# High School Chemistry/The Dual Nature of Light

Developing a theory to explain the nature of light was a difficult task. An acceptable theory in science is required to explain all the observations made on a particular phenomenon. Light, appears to have two different sets of behaviors under different circumstances. As you will see, sometimes light behaves like wave-form energy and sometimes it behaves like an extremely tiny particle. It required the very best scientific minds from all over the world to put together a theory to deal with the nature of light.

## Lesson Objectives

* Explain the double-slit experiment and the photoelectric effect.
* Explain why light is both a particle and a wave.
* Use and understand the formula relating a light's velocity, frequency, and wavelength, c = fλ
* Use and understand the formula relating a light's frequency and energy, E = hf

## The Difficulties of Defining Light Solved by Einstein and Planck

Why was light so hard to understand? Part of the problem was that when scientists performed different experiments on light, they got conflicting results. Some experiments suggested that light was like a wave, while others suggested that light was like a particle! Let's take a look at what's meant by "like a wave" and "like a particle".
Since you're probably familiar with water waves, we'll use water to explain wave behavior. Whenever a water wave is forced through a small opening, such as the space between the two rocks in Figure 5.4, it spreads out into a circular shape through a process known as diffraction. If several of these circular waves run into each other, they can interfere with one another and produce interesting patterns in the water.
Figure 5.5 shows some of these patterns. Look carefully at the red line which defines a cross-section of the pattern (portion (a)). That same cross-section is blown up in portion (b), where you can clearly see how it is composed of alternating "peaks" and "troughs". The peaks are actually extra high points in the waves (hills), while the troughs are extra low points (valleys).
Imagine how surprised scientists were when they shone light through two narrow slits in a solid plate and saw a similar pattern of peaks (bright spots) and troughs (dark spots) on the wall opposite the plate. Obviously, this proved that light had some very wave-like properties. In fact, by assuming that light was a wave and that it diffracted through the two narrow slits in the plate, just like water waves diffract when they pass between rocks, scientists were even able to predict where the bright spots would occur! Figure 5.6 shows how the results of the "double-slit" experiment could be understood in terms of light waves.
What was also obvious from the double-slit experiment was that light could not be understood as a particle. Imagine rolling a particle (like a marble) through a small opening. Would you expect it to diffract? Of course not! You’d expect the marble to roll in a straight line from the opening to the opposite wall, as shown in Figure 5.7. Light traveling as a particle, then, should make a single bright spot directly across from each slit opening. Since that wasn't what scientists observed, they knew that light couldn't be composed of tiny particles.
The wave theory of light seemed to work, at least for the double-slit experiment. Remember, though, that according to the scientific method, a theory should be tested with further experiments to make sure that it's accurate and complete. Unfortunately, the next experiment that scientists performed suggested that light was not a wave, but was, instead, a stream of particles! By shining light on a flat strip of metal, scientists found that they could knock electrons off of the metal surface. They called this phenomenon the photoelectric effect, and they called electrons that were bumped off photoelectrons. Why did the photoelectric effect prove that light wasn't a wave? The problem was that the energy of the photoelectrons struck by a beam of light didn't depend on how bright the light was, but instead depended on the light's color. To see why this was so important, we need to talk a little bit more about waves and light waves in particular.
Suppose you were out boating offshore a windy Florida beach on the Atlantic coast, renowned for its recurrent storms and cyclones. It would be very exciting to find yourself riding a large wave! But if, instead, you rowed to a quieter place and dropped a pebble into the sea to watch the ripples crawl under your boat, you would barely feel a thing. So what's the difference between these two waves? You may say the height of the wave, which is the result of its amplitude - for any wave, amplitude basically tells you how far the wave swings "up and down". Therefore, no matter how quick the tiny ripples come at you, you can barely expect to be rocked by them!
In order to translate this example to that of light waves striking electrons, we will have to make a few analogies. The brighter the light, the greater its amplitude as a wave (the thing that's "swinging up and down" is the electromagnetic field, as we saw earlier; this can be demonstrated experimentally). When an electron gets "rocked" by the light wave, it gains energy and enters an excited state, and we call it a photoelectron. This can also be detected in the laboratory.
Let's try it out and shine some yellow light at the metal. Here comes the twist: no matter how bright or how dim the light is, each individual electron gains the same amount of energy. But when we increase the brightness, where does the extra energy go? It turns out that more light simply energizes more electrons. If light were a wave, it would rock the electrons as hard as it struck them. Instead, it seems to be precisely distributing equal packets of energy to each electron.
Even more interestingly, if we tune up the frequency and turn the yellow light into green or blue light, each electron now receives more energy! Therefore:
* The more intense the light, the more electrons it strikes • The higher the frequency, the more energy received by each electron
So, for the same brightness, the metal is actually absorbing different amounts of energy depending on the color of the light, which can be equated to its frequency/wavelength! (Figure 5.8). Red light is so weak that it fails to produce any photoelectrons, while yellow light produces energized photoelectrons and blue light produces even stronger photoelectrons.
This made the photoelectric effect very puzzling to scientists because they knew a wave with a high amplitude couldn't just "choose" to give less of its energy to each electron or vice versa. Instead, light had to be made up of (or quantized into) numerous particles of the tiniest possible amplitude, or "quanta", known to us today as photons. Each photon then has a frequency, which determines how much energy it can deliver to its target.
What is frequency? Frequency can be a difficult concept to understand, but it's really just a measure of how many times an event occurs in a given amount of time. In the case of waves, it's the number of waves that pass by a specific reference point per unit time. Figure 5.9 shows two different types of waves, one red and one blue. Notice how, in a single second (one full turn of the clock hand), 4 red waves pass by the dotted black line while 16 blue waves pass by the same reference point. We say that the blue waves have a higher frequency than the red waves. The SI unit used to measure frequency is the Hertz (Hz). One hertz is equivalent to one event (or one full wave passing by) per second.
How does the frequency of the light affect the length of the light waves? Take a close look at Figure 5.9 again. What do you notice about the lengths of the blue and red waves? Obviously, the blue waves (higher frequency) have a shorter wavelength, while the red waves (lower frequency) have a longer wavelength. This has to be true, provided that the waves are traveling at the same speed. You can tell that the red and blue waves are traveling at the same speed, because their leading edges (marked by a red dot and a blue dot respectively) keep pace with each other. All light waves travel at the same speed.
The explanation of the photoelectric effect began with a man named Max Planck. Max Planck wasn't actually studying the photoelectric effect himself. Instead, he was studying something known as black-body radiation. Black-body radiation is the light produced by a black object when you heat it up (think, for example, of a stove element that glows red when you turn it on). Like the photoelectric effect, scientists couldn't explain black-body radiation using the wave theory of light either. Max Planck, however, realized that black-body radiation could be understood by treating light like a stream of tiny energy packets, or photons as we saw above.
Albert Einstein applied the theory of quantized light to the photoelectric effect and found that the energy of the photons, or quanta of light, did depend on the light's frequency. In other words, all of a sudden Einstein could explain why the frequency of a beam of light and the energy of a beam of light were related. That made it a lot easier to understand why the number of photoelectrons produced by the light depended on the light's color (frequency). The only assumption that Einstein needed to make was that light was composed of particles.
Wait! Sure the particle theory of light explained black-body radiation and the photoelectric effect, but what about the double-slit experiment? Didn't that require that light behave like a wave? Either the double-slit experiment was wrong, or else the photoelectric effect and black-body radiation were wrong. Surely, light had to be either a wave or a particle. Surely, it couldn't be both. Or could it? Albert Einstein suggested that maybe light wasn't exactly a wave or a particle. Maybe light was both. Albert Einstein's theory is known as the wave-particle duality of light, and is now fully accepted by modern scientists.

## Light Travels as a Wave

You just learned that light can act like a particle or a wave, depending entirely on the situation. Did you ever play with Transformers when you were younger? If you did, maybe you can answer the following question: are Transformers vehicles (cars and aircraft) or are they robots? It's kind of a stupid question, isn't it? Obviously Transformers are both vehicles and robots. It's the same with light – light is both a wave and a particle.
Even though Transformers are both vehicles and robots, when they want to get from one place to another quickly, they usually assume their vehicle form and use all of their vehicle properties (like wheels or airplane wings). Therefore, if you were trying to explain how a Transformer sped off in search of an enemy, you’d probably describe the Transformer in terms of its car or aircraft properties.
Just as it's easiest to talk about Transformers traveling as vehicles, it's easiest to talk about light traveling as a wave. When light moves from one place to another, it uses its wave properties. That's why light passing through a thin slit will diffract; in the process of traveling through the slit, the light behaves like a wave. Keeping in mind that light travels as a wave, let's discuss some of the properties of its wave-like motion.
First, and most importantly, all light waves travel, in a vacuum, at a speed of 299,792,458 m/s (or approximately 3.00×10$^8$ m/s). Imagine a tiny ant trying to surf by riding on top of a light wave (Figure 5.10). Provided the ant could balance on the wave, it would move through space at 3.00×10$^8$ m/s.
To put that number into perspective, when you go surfing at the beach, the waves you catch are moving at about 9 m/s. Unlike light waves, though, which all travel at exactly the same speed, ocean waves travel at different speeds depending on the depth of the ocean, the temperature, and even the wind!
Previously, you learned that light can have different frequencies and different wavelengths. You also learned that because light always travels at the same speed 3.00×10$^8$ m/s, light waves with higher frequencies must have smaller wavelengths, while light waves with lower frequencies must have longer wavelengths. Scientists state this relationship mathematically using the formula
| $c\,\!$ | $=\,\!$ | $f\,\!$ | $\times \,\!$ | $\lambda \,\!$ |
| ${\text{speed of light}}\ (3.00\times 10^{8}\,{\text{m/s}})\,\!$ | | ${\text{frequency}}\ ({\text{Hz or s}}^{-1})\,\!$ | | ${\text{wavelength}}\ ({\text{m}})\,\!$ |
where c is the speed of light, 3.00×10$^8$ m/s, f is the frequency and λ is the wavelength. Remember that the unit we use to measure frequency is the Hertz (Hz), where 1 hertz (Hz) is equal to 1 per second, s$^{−1}$. Wavelength, since it is a distance, should be measured in the SI unit of distance, which is the meter (m). Let's see how the formula can be used to calculate the frequency or the wavelength of light.
Example 1
What is the frequency of a purple colored light, if the purple light's wavelength is 4.45×10$^{−7}$ m?
Solution:
speed of light, c = 3.00×10$^8$ m/s — You always know the speed of light, even if the question doesn't give it to you.
wavelength, λ = 4.45×10$^{−7}$ m
$c\,\!$$=\,\!$$f\times \lambda \,\!$
$(3.00\times 10^{8}\,{\text{m/s}})\,\!$$=\,\!$$f\times (4.45\times 10^{-7}\,{\text{m}})\,\!$
To solve for frequency, f, divide both sides of the equation by 4.45×10$^{−7}$ m.
${\frac {3.00\times 10^{8}\,{\text{m/s}}}{4.45\times 10^{-7}\,{\text{m}}}}$$=\,\!$$f\times {\frac {4.45\times 10^{-7}\,{\text{m}}}{4.45\times 10^{-7}\,{\text{m}}}}$
$6.74\times 10^{14}\,{\text{s}}^{-1}\,\!$$=\,\!$$f\times (1)\,\!$
$f\,\!$$=\,\!$$6.74\times 10^{14}\,{\text{Hz}}\,\!$
The frequency of the purple colored light is 6.74×10$^{14}$ Hz. :
Example 2
What is the frequency of a red colored light, if the red light's wavelength is 650 nm?
Solution:
speed of light, c = 3.00×10$^8$ m/s
$\lambda \,\!$$=\,\!$$650\,{\text{nm}}\,\!$
$\lambda \,\!$$=\,\!$$(650\,{\text{nm}})(1\times 10^{-9}\,{\text{m/nm}})=6.50\times 10^{-7}\,{\text{m}}\,\!$
$c\,\!$$=\,\!$$f\times \lambda \,\!$
$(3.00\times 10^{8}\,{\text{m/s}})\,\!$$=\,\!$$f\times (6.50\times 10^{-7}\,{\text{m}})\,\!$
To solve for frequency, f, divide both sides of the equation by 6.50×10$^{−7}$ m.
${\frac {3.00\times 10^{8}\,{\text{m/s}}}{6.50\times 10^{-7}\,{\text{m}}}}$$=\,\!$$f\times {\frac {6.50\times 10^{-7}\,{\text{m}}}{6.50\times 10^{-7}\,{\text{m}}}}$
$4.61\times 10^{14}\,{\text{s}}^{-1}\,\!$$=\,\!$$f\times (1)\,\!$
$f\,\!$$=\,\!$$4.61\times 10^{14}\,{\text{Hz}}\,\!$
The frequency of the red colored light is 4.61×10$^{14}$ Hz. :
Notice that the wavelength in Example 1, 4.45×10$^{−7}$ m, is smaller than the wavelength in Example 2, 6.50×10$^{−7}$ m, while the frequency in Example 1, 6.74×1014 Hz is bigger than the frequency in Example 2, 4.61×10$^{14}$ Hz. Just as you'd expect, a small wavelength corresponds to a big frequency, while a big wavelength corresponds to a small frequency. (If you're still not comfortable with that idea, take another look at Figure 8 and convince yourself of why this must be so, provided the waves travel at the same speed.) Let's take a look at one final example, where you have to solve for the wavelength instead of the frequency.
Example 3
Scientists have measured the frequency of a particular light wave at 6.10×10$^{14}$ Hz. What is the wavelength of the light wave?
Solution:
speed of light, c = 3.00×10$^8$ m/s
frequency, f = 6.10×10$^{14}$ Hz = 6.10×10$^{14}$ s$^{−1}$ (To do dimensional analysis, it is easiest to change hertz to per second)
$c\,\!$$=\,\!$$f\times \lambda \,\!$
$(3.00\times 10^{8}\,{\text{m/s}})\,\!$$=\,\!$$(6.10\times 10^{14}\,{\text{s}}^{-1})\times \lambda \,\!$
To solve for wavelength, λ, divide both sides of the equation by 6.10×10$^{14}$ s$^{−1}$.
${\frac {3.00\times 10^{8}\,{\text{m/s}}}{6.10\times 10^{14}\,{\text{s}}^{-1}}}$$=\,\!$${\frac {6.10\times 10^{14}\,{\text{s}}^{-1}}{6.10\times 10^{14}\,{\text{s}}^{-1}}}\times \lambda$
$4.92\times 10^{-7}\,{\text{m}}\,\!$$=\,\!$$(1)\times \lambda \,\!$
$\lambda \,\!$$=\,\!$$4.92\times 10^{-7}\,{\text{m}}\,\!$
The wavelength of the light is 4.92×10$^{−7}$ m (or 492 nm, if you do the conversion). :

## Light Consists of Energy Packets Called Photons

We have already seen how the photoelectric effect proved that light wasn't completely wavelike, but rather, had particle-like properties too. Let's return to our comparison between Transformers and light. Transformers travel as vehicles; however, when Transformers battle each other, they fight as robots, not as cars and planes. The situation with light is similar. Light may travel as a wave, but as soon as it strikes an object and transfers its energy to that object, the light behaves as if it's made up of tiny energy packets, or particles, called photons.
Remember, the energy of a wave depends only on the wave's amplitude, but not on the wave's frequency. The energy of a photon, or a light "particle", however, does depend on frequency. The relationship between a photon’s energy and a photon's frequency is described mathematically by the formula
| $E\,\!$ | $=\,\!$ | $h\,\!$ | $\times \,\!$ | $f\,\!$ |
| ${\text{energy}}\ ({\text{J}})\,\!$ | | ${\text{Planck}}'{\text{s constant}}\ (h=6.63\times 10^{-34}\,{\text{J}}\cdot {\text{s}})\,\!$ | | ${\text{frequency}}\ ({\text{Hz or s}}^{-1})\,\!$ |
where E is the energy of the photon, h is Planck's constant (which always has the value h = 6.63×10$^{−34}$ J · s, and f is the frequency of the light. The SI unit for energy is the Joule (J); the SI unit for frequency is the Hertz (or per second, s$^{−1}$); the SI unit for Planck's constant is the Joule-second (J · s). Although this equation came from complex mathematical models of black-body radiation, its meaning should be clear – the larger the frequency of the light beam, the more energy in each photon of light.
* Example 4
* What is the energy of a photon in a stream of light with frequency 4.25×10$^{14}$ Hz?
* Solution:
* Planck's constant, h = 6.63×10$^{−34}$ J · s
* frequency, f = 4.25×10$^{14}$ Hz = 4.25×10$^{14}$ s$^{−1}$ (To do dimensional analysis, it is easiest to change hertz to per second)
* $E=h\times f\,\!$
* $E=(6.63\times 10^{-34}\,{\text{J}}\cdot {\text{s}})\times (4.25\times 10^{-14}\,{\text{s}}^{-1})\,\!$
* $E=2.82\times 10^{-19}\,{\text{J}}\,\!$
* The energy of a photon of light with frequency 4.25×10$^{14}$ s$^{−1}$ is 2.82×10$^{−19}$ J.
Example 5
What is the frequency (in Hz) of a beam of light if each photon in the beam has energy 4.44×10$^{−22}$ J?
Solution:
Planck's constant, h = 6.63×10$^{−34}$ J · s
energy, E = 4.44×10$^{−22}$ J
$E=\,\!$$h\times f\,\!$
$4.44\times 10^{-22}\,{\text{J}}=\,\!$$(6.63\times 10^{-34}\,{\text{J}}\cdot {\text{s}})\times f\,\!$
To solve for frequency (f), divide both sides of the equation by 6.63×10$^{−34}$ J · s
${\frac {4.44\times 10^{-22}\,{\text{J}}}{6.63\times 10^{-34}\,{\text{J}}\cdot {\text{s}}}}=$
${\frac {6.63\times 10^{-34}\,{\text{J}}\cdot {\text{s}}}{6.63\times 10^{-34}\,{\text{J}}\cdot {\text{s}}}}\times f$
$6.70\times 10^{11}\,{\text{s}}^{-1}=\,\!$$(1)\times f\,\!$
$f=\,\!$$6.70\times 10^{11}\,{\text{Hz}}\,\!$
The frequency of the light is 6.70×10$^{11}$ Hz. :

## The Electromagnetic Spectrum

When scientists speak of light in terms of its wave-like properties, they are often interested in the frequency and wavelength of the light. One way that we, as humans, can distinguish between light beams of different frequencies (and thus different wavelengths) is to use their colors. Light that is reddish colored has large wavelengths and small frequencies, while light that is bluish colored has small wavelengths and large frequencies. Humans can't, however, see all types of light. We can only see visible light. In fact, if the light's wavelength gets too small, the light becomes invisible to our eyes. We call this light ultraviolet (UV) radiation. Similarly, if the light's wavelength gets too large, the light also becomes invisible to our eyes. We call this light infrared (IR) radiation.
Believe it or not, there are types of light with wavelengths even shorter than those of ultraviolet radiation. We call these types of light X-rays and gamma rays. We can use X-rays to create pictures of our bones, and gamma rays to kill bacteria in our food, but our eyes can't see either (you can see the picture that an X-ray makes, but you can’t actually see the X-ray itself). On the other side of the spectrum, light with wavelengths even longer than those of infrared radiation are called microwaves and radio waves. We can use microwaves to heat our food, and radio waves to broadcast music, but again, our eyes can't see either.
Scientists summarize all the possible types of light in what's known as the electromagnetic spectrum. Figure 5.11 shows a typical electromagnetic spectrum. As you can see, it's really just a list of all the possible types of light in order of increasing wavelength. Notice how visible light is right in the middle of the electromagnetic spectrum. Since light with a large wavelength has a small frequency and light with a small wavelength has a large frequency, arranging light in order of "increasing wavelength", is the same as arranging light in order of "decreasing frequency". This should be obvious from Figure 5.11 where, as you can see, wavelength increases to the right (decreases to the left), while frequency increases to the left (decreases to the right).
Unlike wavelength and frequency, which are typically shown on the electromagnetic spectrum, energy is rarely included. You should, however, be able to predict how the energy of the light photons changes along the electromagnetic spectrum. Light with large frequencies contains photons with large energies, while light with small frequencies contains photons with small energies. Therefore energy, like frequency, increases to the left (decreases to the right).

## Lesson Summary

* When waves pass through narrow openings, they spread out into a circular shape through a process known as diffraction.
* When circular waves interact, they produce predictable patterns of peaks and troughs.
* When light is passed through two narrow slits, the light appears to interact in a manner similar to two circular waves spreading out from the slits. This suggests that light diffracts into circular waves when it passes through the slits and that these circular waves interact with each other. As a result, many scientists believed that light was wave-like.
* Shining light on a flat strip of metal knocks electrons off of the metal surface through what is known as the photoelectric effect.
* The number of photoelectrons produced by a beam of light depends on the color (wavelength) of the light but not on the brightness (amplitude) of the light.
* Since the energy of a wave should depend on the amplitude of the wave, scientists couldn't understand why a brighter light didn't knock more photoelectrons off of the metal. This led them to question of whether light was truly wave-like.
* Together Max Planck and Albert Einstein explained the photoelectric effect by assuming that light was actually a stream of little particles, or packets of energy known as photons or quanta.
* Scientists now believe that light is both a wave and a particle – a property which they term the wave-particle duality.
* Light travels as a wave. The speed of a light wave is always c = 3.00×10$^8$ m/s. The frequency, f, and wavelength, λ, of a light wave are related by the formula c = fλ.
* Light gives up its energy as a particle or photon. The energy (E) of a photon of light is related to the frequency, f, of the light according to the formula E = hf.
* The relationship between the frequency, the wavelength, and the energy of light is summarized in what's known as the electromagnetic spectrum. The electromagnetic spectrum is a list of light waves in order of increasing wavelength, decreasing frequency, and decreasing energy.

## Review Questions

* Decide whether each of the following statements is true or false:
    (a) Light always behaves like a wave.
    (b) Light always behaves like a particle.
    (c) Light travels like a particle and gives up its energy like a wave.
    (d) Light travels like a wave and gives up its energy like a particle.
* Which of the following experiments suggested that light was a wave, and which suggested that light was a particle?
    (a) the double-slit experiment
    (b) the photoelectric effect
    (c) black-body radiation
* Fill in each of the following blanks.
    (a) The brightness of a beam of light is determined by the ___________ of the light wave.
    (b) The color of a beam of light is determined by the _____________ of the light wave.
* What is the name of the quantity depicted by each of the arrows in the diagram below (Figure 5.12)?Figure 5.12
* Consider light with a frequency of 4.4×10$^{14}$ Hz. What is the wavelength of this light?
* What is the frequency of light with a wavelength of 3.4×10$^{−9}$ m?
* What is the frequency of light with a wavelength of 575 nm?
* What is the energy of a photon in a beam of light with a frequency of 5.66×10$^8$ Hz?

## Vocabulary
### black-body radiation

    Light produced by a black object when the object is heated.

### diffraction

    The tendency of a wave to spread out in a circular shape when passed through a small opening.

### double-slit experiment

    When light is passed through two narrowly separated openings (slits), the light produces a resulting pattern of peaks and troughs that suggests that light behaves like a wave.

### electromagnetic spectrum

    A list of all the possible types of light in order of decreasing frequency, or increasing wavelength, or decreasing energy. The electromagnetic spectrum includes gamma rays, X-rays, UV rays, visible light, IR radiation, microwaves and radio waves.

### photoelectric effect

    The process whereby light shone on a metal surface knocks electrons (called photoelectrons) off of the surface of the metal.

### photon or quanta of light

    A tiny particle-like packet of energy.

### wave-particle duality of light

    Einstein's theory, which concluded that light exhibits both particle and wave properties.
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This material was adapted from the original CK-12 book that can be found here. This work is licensed under the Creative Commons Attribution-Share Alike 3.0 United States License