# 6.1: The Wave and Particle Natures of Light

The physics of electromagnetic radiation is described by Maxwell's equations, Equations 1.6.3 - 1.6.6, and discussed in Sections 1.6.1 and 4.4.1. Optical energy is electromagnetic energy with wavelengths roughly in the range

$400 nm \lesssim \lambda \lesssim 650 nm. \nonumber$

This wavelength range corresponds to the frequency range

$4.6 \cdot 10^{14} \text{Hz} \lesssim f \lesssim 7.5 \cdot 10^{14} \text{Hz}. \nonumber$

We often think of electromagnetic radiation as behaving like a wave. However, it has both wave-like and particle-like behavior.

One way to understand light is to think of it as composed of particles called photons. A quantum is a small chunk, and a photon is a quantum, small chunk, of light. A related quantity is a phonon, which is a quanta, or small chunk, of lattice vibrations. We will discuss phonons in a later section, and they do not relate to light. Although, phonons can perturb light, and that is the basis for acousto-optic devices. The second way to understand light is to think of it as a wave with a wavelength $$\lambda$$ measured in $$nm$$. White light has a broad bandwidth while the light produced by a laser has a very narrow bandwidth.

These two descriptions of light complement each other. A photon is the smallest unit of light, and it has a particular wavelength. The energy of a photon of light with wavelength $$\lambda$$ is given by

$E = hf = \frac{hc}{\lambda}. \label{6.1.1}$

The quantity $$h$$ is called the Planck constant, and it has a tiny value, $$h = 6.626 \cdot 10^{-34} J \cdot s$$. The quantity $$c$$ is the speed of light in free space, $$c = 2.998 \cdot 10^8 \frac{m}{s}$$.

In SI units, energy is measured in joules. However, other units are sometimes used by optical engineers because the energy of an individual photon is tiny compared to a joule. Another unit that is used is the electronvolt, or eV. The magnitude of the charge of an electron is $$q = 1.602 \cdot 10^{-19} C$$. The electronvolt is the energy acquired by a charge of this magnitude in the presence of a voltage difference of one volt [68, p. 8]. Energy in joules and energy in eV are related by a factor of $$q$$.

$E_{[J]} = q \cdot E_{[eV]} \label{6.1.2}$

Equations \ref{6.1.1} and \ref{6.1.2} can be combined to relate the energy of a photon in eV and the corresponding wavelength in $$nm$$.

$\frac{1240}{\lambda_{[nm]}} = E_{[eV]}.$

Sometimes, energy is specified in the unit of wave number, $$cm^{-1}$$, which represents the reciprocal of the wavelength of the corresponding photon. Energy in joules and energy in wave number are related by

$E_{[J]} = \frac{hc}{\lambda}$

$E_{[J]} = \frac{6.626 \cdot 10^{-34} J \cdot s \cdot 2.998 \cdot 10^8 \frac{m}{s} \cdot 100 \frac{cm}{m}}{\lambda_{[cm]}}$

$E_{[J]} = 1.986 \cdot 10^{-23} E_{[cm^{-1}]}.$

The human eye can sense light from approximately $$\lambda = 400 nm$$ to $$\lambda = 650 nm$$. Using the expressions above, we can calculate in different units the energy range over which the human eye can respond. An individual red photon with $$\lambda = 650 nm$$ has energy

$E_{red} = 3.056 \cdot 10^{-19} J = 1.908 eV = 1.538 \cdot 10^4 cm^{-1}$

in the different units. Similarly, an individual blue photon with $$\lambda = 400 nm$$ has energy

$E_{blue} = 4.966 \cdot 10^{-19} J = 3.100 eV = 2.500 \cdot 10^4 cm^{-1}.$

We can calculate the energy of individual photons of electromagnetic radiation at radio frequencies, at microwave frequencies, or in other frequency ranges too. For example, the radio station WEAX broadcasts with a frequency $$f = 88 \text{MHz}$$. This corresponds to a wavelength of $$\lambda = 3.407 m$$. An individual photon at this frequency has energy

$E = 5.831 \cdot 10^{-26} J = 3.640 \cdot 10^{-7} eV.$

As another example, wi-fi operates at frequencies near $$f = 2.4 \text{GHz}$$ which corresponds to the wavelength $$\lambda = 0.125 m$$. Each photon at this frequency has energy

$E = 1.590 \cdot 10^{-24} J = 9.927 \cdot 10^{-6} eV.$

Ultraviolet light has a wavelength slightly shorter than blue light. A photon of ultraviolet light with wavelength $$\lambda = 350 nm$$, which corresponds to frequency $$f = 8.57 \cdot 10^{14} \text{Hz}$$, has energy

$E = 5.676 \cdot 10^{-19} J = 3.543 eV.$

X-rays operate at wavelengths near $$\lambda = 10^{-10} m$$. An x-ray photon with wavelength $$\lambda = 10^{-10} m$$ has energy

$E = 1.986 \cdot 10^{-15} J = 1.240 \cdot 10^{4} eV.$

Why do we talk about radio waves but not radio particles while we treat light as both wave-like and particle-like? A person is around 1.5 to 2 m tall. The wavelength of the radio station broadcast in the example above was $$\lambda_{RF} \approx 3.4 m$$ while the wavelength of blue light was $$\lambda_{blue \; light} \approx 400 nm$$. Both radio frequency and optical signals are electromagnetic radiation. Both are well described by Maxwell's equations. Both have wave-like and particle-like properties. Humans typically talk about the wave-like nature of radio waves because they are on a scale we can measure with a meter stick. However, with the correct tools, we can observe both the wave-like and particle-like behavior of light.

Why is UV light more dangerous than visible light? Why are x-rays so dangerous? Each photon of x-ray radiation has around a thousand times more energy than a photon of green light. This type of radiation is called ionizing radiation because each photon has enough energy to rip an electron from skin or muscles. UV radiation also has enough energy per photon to rip an electron off while red light and blue light do not have enough energy. Photons of radio frequency and microwave electromagnetic radiation contain nowhere near enough energy per photon to do this damage. These types of radiation can still pose a safety hazard if enough photons land on your skin. Microwave ovens are used to cook food. However, they do not pose the hazards of ionizing radiation.