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

Typical Photon Energies

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.