# Video: Matter Waves

In this video, we will learn how to calculate the de Broglie wavelength of massive particles that have a given momentum or velocity.

11:40

### Video Transcript

In this video, we’re talking about matter waves. Matter waves speak to the surprising wavelike nature of objects with mass, like electrons, protons, atoms, and molecules. To see where this idea came from, it’s helpful to go back to the history behind it. In the late 19th and early 20th centuries, scientists were very interested in studying the nature of light. This was partly due to the fact that light, which had been thought of as a wave for hundreds of years, was starting to show signs of behaving like a particle as well. The wave nature of light was supported by observations that light was able to diffract as well as interfere with itself, which are wave properties.

But then, the hypothesis that light also acts as a particle help to explain black body radiation as well as the photoelectric effect. This was confusing, and it came to be called the wave particle duality of light. With all this going on in the study of light, a French physicist named Louis de Broglie came up with an idea of how this might apply to matter. de Broglie claimed that objects with mass behave like particles, but they also behave like waves. At the time, people were used to thinking of matter — like protons, neutrons, and electrons — as particles. But the notion that matter was also a wave seemed very strange. What de Broglie meant when he said that matter is a wave is that it demonstrates wavelike properties, in particular, that it has a wavelength.

Now, if we go back to talking about light, and in particular light as a particle, it was known that these particles, called photons, each possessed an amount of energy. That energy, according to an equation developed by Albert Einstein, was equal to the momentum 𝑝 of the photon multiplied by the speed of light in vacuum 𝑐. And thinking along separate lines, another physicist, named Max Planck, had come up with a separate equation for photon energy. By studying black-body radiation, Planck said that the energy of a photon is equal to a constant, what came to be called Planck’s constant, multiplied by the frequency of the photon. Both of these energy equations, Planck’s and Einstein’s, apply only to photons and nothing else. As a side note, we can see in Planck’s equation the wave as well as the particle nature of light. This relationship applies to particles of light, photons, but it depends on the frequency of this radiation, which is a wavelike property.

Anyway, combining the equations of Einstein and Planck, we can take this resulting equation, divide both sides of it by the momentum of the photon times its frequency. This causes the momentum 𝑝 to cancel on the left and the frequency 𝑓 to cancel on the right. And we find that 𝑐 over 𝑓, the speed of a photon divided by its frequency, is equal to Planck’s constant ℎ divided by the momentum of the photon. Then, we can recall that for a wave traveling at the speed of light 𝑐, that speed is equal to the frequency of the wave times its wavelength. Or if we divide both sides of the equation by the frequency, we find that wavelength 𝜆 is equal to 𝑐 over 𝑓. Comparing this equation with the equation we generated by combining Einstein’s and Planck’s expressions for photon energy, we can see that the wavelength 𝜆 can be substituted in for 𝑐 over 𝑓. And so wavelength 𝜆 is equal to Planck’s constant ℎ over 𝑝, momentum.

Now, recall that this equation applies specifically to photons. It’s based on energy equations for that particle. But de Broglie, in his work, made the radical assumption that this same equation could apply to particles with mass. Specifically, he claimed it applied to electrons, that is, that electrons have some wavelength. And this was how the idea that matter is both a wave as well as a particle was introduced. Now, it was important that this idea, that particles with mass also have a wavelength, could be tested. The best way for doing that, it was decided, was to see if these particles, electrons, would diffract. If they did, they would be demonstrating wavelike behavior. Running an experiment to test this idea, though, was not as easy as it might seem. That was because these wavelengths that de Broglie predicted electrons would have were very, very small.

Based on the known value of Planck’s constant, as well as the known value of the mass of an electron and giving an average value for its velocity, the wavelength of the electron was predicted to be on the order of 10 to the negative 10 meters. Now, wave diffraction is best observed by passing a wave through an opening about as wide as the wavelength. 10 to the negative 10 meters is roughly the diameter of an atom, which means that, experimentally, in order for the diffraction of these electrons to be demonstrated, an apparatus with gaps separated by about this width, the width of an atom, would be necessary. This was no small feat, but eventually it was accomplished. And when it was, in support of de Broglie’s hypothesis, a diffraction pattern of the electrons was revealed.

So people began to think that indeed matter, not just electrons, but even larger objects, does behave like a wave. This may raise the question though, why don’t we experience that in our everyday life? For example, why don’t we see things like, say, tennis balls diffracting? The answer goes back to this equation for what’s called the de Broglie wavelength of a particle. Knowing that momentum 𝑝 is equal to an object’s mass times its velocity, if we plug in what we could consider standard values for the mass and velocity of a tennis ball. Let’s say that our ball’s mass is 60 grams, or 0.60 kilograms, and that it’s moving at 10 meters per second. Then if we divide this product into Planck’s constant, which is a very small number, 6.63 times 10 to the negative 34th joule-seconds, we get a result for the de Broglie wavelength of a tennis ball, which is on the order of 10 to the negative 33 meters.

That is, to see a tennis ball diffract, and thereby behave like a wave and provide evidence of its de Broglie wavelength, we would need to use an opening or a slit or a gap of about this distance wide. This is many orders of magnitude smaller than even the radius of an atomic nucleus. In our everyday life then, we simply wouldn’t observe this diffracted behavior. Nonetheless, researchers have worked to try to demonstrate diffraction of increasingly larger objects, electrons to protons to atoms and even to molecules. So even though we can’t see it, it’s now understood that matter, like light, behaves like a wave and like a particle. Knowing this, let’s get some practice now working with the de Broglie wavelength of various objects.

What is the de Broglie wavelength of an electron that has a momentum of 4.56 times 10 to the negative 27th kilograms meters per second? Use a value of 6.63 times 10 to the negative 34th joule-seconds for the Planck constant. Give your answer to three significant figures.

All right, so in this example, we have an electron. And an electron is moving along with some speed. Since the electron has mass and is in motion, that means it has some amount of momentum. And we’re told just how much that amount is. So knowing the mass of an electron, we could work back, if we wanted to, to solve for its velocity. But rather, what we want to do is calculate the de Broglie wavelength of this electron. To do that, we can recall that this wavelength, we’ll call it 𝜆, is equal to Planck’s constant ℎ divided by 𝑝, the momentum of the object. Using a value of 6.63 times 10 to the negative 34th joule-seconds for ℎ, we’re able to use that value along with the momentum 𝑝 of our electron to solve for its de Broglie wavelength. So then, here we have our equation for the de Broglie wavelength of the electron. But before we calculate this fraction, let’s consider the units in this expression.

We see that in the numerator, we have units of joule-seconds. A joule is equal to a newton times a meter. And a newton is equal to a kilogram times a meter divided by a second squared, which means we can replace the joule in our units with kilograms meter squared per second squared. When we do this, we now see that we can cancel out some of these units. In the numerator, one factor of seconds will cancel out. And then we can see that both numerator and denominator have a factor of one over seconds. So that cancels as well. And along with this, our units of kilograms cancel one another out, as does one factor of meters.

In the end, the only unit left is a single factor of meters. That’s a good sign because we’re calculating a distance, a wavelength. With each one of the values in our fraction given to three significant figures, we also want to give our answer to that precision. That works out to be 1.45 times 10 to the negative seventh meters. Expressed in units that may be more familiar, this is equal to 145 nanometers. That’s the de Broglie wavelength of this electron.

Let’s look now at a second example exercise.

A proton has a rest mass of 1.67 times 10 to the negative 27th kilograms. At what speed would a proton have to move in order to have a de Broglie wavelength of 8.82 times 10 to the negative ninth meters? Use a value of 6.63 times 10 to the negative 34th joule-seconds for the Planck constant. Give your answer to three significant figures.

Okay, so in this example, we have a proton. And we’re told that this proton has a rest mass, we’ll call it 𝑚 sub zero, of 1.67 times 10 to the negative 27th kilograms. Now, an object’s rest mass is the mass it possesses when it’s being measured by an observer at rest relative to the object. That is, compared to the observer, the object is not in motion. What we want to consider is how fast does a proton with this rest mass have to move in order to have this specific de Broglie wavelength, 8.82 times 10 to the negative ninth meters.

So let’s say that we want to solve for the velocity 𝑣 of this proton. And in support of finding this, we can say that the given de Broglie wavelength of the proton is represented by 𝜆 sub B. At this point, we can recall that the de Broglie wavelength of any object is equal to Planck’s constant ℎ divided by the momentum of that object. In other words, it’s equal to ℎ divided by 𝑚 times 𝑣, the object’s mass and its velocity.

Now, in our case, it’s not the de Broglie wavelength of the proton we want to solve for, but rather its speed. So let’s rearrange this equation to solve for 𝑣. To do this, let’s multiply both sides by 𝑣 divided by 𝜆 B. This cancels out the de Broglie wavelength on the left. And it cancels out the speed 𝑣 on the right. So then, the proton speed we want to solve for is equal to Planck’s constant divided by the mass of the proton times its de Broglie wavelength. Regarding these values, we’re to use a value of 6.63 times 10 to the negative 34th joule-seconds for ℎ. We’re given the de Broglie wavelength 𝜆 B of the proton. And for its mass, we’ll use its rest mass, 𝑚 sub zero, given as 1.67 times 10 to the negative 27th kilograms.

With all these values plugged in, note that all three are given to a precision of three significant figures. Our final answer then will have that same number. Calculating this result, we find an answer of 45.0 meters per second. This is a fairly low speed, certainly nonrelativistic. And it’s the speed the proton would need to have in order to have the given de Broglie wavelength.

Let’s now summarize what we’ve learned about matter waves. Starting out, we saw that matter waves, which are also called de Broglie waves, refer to the wave nature of objects with mass, for example, electrons. We saw further that objects with mass have a wavelength that depends on their mass and speed. This wavelength is known as their de Broglie wavelength. And it’s equal to Planck’s constant divided by the object’s momentum or, equivalently, Planck’s constant divided by the object’s mass times its speed. Lastly, we saw that the de Broglie wavelength of a common-sized object is very small. This helps us understand why the wavelike nature of matter is not readily apparent.