Solar Energy

Quantum Light

In the last quiz we learned about the properties of electromagnetic radiation, a.k.a. light. We also learned that classically, light was thought of as being carried by waves. The wave model of light became favored in the early 1800s, thanks to Young's experiment. He showed that light shining through a series of screens with thin slits resulted in a final light pattern with alternating light and dark stripes. This pattern wouldn't appear if light was carried by particles that only moved in straight lines, but could be explained by light being carried by waves.

Despite the persuasiveness of Young's experiment, the wave model could not explain all the observed behavior of EM radiation. The full picture of light is more complex, and has particle-like properties as well, which we will learn about in this quiz.

Quantum Light

Some materials will emit electrons when light is shined on them - this is called the photoelectric effect. The photoelectric effect occurs because energy delivered by the light can be absorbed by electrons, and once an electron has gained enough energy, it can escape from its host material. The photoelectric effect is important to our understanding of EM radiation because it can't be fully explained by the wave model of light.

Suppose we're measuring the photoelectric effect in lead, and we expose a sample of lead to two different conditions. In the first case, we illuminate the sample with EM radiation with a wavelength of \(\SI{300}{\nano\meter}\) and an intensity of \(\SI[per-mode=symbol]{500}{\watt\per\meter\squared}\). In the second case, we illuminate the sample with EM radiation with a wavelength of \(\SI{600}{\nano\meter}\) and an intensity of \(\SI[per-mode=symbol]{1000}{\watt\per\meter\squared}\). If light was completely described by the wave model, which situation would lead to more electrons being emitted from the lead sample by the photoelectric effect?


Quantum Light

In the wave model of light, energy delivered by incident radiation only depends on intensity. If the wave model explained everything, the wavelength of incident radiation should be irrelevant to the photoelectric effect, and higher intensity radiation should always lead to more emitted electrons .

However, this is not how the photoelectric effect actually works. For a given material, when the wavelength of incident light is increased beyond a particular value, electrons cease being emitted, no matter how high the intensity. In lead, this cutoff wavelength is around \(\SI{300}{\nano\meter}\), so radiation of longer wavelengths won't emit electrons, regardless of intensity.

This observation is inconsistent with the wave picture, since waves continuously deliver energy. If light was purely a wave phenomenon, electrons should always be emitted once enough energy has been absorbed, regardless of wavelength.

Quantum Light

Einstein solved the dilemma of the photoelectric effect with the introduction of the "photon," a quantized packet of electromagnetic radiation.* Here the idea of quantization means that something cannot be separated into arbitrarily small pieces: eventually, you reach a piece small enough that it can no longer be divided. Light being quantized into photons is similar to how matter is quantized into atoms.

A photon has many of the same properties we used to describe light when we modeled it as a wave: photons still have a direction, wavelength and frequency. Since photons have wavelengths and frequencies (which are still related through the speed of light), photons can still behave like waves. Einstein's work also established that photons each carry a certain amount of energy, and that energy depends on the photon's wavelength.

The quantization of light into photons resolved the dilemma of the photoelectric effect. How can the idea of photons explain the photoelectric effect's dependence on wavelength?

*While Einstein's theories of special relativity and mass-energy equivalence are much more famous, it was his work on photons that would lead to him winning the 1921 Nobel Prize in physics.


Quantum Light

Since photons are discrete packets of light, they have an associated energy. This is in contrast to the wave model, where light has an intensity and delivers energy through this intensity continuously.

Photon energies solve the photoelectric effect dilemma. Photon energy is related to wavelength, with longer wavelengths corresponding to lower energy. In order for an electron to be emitted by the photoelectric effect, an individual photon needs enough energy for an individual electron to escape the material. Photons with long wavelengths have insufficient energy to eject an electron from a material, and increasing intensity doesn't help. Increased intensity would lead to more photons incident on the material, but more photons won't lead to more emitted electrons if each individual photon still has insufficient energy to eject an electron.

The plot below shows how many electrons would be emitted from a material with a cutoff wavelength of \(\SI{300}{\nano\meter}\) for a fixed photon flux. This means the number of photons hitting the material is constant, but their wavelength can vary. If we idealize the behavior, all photons with a wavelength of \(\SI{300}{\nano\meter}\) or shorter will eject an electron, and all photons with a longer wavelength won't eject an electron. In reality, the transition is more gradual.

Quantum Light

The energy of a photon is given by the Planck relation, \[E = hf,\] where \(h\) is the Planck constant (\(\SI{4.136e-15}{\electronvolt\second}\)), and \(f\) is the frequency of the photon. In the previous quiz we learned that the product of wavelength and frequency is the speed of light, so \(c = \lambda f\), and an equivalent expression of the Planck relation in terms of wavelength is given by \[E = \frac{h c}{\lambda}\]

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 \(\SI{1.6e-19}{\joule}.\)

Suppose we are investigating the photoelectric effect in a material whose electrons need \(\SI{2.5}{\electronvolt}\) in order to be emitted. What is the approximate wavelength of a photon with an energy of \(\SI{2.5}{\electronvolt}?\)

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 \(\SI{1240}{\electronvolt\nano\meter}.\)


Quantum Light

In the material we are investigating, electrons require \(\SI{2.5}{\electronvolt}\) of energy in order to be emitted. If a photon with at least \(\SI{2.5}{\electronvolt}\) of energy is absorbed by an electron in the material, that electron will be emitted.

If a photon with exactly \(\SI{2.5}{\electronvolt}\) of energy has the wavelength \(\lambda^*\), what photon wavelengths will lead to the photoelectric effect (i.e., emitted electrons) in our material?


Quantum Light

The main difference between the wave picture and photon picture of light is how energy is delivered. In the wave picture, energy is delivered continuously, while in the photon picture, energy is delivered in discrete packets.

This is analogous to spraying water with a hose versus throwing water balloons. When you spray a hose, you're continuously delivering water, similar to how in the wave picture light is continuously delivering energy. When you throw water balloons, the water comes in discrete packets, and you can increase the amount of water you deliver by throwing more water balloons OR by throwing bigger water balloons. This is more like the photon picture, since with photons we can deliver more energy by increasing the number of photons or increasing the energy per photon.

In many cases, it doesn't matter whether you think of light in the wave picture or the photon picture. Like our water analogy, in most cases it doesn't matter whether you're spraying a hose or throwing water balloons: you'll still get your target wet! However, there are a few cases where the difference between light as a wave and light as a photon matter, and the photoelectric effect is one of them. Tying this back into our hose vs. water balloons analogy, the photoelectric effect is like a game where you're trying to fill a leaky bottle past a certain line. With a hose, the water you add is always leaking out, so you can't reach the line. With a water balloon, you deliver all the water at once, so right after the balloon pops, the water will reach above the line. If the balloons you're throwing are too small, however, you'll never be able to fill the bottle past the line.


Quantum Light

While photons share properties with the wave model of light like direction, wavelength, and frequency, individual photons don’t have an intensity. Instead, they have an energy, but we can recover the idea of intensity in a steady stream of electrons. The intensity of a photon stream upon a surface is equal to the energy of each photon times the number of photons that hit the surface per unit time (called the photon flux, \(\Phi\)): \[I = \frac{\textrm{energy}}{\textrm{area}\times\textrm{time}} = \frac{\textrm{photons}}{\textrm{area}\times\textrm{time}} \times \frac{\textrm{energy}}{\textrm{photon}} = \Phi E\]

Thus, in the photon picture, intensity depends on photon energy and how many photons are present. Approximately how many photons will hit a \(\SI{1}{\meter\squared}\) collector exposed to sunlight for \(\SI{1}{\second}\)?

Assume that the total intensity of sunlight is \(\SI[per-mode=symbol]{1000}{\watt\per\meter\squared}\), and that the photons in sunlight have an average energy of \(\SI{1}{\electronvolt}\), equivalent to \(\SI{1.6e-19}{\joule}\).


Quantum Light

An individual photon carries a very small amount of energy, and is very difficult to detect by itself. This isn't a problem, since we're usually concerned with measuring the overall energy delivered by radiation, not measuring a single photon. Your skin absorbs trillions of photons after being in the sun for a fraction of a second, so it would take much more than one extra or missing photon for you to notice a change in energy.

Since individual photons carry so little energy, energy being delivered by photons can often be approximated as a continuous delivery of energy, and in these cases the distinction between the wave and photon picture is not important. In most solar-thermal systems, for example, it doesn't matter whether we think of light as being delivered by photons or waves. In some cases, however, the details of the photon are important. Understanding how PV cells work, as well as the limit of PV performance, depends on the photon picture of light. As we learn about solar energy moving forward, it will often be important to keep the photon picture in mind.


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