22.1 Energy and Momentum of a Photon

---

1. The Particulate Nature of Electromagnetic Radiation

Electromagnetic radiation (EM radiation) includes all types of light — visible light, X-rays, ultraviolet, infrared, radio waves, and so on. For a long time, scientists believed EM radiation behaved purely as a wave. Evidence like diffraction (waves spreading around corners) and interference (waves adding up or cancelling) supported this idea.

However, certain experiments — most famously the photoelectric effect — could not be explained by wave theory alone. These experiments showed that EM radiation sometimes behaves as if it is made up of tiny particles, not just waves. This is called the particulate nature of EM radiation.

This was a revolutionary idea: light is not just a wave — it also comes in small, separate "packets" of energy.

---

2. What Is a Photon?

A photon is one of those tiny packets of electromagnetic energy. Think of it like a small "bullet" of light.

Key points about photons:

• A photon is a quantum of electromagnetic energy. Quantum just means the smallest possible discrete (separate) amount of something — you cannot have half a photon.
• Photons have no mass.
• The energy carried by a photon is fixed — it depends on the frequency of the radiation.
• Energy is not transferred as a continuous stream; instead, it is delivered in these individual packets, one photon at a time.

---

3. The Energy of a Photon: E = hf

The energy of a single photon can be calculated using:

$E = hf$

Where:

• E = energy of the photon, measured in joules (J)
• h = Planck's constant = 6.63 × 10⁻³⁴ J s
• f = frequency of the electromagnetic radiation, measured in hertz (Hz)

Planck's constant is a fixed number in nature, named after the physicist Max Planck. It tells us how big each quantum of energy is.

Alternative version of the formula

Since frequency and wavelength are related by the wave equation c = fλ, you can write f = c/λ and substitute:

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

Where:

• c = speed of light = 3.0 × 10⁸ m s⁻¹
• λ = wavelength of the radiation, measured in metres (m)

This tells you: the longer the wavelength, the lower the energy of each photon.

Worked Example 1

Calculate the energy of a photon of ultraviolet radiation with wavelength λ = 300 nm.

Step 1: Convert the wavelength to metres. λ = 300 nm = 300 × 10⁻⁹ m

Step 2: Use E = hc/λ.

$E = \frac{(6.63 \times 10^{-34}) \times (3.0 \times 10^8)}{300 \times 10^{-9}}$

$$E = 6.63 \times 10^{-19} \text{ J}$$