

Key Concepts

• Photons are quanta of electromagnetic radiation and have no rest mass.
• The energy of a photon is directly proportional to its frequency: $$E = hf$$ where $h = 6.626\times10^{-34}\ \text{J·s}$ (Planck constant) and $f$ is the frequency.
• Since $c = \lambda f$, the energy can also be expressed in terms of wavelength: $$E = \frac{hc}{\lambda}$$
• The momentum of a photon is related to its energy by: $$p = \frac{E}{c} = \frac{h}{\lambda} = \frac{hf}{c}$$
• Photon momentum, though small, can produce measurable effects (e.g., radiation pressure, photon recoil).

Useful Constants

Constant	Symbol	Value	Units
Planck constant	$h$	6.626 × 10⁻³⁴	J·s
Speed of light in vacuum	$c$	2.998 × 10⁸	m·s⁻¹
Elementary charge	$e$	1.602 × 10⁻¹⁹	C

Typical Photon Energies

Radiation Type	Wavelength $\lambda$ (nm)	Frequency $f$ (THz)	Energy $E$ (eV)
Radio (AM)	≈ 10⁶	≈ 0.3	≈ 1.2 × 10⁻⁶
Microwave (2.45 GHz)	≈ 122 mm	2.45	≈ 1.0 × 10⁻⁵
Infrared (10 µm)	10 000	30	≈ 0.124
Visible (green, 550 nm)	550	545	≈ 2.25
Ultraviolet (200 nm)	200	1500	≈ 6.20
X‑ray (0.1 nm)	0.1	3 × 10⁶	≈ 12.4 × 10³
Gamma (0.01 nm)	0.01	3 × 10⁷	≈ 124 × 10³

Derivation of Photon Momentum

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

$$E = hf,\qquad c = \lambda f$$

Eliminate $f$:

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

Since momentum $p$ for a massless particle satisfies $E = pc$, we obtain:

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

Example Problems

1. Energy of a green photon
   A photon has a wavelength of $550\ \text{nm}$. Find its energy in joules and electron‑volts.
   

   Solution:

   $$E = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{550\times10^{-9}} \approx 3.61\times10^{-19}\ \text{J}$$

   Convert to e \cdot using $1\ \text{eV}=1.602\times10^{-19}\ \text{J}$:

   $$E \approx \frac{3.61\times10^{-19}}{1.602\times10^{-19}} \approx 2.25\ \text{eV}$$
2. Momentum of an X‑ray photon
   An X‑ray has a wavelength of $0.1\ \text{nm}$. Determine its momentum.
   

   Solution:

   $$p = \frac{h}{\lambda} = \frac{6.626\times10^{-34}}{0.1\times10^{-9}} = 6.63\times10^{-24}\ \text{kg·m·s}^{-1}$$
3. Radiation pressure on a perfectly absorbing surface
   A laser beam of power $5\ \text{W}$ strikes a black surface perpendicularly. Find the force exerted by the photon momentum.
   

   Solution:

   For a completely absorbing surface, $F = \dfrac{P}{c}$ where $P$ is the power.

   $$F = \frac{5}{2.998\times10^{8}} \approx 1.67\times10^{-8}\ \text{N}$$

Common Misconceptions

• Photons have zero rest mass but still carry momentum; momentum is not solely a function of mass.
• The energy–frequency relation applies to all electromagnetic radiation, not just visible light.
• When converting between wavelength and frequency, always keep track of units (nm ↔ m, THz ↔ Hz).

Suggested Classroom Activities

1. Calculate the photon energy for the colours of the visible spectrum and plot $E$ versus $\lambda$.
2. Demonstrate radiation pressure using a light‑mill or a solar‑sail model.
3. Use a spectrometer to measure the wavelength of a laser and then compute its momentum.