Mass Spectrometer - Online Tutor, Practice Problems & Exam Prep

Mass spectrometers measure the mass of charged particles through four key steps: ionization, acceleration, velocity selection, and deflection. Ionization creates charged particles, which are then accelerated through a potential difference, leading to kinetic energy changes described by the equation $q^{\mathrm{\Delta v}}=\mathrm{\Delta k}$. The velocity selector filters particles based on speed using electric and magnetic forces, while the final deflection occurs in a magnetic field, resulting in circular motion described by $r=\frac{\mathrm{mv}}{\mathrm{qB}}$.

Mass Spectrometers

Hey, guys. So in this video, we're going to talk about mass spectrometers, which are instruments used to measure mass. Now this can be overwhelming at first because there are 3 different parts and there are 3 different equations you can use. But I will break it down really simply for you. And hopefully, by the end of this video, you'll agree with me that this is totally manageable. Let's check it out.

Alright. So, as I said, mass spectrometers are instruments that measure the mass of a known charge, meaning that experimentally you would know your Q and be looking for M. Now, in a physics problem, you may actually be given the M and asked for a Q, right? There are 4 steps to a mass spectrometer that you should know. I merge the first two steps because they're pretty basic. So steps 1 and 2 are ionization and acceleration. Ionizing something means that you are adding or removing electrons so that it becomes either positively or negatively charged. And the reason you want to do this is that particles need to have a charge, an excess of electrons or protons, so that they actually feel a force. So, you do ionization so that the particles feel a magnetic force and an electric force. If you don't have a charge, you don't feel those two forces. So right here you have something that's called an ionizer. So let's just draw that there. And what it does, it's got some particles chilling here and somehow it magically strips them of electrons so that they become positively charged. It then sort of shoots them in this direction so that these bunch of positive charges are hanging out over here.

Okay. We're going to do this with positive charges, but it could be with negative charges as well. Now, that's ionization. We ionize them and then we throw them over here. Now they're going to go through acceleration. They're going to be accelerated through a potential difference, meaning you can have a positive potential here and a negative potential here. Electric fields are created whenever you have a situation like this, always going from high to low. So this is the direction of the electric field. Remember that positive charges feel a net force in the direction of the electric field. So these guys will be pushed this way by an electric force, which will result in them accelerating this way. By the way, the initial velocity here will be 0.

Okay. I'm dropping a ton of information on you, but when we do an example, you'll see that this is pretty straightforward. So they are going to get over here with some sort of final velocity. Okay. So between here and here, they're actually going to have a constant velocity. Okay.

Now for this first part, the only equation you need to know, is, you have one equation per part. You remember that the work done on a charge as it moves through a potential difference is \( q \Delta V \) where \( \Delta V \) is the potential difference between the two plates, and that's typically given to you or you're asked to calculate it. Remember also that, work is also the change in kinetic energy. Okay. So because these 2 are both work, I can just say that one equals the other. Okay. So I can say that \( q \Delta V = \Delta K \). Now remember, \( \Delta \) means change and \( \Delta K \) is \( K_{\text{final}} - K_{\text{initial}} \). But this is almost always or probably always going to be 0, the initial velocity. So what you really end up with is the final kinetic energy, which is \( \frac{1}{2} m v_{\text{final}}^2 \). So this is the first equation that you have for this problem, and there's 2 more.

Okay. Now if you don't want to remember all these letters, you could remember this part and sort of work your way there.

Cool. So that's this first part. This thing gets accelerated. It moves through with a constant speed, and now it's going to go through what's called a velocity selector. So the second part is velocity selection to the velocity selector. And the idea is that you want to filter desired speeds. So let's talk about that. The only way to calculate a mass on a mass spectrometer is if you basically know everything else. And I'll show you this later with the equations. You have to know everything else, meaning you have to know the velocity even to be able to calculate it.

So what's going to happen here is that these guys will have different masses. This is mass 1. This is mass 2, etc. Because they have different masses, they're going to have different speeds here. And this is because of just basic \( F = ma \). Right? Acceleration is force over mass. So if you have different masses, you're going to have different accelerations. Therefore, you're going to have different speeds. But remember, we have to control the speeds. So what we're going to do is we're going to make sure that all these charges run through this device so that we can get rid of the ones that are not the right speed. And the way you do this is by running them through an electric field again. So notice that this is going from positive to negative here. So there's an electric field that's pointing down. These are all positive charges. So when these charges are here, they're going to feel a force this way, force electric.

Okay? And they're going to potentially get deflected, but you want the right ones you want the particles with the target's desired ideal speed to make it.

K? You want them to make it all the way through the end and come out of the gate over here into this green area that we're going to talk about. But if you have an electric field here, they're going to be pulled down, which means they're going to hit the wall, and that's bad news. Right? Slow ones are going to hit it over here. Fast ones are going to hit it over here. So what you do is you want to try to cancel that electric force with another force, and we're going to try to cancel with the magnetic force.

So over here, we're going to try to have a magnetic force that cancels that electric force. So the next equation you can write is that \( F_E = F_B \).

Now let's expand this real quick. Electric force is \( qE \) and magnetic force is \( qvB\sin(\theta) \). The angle here between \( v \) and \( B \) is going to be 90. We'll talk about that. So this is just going to go away. Okay. I can cancel the \( q \)'s and end up with this relationship here, which is that, I can simplify or not simplify, I can make this more standard looking by moving the \( B \) around. So it's going to look like this. This is the second equation that you can use or that you will need to use in a lot of these questions.

\( V = \frac{E}{B} \)

Okay. So let's talk about the direction of this thing. So I have a positive charge. I'm going to use the right-hand rule, right, with my right hand and I want the force to be up. How do we get the force to be up? Well, my palm has to be going up because this is force. So it's something like this. Right? Something like this or something like this. Well, it's actually in this orientation because I want my velocity to be going to the right if you look at the diagram. Right? So velocity to the right, force up means that my hand is going away from me and into the page. Right? Don't just look at the video. Do this yourself away from you into the page.

So that means that to accomplish this cancellation of forces, my magnetic field needs to be into the page here. Okay. So that's the direction.

Now if you look at this equation, this equation ties E, B, and V. And the idea is that this only is going to cancel the charges that are going to have the target speed. Because if you have a different speed, if you look at \( F_B \), \( F_B = qvB \). If you have the wrong speed, you're going to have the wrong force that isn't going to exactly cancel \( E \). So what you're going to end up doing is you're going to either hit over here, hit over here. Right? So this is sort of a filter, and only the forces only the particles with the right speed will make it over here.

Okay. Now we're going to get to the last part. When you get here, the way that these devices are built is that not only is there an electric and magnetic field here, inside of the velocity selector, there's also the same exact magnetic field out here into the deflection area. Okay. Now what's going to happen that's a little bit different here is when you come out of the gate here, right? You still have the magnetic force because you still have a magnetic field, but you no longer have an electric force. So I want you to write it and then scratch it out. You no longer have an electric force because there's no longer an electric field.

Right? You only have that between the parallel plates. So what happens is you're now going to move in a circular path like this, and that's a terrible semicircle. It's pretty good.

So remember if you're going into if you have if you are charged moving into a constant magnetic field, you have circular motion. Okay. And this circular motion is going to have a radius \( r \). And you may remember our circular motion equation, which is the 3rd equation we're going to use, which is \( F_c = \frac{mv^2}{r} = qvB \). We talked about this in the previous video. You have to memorize that equation, not just, for those simpler problems, but also because you're going to need it here. Okay.

One last thing that I want to mention about equations is that sometimes you were given or asked for not the radius, but this distance here. Okay. This distance, I'm going to draw it over here and hopefully you'll see that this distance is just one radius, 2 radius. Okay. So I'm going to erase this so that it's not very messy. Hopefully, you believe me. Distance is true r. That's another one. It's not really an equation. It's just sort of like an accessory that you may need to use.

Cool. So if you know the velocity because you filtered only the good particles with the right velocity, and you know the magnetic field because you control the device so you can adjust the magnetic fields and you know the charge, you know the charge because you know how much charge you gave these things over here on the ionizer. And you and you can, you can measure the distance, right? You take a picture of this and you see, Oh, crap. They're all hitting over here. Now I can measure \( d \) and from \( d \), I can find \( r \), right? Now, you know, all 3, all 4 variables, which means you're able to solve for mass. Now, if you want, you can rewrite this and say, you know, mass equals \( \frac{qB_r}{v} \), so that it's a more straightforward equation. But now you gotta remember 2 equations. I don't think it's worth it. I wouldn't really do this. I would just leave it, as the standard rotation equation circular motion equation.

Cool.

Alright. So there's a lot of a lot of crap, but let's solve a problem here, and I think you'll see that it's not that bad. This is quite a long problem because I wanna hit up a lot of different things. So here, and by the way, a lot of the text here is just describing how a mass spectrometer works so you get familiar with the language. A charge of 2, so charge positive 2 c. Positive means we're going to use the right-hand rule, not the left. It's accelerated through an x, accelerated in the positive x-axis, which is to the right. So we have a positive charge. So if you want to draw, you don't have to draw, but I'm just going to draw here. You got your little \( q \)'s. By the way, there's, so you got your \( q \) here, and you are going to have a little negative there. Right? It gets accelerated through a potential difference \( \Delta V \). So the potential difference here is \( \Delta V \). \( \Delta V \) is not given to us. And if you look at question d, we're being asked for what is \( \Delta V \). So that's coming.

We wanna know what that is later. It then passes through horizontal plates, so you get to accelerate here and then it goes here with a constant speed. And then here it says accelerated, and it's going to pass through horizontal plates right here. Horizontal plates that have an electric field 3 Newtons per Coulomb. So electric fields over here, \( E = 3 \), that points up. By the way, if it points up, it means that you're positive here and negative here, always high to low. It has a magnetic field, magnetic field \( B = 4 \) Tesla that also exists between the plates. So inside of here, there's a magnetic field. I don't know the direction yet, so let's not draw that. And remember here you have sort of the deflection zone because this guy will go straight through here. Right? It's going to go straight through here and it's going to deflect either this way or that way. We don't know yet, so let's not write that yet.

And by the way, the same magnetic field 4 is also going to exist in this deflection zone over here. Okay. That's what it says here. This magnetic field also exists outside of the plates and causes the charge to deflect in the circular arc of radius 5 centimeters. So this means that the radius of this deflection is going to be 0.05 meters. Okay.

Alright. Question. What must the what must the direction of the magnetic field be? So the direction of the magnetic field, you will always determine that by looking here. Okay. So you have a positive charge, which means the electric force will be up because the electric force for a positive charge goes in the direction of the electric field. It would be backward if you have a negative charge. Therefore, you want your magnetic charge to be going down. That's the step.

Now we're going to use a right-hand rule to determine the direction that the magnetic field has to have so that you actually get a force down. So force down means palm down and it could be this, but actually my charge is supposed to be, going to the right. So you end up with something like this. Okay. Actually that's bad because now my palm is up. So you actually have to do this. Okay? So your fingers are pointing towards your face, your thumb is to the right. Please do this. Now my force is down, which is what we want. What this means is that my fingers are coming at my face, which means they're popping out of the page into me, which means that this is the direction of out of the page, which is given by a dot. So the direction of the magnetic fields is little dots everywhere. Right? The little dots everywhere. Lots of little dots. So it's going to look like this. So what is the direction? The direction is out of the page and we got the first part done. Sketch the deflection that the charge will experience. There are 2 options here and it's either going to be like this or like this. Right? And that will depend on, the direction of the force. So the direction of the magnetic force, the magnetic force will be the same out here. And because this is being pulled down, it's going to arc down. Okay. So that's the sketch. We got the sketch done. They could have asked, is this going to arc up or down? It's arcing down. Okay. And then now we want to calculate the mass of the charge. Right? So now we're actually going to start calculating stuff. How do we do this? Well, we have 3 equations. You can think of it as sort of a menu, and I'm going to write them here. 1 is \( q \Delta V = \Delta KE \) or you can write \( q \Delta V = KE_{\text{final}} \), squared, that's equation 1. The second equation is, \( V = \frac{E}{B} \), \( e \) over \( b \). And the third equation is the radius one \( m = \frac{qBr}{v} \). So that's it. Those are the 3 equations. That's all you gotta play with. Now this is just an algebra problem and I'm going to get out of the way. So we're looking for mass. Where do you see mass? There's mass here and there's mass here. So let's see. Do I know \( \Delta V \)? I don't know \( \Delta V \). Do I know the velocity? We don't know the velocity either. And I know \( q \). So this equation is kind of ugly. I have 2 unknowns and I'm looking for \( m \). I have 3 unknowns. That's terrible. What about this equation? Do I know \( v \)? I don't know \( v \), But I know \( q \), I know \( b \). Right? I know \( q \), \( q \) is 2. I know \( b \). \( B \) is 4. And I know \( r \) because the radius is 5 centimeters. So I know this as well. So this equation is still not totally ready, but at least here I have 3 unknowns, which is bad news. And here I only have 2. So this equation is not as bad, but before I can find \( m \), I'm going to have to find \( v \). Well, there's a \( V \) here and there's a \( V \) here. Right? And this is just problem-solving skills. This equation we already said was bad. So let's try to get the \( v \) from here. Do I know \( e \) and \( b \)? And I do. So the first thing we're going to do is actually do \( v = \frac{e}{b} \

A negative charge in a spectrometer is accelerated in the negative x-axis. It is later deflected and collides some distance ABOVE velocity selector. What are the orientations of the electric and magnetic fields, respectively, inside the selector?

• A. up and out of the page
• B. up and into the page
• C. down and out of the page
• D. down and into the page

A 2 kg, −3 C charge is accelerated through a potential difference of 4 V. The velocity selector has an electric field of magnitude 5 N/C. How far from the velocity selector will the charge collide against the spectrometer "wall"?

• A. 0.117 m
• B. 0.267 m
• C. 1.60 m
• D. 3.20 m
• E. 6.69 m
• F. 38.4 m

Find Mass-to-Charge Ratio in Spectrometer

Hey guys. So some questions will ask you to find the mass to charge ratio of particles going through a mass spectrometer. And that's what we're going to do here. So let me show you. So here it says a mass spectrometer has a velocity selector, electric field of magnitude 20. So remember you get accelerated and then you go through a velocity selector, that's going to have an electric field and we're telling you here that it has a magnitude of 20. It says when a certain charge is accelerated to a constant 30, so it gets accelerated over here, it goes through the little hole and then it's now going to move here with a constant 30. It collides 40 meters away from the velocity selector. What does that look like? Well, it means it's going to curve this way and it's going to hit this wall here at a distance d equals 40 meters. Remember also that distance is twice the radius because you have a radius and radius, which means if the distance is 40, this means that the radius is 20. And the reason we want to change it to radius is that our other equations, or the one other equation we're going to need here, is in terms of radius and not in terms of distance. I drew it down even though it could have gone up, we don't actually know. We are not being given enough information to figure out which way this is going but it doesn't matter. I just picked one for the sake of illustration. Okay? In this question we're going to find the mass to charge ratio. Okay? Mass to charge ratio which is m/q. By the way, if I hadn't told you that we were looking for m/q, you can just interpret that from the question. The mass to charge ratio, you would do this and say, hey, I'm looking for this. Which means you're not looking for m or q. You want to leave this as a fixed unit and then solve it. So how the heck are we going to do that? Well, we got these 3 equations and one of them hopefully

q ∆ V = 1 2 m v 2 .

By the way, this is a potential difference or voltage, and this is a velocity, 2 different things. The other equation is that

v = E B .

And the other equation is that

r = m v qB .

And we're looking for the ratio m/q. Luckily, this is actually really straightforward. Because if you look here, you'll find one of these equations has an m and a q and it's right here. And not only do they have it, but they're already sitting next to each other which is awesome. So all you got to do is move stuff around in such a way that the m and the q stay exactly where they are. So we're going to move the B to the other side. So I'm going to get BR/v. Do I have b? Yep. Do I have b? No, I don't have b. Got excited there. For a second, do I have r? Yep. R is 20. Do I have the velocity? The velocity is 30. So we don't have b. We got to get it. Can we get b? This equation here seems to be going to work. So

v = E B .

Therefore, b is E/v. The e is 20, electric field strength and the velocity is 30. So b is 0.67. Okay. 0.67 Tesla. So that's what's going to go over here. 0.67 Tesla. And if you do all of this, you're going to get that ratio is 0.44. The units are kilograms per coulomb. Okay. So that's how you could do this. The second you got stuck here because you didn't have v, you just go to a different equation and you get it. Alright? That's a bonus one. Let's get going.

Here’s what students ask on this topic:

What are the main components of a mass spectrometer?

A mass spectrometer consists of four main components: ionization, acceleration, velocity selection, and deflection. In the ionization stage, particles are charged by adding or removing electrons. During acceleration, these charged particles are accelerated through a potential difference, gaining kinetic energy. The velocity selector filters particles based on their speed using electric and magnetic fields. Finally, in the deflection stage, particles move in a circular path within a magnetic field, allowing for the measurement of their mass based on the radius of their trajectory.

How does a mass spectrometer measure the mass of a particle?

A mass spectrometer measures the mass of a particle by first ionizing it to create a charged particle. This particle is then accelerated through a potential difference, gaining kinetic energy. The velocity selector ensures that only particles with a specific speed pass through. Finally, in the deflection stage, the particle moves in a circular path within a magnetic field. The radius of this path is given by the equation $r=\frac{mv}{qB}$, where $r$ is the radius, $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength. By knowing the other variables, the mass can be calculated.

What is the role of the velocity selector in a mass spectrometer?

The velocity selector in a mass spectrometer ensures that only particles with a specific velocity pass through to the deflection stage. It uses both electric and magnetic fields to filter particles. The electric force $F=qE$ and the magnetic force $F=qvB$ are set to cancel each other out for particles with the desired velocity, given by $v=\frac{E}{B}$. Particles with the correct velocity pass through, while others are deflected and filtered out.

How do you calculate the radius of the circular path in a mass spectrometer?

The radius of the circular path in a mass spectrometer is calculated using the equation $r=\frac{mv}{qB}$, where $r$ is the radius, $m$ is the mass of the particle, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength. By knowing the charge, velocity, and magnetic field, you can determine the radius of the particle's path, which is crucial for calculating its mass.

What is the purpose of ionization in a mass spectrometer?

Ionization in a mass spectrometer is the process of adding or removing electrons from particles to create charged ions. This is essential because only charged particles can be manipulated by electric and magnetic fields, which are used in subsequent stages of the mass spectrometer. Ionization allows the particles to be accelerated, filtered by velocity, and ultimately deflected in a magnetic field, enabling the measurement of their mass.