Video: Electron Beams

In this lesson, we will learn how to calculate the electrical properties of an electron beam and describe the uses of electron beams in imaging.

14:20

Video Transcript

In this video, we’re talking about electron beams. Like we might expect, these beams are made up of a stream of electrons, individual charged particles. And as we talk about electron beams, that’s a great place to start, with the electron.

Now, we may already be familiar with electrons, knowing that they’re one of the ingredients that make up atoms. Atoms consist of a nucleus, a central core, made up of particles called protons and neutrons. And then, for most atoms, there is also a collection of orbiting electrons; that’s these guys. If we consider an individual electron, there are certain properties that it has. Electrons are fundamental charged particles. When a particle is fundamental, that means, as far as we know, it’s not made up of any particles smaller than it. And the fact that electrons are charged means they carry a nonzero electric charge. That charge, which we sometimes label as 𝑞 sub 𝑒, is equal to negative 1.6 times 10 to the negative 19th coulombs. And along with its charge, electrons also have mass, which we can refer to as 𝑚 sub 𝑒. That mass is approximately 9.1 times 10 to the negative 31st kilograms, a very, very small particle.

Now, the fact that electrons are charged particles means that if we bring two of them close together, they’ll want to repel each other. They’ll want to push one another away. That’s because they have the same sign of charge, in this case a negative sign. But, nonetheless, even though electrons push one another away, it’s still possible to collect a lot of electrons and have them moving in the same direction, in order, as a beam. Here is how this can work.

Let’s say that we take a chunk of metal. And we shape it so that it has a very sharp point to it. Since this material is made of metal, that means it has a lot of electrons in it which we consider mobile. They’re very capable of moving around. Unless they further that along with our pointed metal, we had this very positively charged plate some distance away from the point. This positively charged plate creates an electric field around itself. And the tendency of this field is to push away any particles that have a positive charge but to pull in any particles that have a negative charge, such as the electrons in our metal.

Now, if there is enough positive charge collected on this plate, then that will make the filed strong enough so that it’s able to strip electrons off of the point of our metal object. Even though the electrons mutually repel or push one another away, the attraction to this positive plate can be strong enough that that repulsion is overcome. When that happens, many, many electrons start to stream off of the tip of this metal towards the positive plate. What we’ve created is an electron beam.

Considering this beam, let’s say we were to stand alongside the beam and count the number of electrons that pass by us over some unit of time. As we do this, since each individual electron has a charge of negative 1.6 times 10 to the negative 19th coulombs, then if we counted that number of electrons that pass by us over sometime, let’s call that capital 𝑁, and then multiply that number by the charge each individual electron has, we would wind up with a total charge, we can call it capital 𝑄, that has passed by us over this time interval. In other words, what we have is some amount of charge, 𝑄, over some amount of time, we can call it 𝑡. But now, looking at this expression, we may start to notice something.

Charge 𝑄 divided by time 𝑡 is the definition of current, capital 𝐼. That is, our electron beam is also a current. It’s an amount of charge that passes some point over some amount of time. That’s important enough to write down. An electron beam is a current of electricity. Now, going back over to our sketch of the positively charged plate and our wedge of metal, we know that, thanks to the electric field created by our positive plate, a potential difference is established between the plate and the tip of metal right here. This difference in electric potential has to do with the fact that the concentration of charges at these two points, right over here where the plane is and right over here where the tip of our metal is, are not the same.

So we can say then that spanning this distance is some potential difference that we can call 𝑉. It’s that potential difference which is responsible for pulling these electrons off of the tip of our metal object and starting them accelerating in a beam. If we follow an electron in the beam as it travels across this entire gap, we know that once the electron reaches this positively charged plate, it will have moved through this entire potential difference of capital 𝑉. And all during that time, the electron is accelerating. It’s speeding up faster and faster and faster. So over at the tip of our metal object, where the electrons start out, we can say that they start with a speed of zero. On average, they’re not moving to the left or to the right or in any direction at all.

But then, thanks to the influence of this potential difference, the electrons start to accelerate towards the left, towards the positive plate. And by the time they reach that plate, they’re moving at the maximum speed they’ll reach. We could call it 𝑉 sub 𝑚. Alright, let’s think about this. These electrons, which we know have some amount of mass to them, are accelerated from a speed of zero to a speed we’ve called 𝑉 sub 𝑚. If an object has both a mass and a speed, then that means we know it also has a kinetic energy. We can recall that an object’s kinetic energy is equal to one-half its mass multiplied by its speed squared.

So as soon as the electrons in our beam speed up past an average speed of zero, they acquire a kinetic energy. And they achieve a maximum kinetic energy right before they hit this positive plate. It’s not that their mass changes anywhere along the journey. But their speed changes; it keeps increasing. Now it turns out that there is a helpful relationship between the kinetic energy of a charged object and the potential difference that that object moves through. The kinetic energy of a charged object, where that object has a charge we can call 𝑞, is equal to its charge 𝑞 multiplied by the potential difference that the charge moves through. And we can see that this relationship connects with electron beams because the electrons that make up the beam each have a charge to them.

If we apply this equation to an electron beam, then the charge we would substitute in is 𝑞 sub 𝑒, the charge of an electron. Now, not only does putting a charged particle through a potential difference give that particle kinetic energy, it can also affect the direction of the particle’s motion. Imagine this, let’s say that we have two parallel plates and that these plates have opposite electric charges on them. The top plate is positive, and the bottom plate is negative. Just like what happened down here with our positively charged plate, these parallel plates will also create an electric field in between them. And that field will point from the positive to the negative plate.

Now, if these electrons here are part of a beam that’s moving from the left to the right through the plates, when they enter that region, the direction they travel in will start to change. The electrons will be pushed away from the negative plate and drawn towards the positive plate. So instead of moving left to right, they’ll curl towards that positive plate. Now, let’s imagine that the charge amounts on these two plates, the positive and the negative, were very carefully tuned. With just the right amount of charge, this electron beam wouldn’t run into the positive plate. But it would be diverted toward it by the time the beam leaves this region. And then, once it’s out from between the plates, this electron beam will continue to travel in that same direction.

So we’re seeing that it’s possible to change the direction of an electron beam by passing it through parallel charged plates. And in fact, this is how old televisions, which worked using a technology called a cathode-ray tube, operated. The idea with a cathode-ray tube is that we would send a beam of electrons toward a screen. And then, in order to control where that beam hits the screen, we would pass it through two sets of parallel plates. The electric charge on these parallel plates would be carefully controlled so that the beam’s direction, both horizontal and vertical, could be affected. Then once the beam had been directed towards a certain spot on the screen, the electric fields created by these parallel plates would rapidly change so that the beam would move. It was using this method that it was possible to scan this beam of electrons across the screen many times a second. And that is how the television image was formed. Now that we’ve learned some about electron beams, let’s get some practice with these ideas through an example.

A block of an unknown material emits charged particles. These particles are accelerated in a vacuum across an electric potential difference of six volts. The kinetic energy of each particle is found to be 1.92 times 10 to the negative 18th joules once it has crossed this potential difference. What is the charge of each particle?

Okay, let’s say that this is our block of material. And we’re told that this block is emitting particles. And along with that, we’re told that there is a potential difference of six volts, which accelerates these particles through a vacuum. If we follow one of these particles as it moves across this six-volt potential difference, once it reaches the barrier of that six-volt potential difference, we’re told its kinetic energy, we can call it 𝐾𝐸, is 1.92 times 10 to the negative 18th joules. Knowing all this, we first want to answer this question. What’s the charge of each one of these particles?

Now, notice that we don’t know what the particles are. And we don’t know what their mass is. Nonetheless, given what we’re given, it’s still possible to answer this question. We’ll do it by recalling a mathematical relationship between particle kinetic energy, potential difference, and particle charge. According to this relationship, if we have a particle with charge 𝑞 and that particle moves through a potential difference 𝑉, then if we take the product of those two values, we’ll find the kinetic energy that the particle acquires. Now, in this scenario, we’re told that kinetic energy. And we’re also told the potential difference 𝑉. And we want to solve for the charge 𝑞. So let’s do that. Let’s rearrange this equation so that we have 𝑞 by itself on one side.

To do that, we can divide both sides by the potential 𝑉, which means that cancels out on this right-hand side. And we see that particle charge 𝑞 is equal to kinetic energy divided by potential difference. Our next step is to substitute in the given values for kinetic energy and potential. And once we do that and calculate this fraction, we find a result of 3.2 times 10 to the negative 19th coulombs. So that’s the answer we’ll give to this first question of what is the charge of each of the particles emitted from this block of unknown material. Each of these particles has a charge of 3.2 times 10 to the negative 19th coulombs. Now let’s move on to the second part of our question.

Our second question says, using a value of 1.6 times 10 to the negative 19th coulombs for the charge of a proton, what is the relative charge of each particle?

And talking about each particle, we mean the particles emitted from this block of unknown material, the particles whose charge we’ve solved for in the earlier part. To answer this question, we need to know that the relative charge of a proton, we’ll symbolize it with 𝑃, is positive one. So rather than 1.6 times 10 to the negative 19th coulombs, which is a somewhat clumsy number being used for that charge, as a shorthand, we can say that that amount of charge, 1.6 times 10 to the negative 19th coulombs, is equal to one charge unit. And the question is, based on the fact that a proton has a relative charge, we could call it a plus one, we want to know the relative charge of these unknown particles.

Here’s how we can figure that out. We can take the charge of our unknown particles that we solved for earlier, that charge is 𝑞. And then we’ll divide that charge by the charge of a proton, not the plus one but by the true charge of a proton. That is, its charge in units of coulombs. So to figure out the relative charge of these unknown particles, we’ll substitute in the value we found for 𝑞. And then, when we perform this division, we find that the answer is simply positive two. And that this, therefore, is the relative charge of these unknown particles. They have a charge magnitude twice as big as the charge of a proton and the same sign charge, positive. Knowing all this, let’s answer a third and final question about this scenario.

It is found that the charged particles are being emitted by radioactive decay of the material in the block. What kind of particles are being emitted by the block?

Okay, so back to our block, we’re told that the particles emitted by it, which right now we don’t know what they are, are being emitted, thanks to radioactive decay of some material, in the block. Now, based on that, that narrows our options for what kind of particles these particles being emitted by the block are. The particles could be neutrons or beta particles, electrons or positrons. Or, they could also be alpha particles. The way we’ll figure out which one of these particle types is actually being emitted will be based on the charge value of these particles that we’ve solved for previously. We know that whatever these particles are, they have a charge of 3.2 times 10 to the negative 19th coulombs. In other words, a relative charge of plus two, two times the charge of a proton.

Now if these particles have an overall charge of two times the charge of a proton, then it makes sense that we would imagine that two protons are contained within these particles. And it’s at this point that we can recall that an alpha particle, also known as a helium nucleus, is a particle that has two protons as well as two neutrons in it. Now, since these neutrons, the green dots, have no electric charge, they don’t contribute anything to the overall charge of this particle. It still has a charge of plus two. So this is an alpha particle, two protons plus two neutrons. And the charge of an alpha particle, we see, is the same as the charge we solved for, 𝑞. This charge is different from the charge of a beta particle and, of course, different from the charge of a neutron, which has no charge. This then is our answer to this final question. The particles being emitted by this block of unknown material are alpha particles.

Let’s take a moment now to summarize what we’ve learned in this lesson. Starting off, we saw that electron beams are made up of individual electrons and that each electron has these given charge and mass values. We saw further that electron beams are created by accelerating individual electrons through an electric potential difference. That’s how the mutual repulsion between electrons is overcome. Along with this, it’s the case that electrons in the beam acquire kinetic energy, that that kinetic energy is equal to the charge of the electrons multiplied by the potential they’re accelerated through. Lastly, we learned that charged parallel plates can be used to redirect electron beams. For example, this is how cathode-ray tubes, which are in old TV sets, work.

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