The previous learning objective, we talked about how electromagnetic radiation, or light, can behave as a wave but also as a particle, it has a dual nature. In this learning objective, we're also going to see dual nature that is applying to electrons. We're going to see why this is a, it was proposed as an explanation for electronic structure. We are going to see the evidence that supports the fact that electrons have this dual nature, and we're going to learn how to do calculations of wavelengths of these particles. Now we know that the electrons, according to Bohr's model, could only occupy certain fixed locations around the nucleus. The question was why? Why can an electron be traveling in a region where the n equals 1, and in a region where n equals 2, but it cannot travel anywhere in between those two? Well de Broglie was the one who provided a solution, by proposing this dual nature. By proposing that electrons not only have the particle nature, they have mass, we know they are particles, but also behave as waves. Now these two diagrams show an allowed state for the electron and a non-allowed state for the electron, when you're thinking about them as a wave. If the red represents the distance from the nucleus that the electron can exist at, so that's an n equals 1 or a n equals 2 or a n equals 3 region. The electron will have a wavelength such that it is an integral number of wavelengths around that area. So that when the electron travels as a wave, the wave will line up on itself, and it can exist there. But if you moved a little bit different distance away, and you were traveling at that same wavelength, then at this distance, the down and the up trough, or the down trough and the up peak of the wave would make it so that the wave could not exist there. So this is why de Broglie proposed that electrons have this wave nature that, that would help explain why an electron can be here, but it cannot be here. So proposing that it has a dual nature is one thing. Seeing evidence that they have this wave nature is something else. Now if an electron were a particle, and you were to pass the particles through the slits, what you would see is those particles would travel straight from the slit and strike the surface of this plate that is detecting electrons. And it would travel, and it would do straight to this slit through there, and you would see it lighting up only in two spots that correspond to the slits. But this is not what happens. If you were to pass these electrons through those slits instead you would see a defraction pattern much like light does. When light travels through the slits you have a, an alternating bright and dark patches on the photo-electric paper. Well, these are detecting electrons instead of just two beams. What we see is a sequence of dark and light and dark and light. What that means is those electrons have a wave nature and those waves are being bent by the slits. And as they interact, and they have a peak and a peak, you'll see a bright spot and a trough and a peak. You'll see a dark spot. And we see this very definite diffraction pattern occurring of constructive and deconstructive interference of those waves. Now, if it behaves as a wave, it would have a wave length. And they were actually able to, with this equation, get a relationship between the wavelength of the particle and the mass and the velocity of the particle as it travels. So that of course is wavelength. This is mass. And the mass is going to need to be put in, in kilograms. This is velocity. And velocity was in meters per second. And we've seen Planck's constant before, okay? That's Planck's constant. And Planck's constant, you will be provided with that when you work problems, but it's 6.626 times 10 to the minus 34. And it's joules times seconds. But, you need to watch your units when you work this. Now I've given what they need to be in, but why do they need to be in this? It is because of what a joule is. A joule if you break it down into its base units, is a kilogram, meter squared per second squared. And then that would be time second. So this would be the units of Plank's constant. If this is in kilograms, it's going to be canceling with this kilograms nicely. And if these are meters per second, they're going to cancel with the meters per second. So we're going to be left with one of these meters left over, and that'll give off our wavelength out, in meters. So let's work a problem. Now this one is a silly problem, dealing with a ping-pong ball. But any particle can have this wave nature. We just don't really observe that wave nature with something as large as a tin, a ping-pong ball. And let's see why that is. Wavelength would be equal to Planck's constant, 6.626 times 10 to the minus 34. Joules times second and joule is a kilogram, meters squared, per second squared and then times seconds. That's the Planck's Constant units. Divided by mass, now it's a 2.5 gram ping-pong ball, so 2.5 grams. We don't want it in grams because of this kilogram here. So we'll convert it to kilograms, a kilogram is 1,000 grams, so that'll give me my mass in kilograms, and velocity. It's 15.6 meters per second. So now let's have our units cancelling here, the grams are certainly canceling. The kilograms are cancelling. One of these seconds is cancelling one of these seconds, and this second is cancelling this one. And this meter is cancelling with one of those, so we're left with meters. Now it's asking for it in nanometers, we'll worry about that later. And when we divide all of these values out we get 1.7 times 10 to the negative 37. Sorry, times 10 to the negative 32 meters. 1.7 times 10 to the negative 32 meters. This is tiny, tiny, tiny, too tiny to even observe kind of wavelength. Let's go ahead and convert it to nanometers cause that is what wavelength is usually reported at, especially when we're dealing with electromagnetic radiation. So a nanometer is 10 to the minus 9 meters and that's going to give me 1.7 times 10 to the minus 23 nanometers. So any particle, if you know its mass and its speed, can have a calculated wavelength. But it's only really the wavelength of the very, very tiny, tiny particles that give you if you have a mass of very, very small, very small mass. Then you're going to have a wavelength that is actually measurable and observable. So this is the end of learning objective five. We see that electrons not only have the particle nature that we would expect. But also have a wave nature, and you can calculate the wavelength of any particle if you know the mass and the velocity.
