Main content
Current time:0:00Total duration:11:30

Video transcript

- [Voiceover] Uranium consists of two major isotopes. So, we have Uranium-235, and Uranium-238. Let's say you wanted to separate the two isotopes from each other. One way to do it is to use a mass spectrometer. A variation of this was done in World War II, to separate Uranium isotopes. You can separate them because they have different masses. Uranium-238 has more neutrons than Uranium-235. Therefore, it has more mass. Let's see how my mass spectrometer works. The first step is ionization. We're going to assume that we knock off one electron from each one of our isotopes, here. So, we knock off an electron from each of these isotopes. We're going to form a positive charge. We're going to form a positive ion. I'm going to draw in a positive charge, right here. The next step is to accelerate the ions over a potential difference. So, we're going to accelerate the ions over potential difference, deltaV. I should warn you, there's going to be a whole lot of physics in this video. Hopefully, you are comfortable with that. So, we accelerate the ions over a potential difference. That means we're going to get a final velocity of those ions. So, final velocity, v. Let's see how to calculate that final velocity. Well, the amount of work that's done in order to accelerate the ions, is equal to q, which is the charge on the ions, times the potential difference. So, q, on the ion, times deltaV, which is the potential difference. That's equal to the kinetic energy of the ion. That's 1/2 m, v-squared, where m is the mass, and v is the velocity. Let's solve for the velocity of the ion. Let's solve for v, just do some algebra. So, v-squared would be equal to ... This would be two-q, deltaV ... divided by m, so we can just take the square root of both sides, so the velocity is equal to the square root of two, times the charge. This is really deltaV, but I'm just going to write capital, big V, here. Now, potential is different from velocity. So, this V is different from this v. This is divided by the mass. That's the final velocity of the ion. So, when the ion enters this portion of the mass spectrometer, it's moving in this direction, with a velocity, v. This region of the mass spectrometer has a magnetic field in it. Let's say, we have a uniform, magnetic field that's pointing directly out of the page. So, it's coming straight at you. This is meant to represent, like if you're looking at the tip of an arrow. If you point an arrow directly at your eye, you would see the tip of the arrow pointing at you. This is the magnetic field. In physics, we represent magnetic field with B. The moving ion is going to experience a magnetic force, due to the presence of that magnetic field. That magnetic force is equal to q, which is the charge of the ion, once again. v cross B. Let's run through these things. q, is the charge in the ion. v, is the velocity of the ion. B, is the magnetic field. The first thing that we're going to do is figure out the direction of the magnetic force on that ion. To do that, we use one version of the Right-hand rule. So, v, is our first vector, and that's the velocity vector. The velocity vector is in the plane of the page, directed up. So, you can see, I have my fingers pointing in the direction of that velocity vector. It's flat, in the plane of the page, but it's pointing towards the top. Next, we think about our second vector. That's our magnetic-field vector, which is coming directly out at us. It's pointing at us. So, it's straight out of the page. I curl my fingers in the direction of that second vector. If you think about my finger, here ... think about that being the tip of my finger. So, we're curling it in the direction of this vector, which is pointing out of the page. Finally, once we've finished curling our fingers, the thumb ... Your thumb, of your right hand, is forced to point in the direction of the magnetic force, on a positive charge. My thumb has to point to the right, here. So, that's the direction of the magnetic force on this ion. When the ion enters the magnetic field, it's going to experience a magnetic force pointed to the right. The magnetic force is in the plane of the page. Notice, there's a 90-degree angle between the velocity vector, and the magnetic field vector. If the magnetic field weren't there, the ion would just continue moving in this direction. But, since ... Straight up, but since there is a magnetic force, it's going to cause the ion to deflect. It's going to cause the ion to move in a circle. The ion is going to move in a circle. It's going to move in this direction. I'm going to attempt to sketch in a semi-circle, here, so you get the idea. There's my semi-circle. A little bit of intuition about why the ion is going to move in this circular path ... If I think about the ion, at this point, the velocity would be in this direction. If you use your Right-hand rule, you would see that the magnetic force would be pointing in this direction. Once again, 90 degrees to your velocity. The magnetic force always points towards the center of this circle. That's a centripetal force. We have a circle, here ... Or, a semicircle is what I've drawn ... of radius, R. This is a certain radius, R. Let me go ahead, and make that ... Let me just do a better radius than that. Let's sketch that in. This is radius. I'll make it lowercase r. We can calculate what that radius of the ion should be, by using our equation for magnetic force. Let's get some more space, down here. Let's rewrite our equation for magnetic force. Magnetic force is equal to q, v, cross B, which is the same thing as q, v, B, sin theta, where theta is the angle between your velocity vector and your magnetic field vector. Let's go back up here. The velocity vector is in the plane of the page, pointing up. The magnetic field vector is coming straight out at us. That's 90 degrees between those two vectors. Let's get some more space, down here. We know that sin of theta, that'd be sin of 90 degrees. Sin of 90 degrees, is equal to one. This would just be q, v, times B, times one. Well, force is equal to mass times acceleration. So, Newton's second law. We know the ion is moving in a circular path, so this would be the centripetal acceleration. We have q,v, B, is equal to mass times the centripetal acceleration. q, v, B, is equal to the mass. The centripetal acceleration is equal to v-squared over r. We can cancel one of our v's. So, we get q, B is equal to m, v, over r. So, we can solve for r. r would be equal to m, v ... This is velocity divided by q, b. Now we have the radius of the circle, and we can go a little bit further. We can take the velocity, that we solved for earlier, and we can plug it in, here. Let's go ahead and do that. This would be equal to m over q, B, times the velocity, which is square root of two, q ... Remember, this was the potential difference ... Over m. To get rid of that square root, we would have to square both sides. So, we square r, so we get r-squared. Then, we square all of this. This would be equal to m-squared, over q-squared, B-squared. This would be two q, v, over m. Now, we can cancel a few things. We can cancel on of the m's. And, we can cancel one of the q's. So, we get r-squared is equal to m, two, times the potential difference, divided by q times b-squared. All right, so take the square root to find the radius. We take the square to both sides. So, we get r is equal to the square root of two, m, times the potential difference, v, divided by q, B-squared. Finally, we have the radius of our circle, here. Let's think about these things. The magnetic field is constant. There's no change in the magnetic field. There's no change in the potential difference. If we assume that the charge on both ions is the same, the only thing that's different between those two ions is the mass. We can say that if we increase this number ... I could increase the mass. We're going to increase the radius. If we increase the mass, just looking at our equation, we're going to increase the radius. I should point out that we have, hiding in here, an m-over-q ratio. So, m-over-q ratio. So, m over q is the mass-to-charge ... mass-over-charge ratio, here. You'll see this written as m over z in a lot of mass spectrometry examples. m over z is the mass-to-charge ratio. For our purposes, we're just trying to think about how the mass affects the radius of the circle that the ion will move in. So, we've seen, if you increase the mass, you increase the radius. Let's go back up here, and let's look at our isotopes again. Let's look at the circular path that we drew. If I wanted to draw the path for ... Let me go ahead, and just label this one, right here. Let's say that this hits ... Where this ion hits, this represents the U-235 ion, the one with the smaller mass. If we represent the one with the larger mass ... So, the U-238 has more mass ... That means that the radius of the circle is going to be greater. Let me use blue, here. We have a lot of things going on. We have blue. I'm going to draw a path with a bigger radius. I draw a path with a bigger radius. Again, not the best drawing, but we can see, that with a bigger radius, this represents the U-238 ion. This is where those ions would hit in your mass spectrometer. This allowed us to separate our ions, based on mass. This final stage, here, the detection stage ... There's a lot that goes into detecting these things. In modern mass spectrometers, you're not going to separate isotopes. There are better ways to do that these days. But, this is just a nice way to introduce how a mass spectrometer works. In modern mass spectrometers, you're going to use it to get very-accurate masses. In some other videos, we can talk about that.