Everyone is familiar with sunlight: it lets us see during the day, it keeps us warm, and it's an integral part of an enjoyable day at the beach.

While we're all familiar with sunlight, the scientific specifics are a bit more complicated. As it turns out, sunlight is electromagnetic radiation from the sun.

This quiz will explore sunlight and electromagnetic radiation in more detail, and provide us with some of the background understanding we need in order to learn how solar energy technologies work and how we can improve their performance.

Sunlight is electromagnetic radiation, as is the beam from a laser pointer and the light from a lamp or a candle. While these different types of light are all fundamentally the same in that they are all electromagnetic radiation, there are some properties that differ between them and allow you to tell them apart in everyday situations.

The important properties of electromagnetic radiation are **intensity**, **direction**, **wavelength** and **frequency**. This quiz will introduce these properties and give a foundation for understanding the science behind sunlight.

One big difference between sunlight and the light from a lightbulb is their **intensity**. Colloquially, intensity refers to how bright or how strong a source of electromagnetic radiation is. Scientifically, intensity refers to the density of radiant power carried by the electromagnetic radiation, typically measured in \(\si[per-mode=symbol]{\watt\per\meter\squared}\).
As you'll see in the following quiz, the intensity of sunlight during the day is on the scale of \(\SI[per-mode=symbol]{1000}{\watt\per\meter\squared}.\)

What would be the intensity, \(I_\textrm{bulb},\) (in \(\si[per-mode=symbol]{\watt\per\meter\squared}\)) on your skin from a \(\SI{40}{\watt}\) light bulb that's \(\SI{3}{\meter}\) away? Assume that the bulb emits equally in all directions, and that all \(\SI{40}{\watt}\) of input power are converted to electromagnetic radiation.

**Hint**: if you could put a sphere of light collecting material around the lightbulb, you'd collect \(\SI{40}{\watt}\) of power, no matter how big the sphere. The surface area of a sphere of radius \(r\) is \(A = 4\pi r^2.\)

**Direction** is another property of radiation. As the name suggests, it refers to the direction in which electromagnetic radiation is traveling. Radiation will continue to travel in the same direction until it interacts with matter, which might reflect or scatter it, changing its course.

Sunlight reaching Earth's surface has a relatively uniform direction: the direction is aligned with the vector pointing from the Sun to Earth. Why do you cast a shadow on a sunny day, but not on a cloudy day?

Like all waves, electromagnetic radiation has an associated **wavelength**. Wavelength is an important property of radiation, and radiation with different wavelengths can interact with matter in very different ways.

Radiation with wavelengths from \(\SI{390}{\nano\meter} - \SI{700}{\nano\meter}\) is considered visible light because the human eye responds to those wavelengths. Each wavelength corresponds to a specific color, with the shortest wavelengths being violet light and the longest wavelengths being red light.

Why does a campfire appear to produce a different color of light than the Sun?

Since electromagnetic radiation is a wave, that means we can relate its wavelength \(\lambda\) and **frequency** \(f\) through a velocity \(v\):
\[v = \lambda f\]
All electromagnetic radiation travels at the speed of light (\(\SI[per-mode=symbol]{3e8}{\meter\per\second}\)), which you may have suspected since visible light is one type of electromagnetic radiation. Thus if we know either the wavelength or the frequency of electromagnetic radiation, we can get the other through the equation above.

What is the frequency of electromagnetic radiation with a wavelength of \(\lambda = \SI{500}{\nano\meter}\) (\(\SI{5e-7}{\meter}\))?

Classically, electromagnetic radiation is thought of as wave energy but Einstein famously developed the theory of photons, which are elementary particles carrying light energy. Photons have some wavelike properties and some particle like properties.

For some solar energy conversion technologies (like photovoltaic cells), it is important to think of solar energy as being carried by photons.

The most important distinction between the wave and photon models is how energy is delivered. In the wave model, energy is delivered continuously (e.g. like the beam from a flashlight), whereas in the photon model, energy is delivered by the photons as discrete packets.

The energy of each photon depends on its frequency \(f\) (or equivalently its wavelength \(\lambda\)), and does not depend on the intensity of radiation. Thus, higher intensity radiation of a particular wavelength corresponds to more photons, rather than photons with higher energy.

The energy of a photon is given by the Planck relation, \[E = hf,\] where \(h\) is the Planck constant, and \(f\) is the frequency of the photon.

An equivalent expression in terms of wavelength is given by \(E = h c/\lambda\) where \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon.

Photon energy is typically reported in units of electron volts (\(\si{\electronvolt}\)), which is the energy required to move an electron across a potential difference of \(\SI{1}{\volt},\) therefore \(\SI{1}{\electronvolt}\) corresponds to approximately \(\SI{1.6e-19}{\joule}.\)

How much energy in \(\si{\electronvolt}\) is carried by a photon with a wavelength of \(\SI{500}{\nano\meter}\)?

**Note**, when using the Planck relation to find photon energy in terms of wavelength, it can be convenient to consider \(hc\) as a single quantity, whose value is given approximately by \(\SI{1240}{\electronvolt\nano\meter}.\)

Approximately how many photons will hit a \(\SI{1}{\meter\squared}\) collector exposed to sunlight in \(\SI{1}{\second}\)?

**Assume** that the total intensity of sunlight is \(\SI[per-mode=symbol]{1000}{\watt\per\meter\squared}\), and that the photons in a beam of sunlight have an average wavelength of \(\bar{\lambda} = \SI{500}{\nano\meter}\).

**Hint**: recall that \(\SI{1}{\watt} = \SI[per-mode=symbol]{1}{\joule\per\second}.\)

An individual photon carries a very small amount of energy, and is almost impossible to detect by itself. Even so, as you will learn in later chapters, thinking of light as photons is important to understanding some solar energy technologies.

Additionally, understanding the properties of intensity, direction and wavelength lays the foundation for learning how sunlight can be converted to other forms of energy. Understanding these properties is critical for investigating how these conversion processes can be improved.

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