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Mass Spectrometer: How to Get the Mass of a Particle?

Important Formula

Formula: Mass Spectrometer

$$\class{brown}{m} ~=~ \frac{ q \, r \, d \, \class{violet}{B}}{ \class{green}{U} }$$ $$\class{brown}{m} ~=~ \frac{ q \, r \, d \, \class{violet}{B}}{ \class{green}{U} }$$ $$q ~=~ \frac{ \class{brown}{m} \, \class{green}{U}}{r \, d \, \class{violet}{B} }$$ $$r ~=~ \frac{ \class{brown}{m} \, \class{green}{U}}{q \, d \, \class{violet}{B} }$$ $$d ~=~ \frac{ \class{brown}{m} \, \class{green}{U}}{q \, r \, \class{violet}{B} }$$ $$\class{green}{U} ~=~ \frac{ q \, r \, d \, \class{violet}{B}}{ \class{brown}{m} }$$ $$\class{violet}{B} ~=~ \frac{ \class{brown}{m} \, \class{green}{U}}{q \, d \, r }$$

Rearrange formula

What do the formula symbols mean?

Mass

$$ \class{brown}{m} $$ Unit $$ \mathrm{kg} $$

Mass of the charged particle exiting behind the aperture of the mass spectrometer.

Electric charge

$$ q $$ Unit $$ \mathrm{C} = \mathrm{As} $$

Charge of a particle shot into the mass spectrometer.

Radius of the circular path

$$ r $$ Unit $$ \mathrm{m} $$

Radius of the semicircular orbit of the particle created behind the aperture.

Plate distance

$$ d $$ Unit $$ \mathrm{m} $$

Distance between the two electrodes of the capacitor.

Voltage

$$ \class{green}{U} $$ Unit $$ \mathrm{V} = \frac{ \mathrm J }{ \mathrm C } = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{A} \, \mathrm{s}^3 } $$

Voltage between the capacitor plates.

Magnetic flux density (B-field)

$$ \class{violet}{B} $$ Unit $$ \mathrm{T} = \frac{\mathrm{kg}}{\mathrm{A} \, \mathrm{s}^2} $$

External magnetic field in which the plate capacitor is placed. This magnetic field is also outside the capacitor behind the aperture. This field is responsible for the charge to perform a circular motion.

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Table of contents

Important Formula
Basic setup
Exercises with Solutions

Using a mass spectrometer, it is possible to determine the unknown mass $ \class{brown}{m} $ of a charged particle. This makes mass spectrometers applicable in a wide range of specific applications:

Mass spectrometers allow the investigation of the chemical composition of samples to detect impurities or harmful substances.
Mass spectrometers help in the analysis of food samples, for example, in the identification of contaminants, the analysis of soil, water and air samples, and the analysis of pesticides and toxins in food.
Mass spectrometers are used in forensic analysis. They help to identify drugs and toxins and thus to solve crimes.
Mass spectrometers are used in archaeology to determine the age of rocks, fossils, and archaeological artifacts.

The examples mentioned are only a fraction of the possible applications of a mass spectrometer. If you want to work in one of these fields later on, it is therefore important to understand how a mass spectrometer works.

Basic setup

A simple mass spectrometer consists of a particle source, a velocity filter, a detector plate, and an external magnetic field.

For a mass spectrometer you need the following things:

Particle source: It provides us with the charged particles to be analyzed.
Velocity filter (WIEN filter): This device, as the name suggests, is used to filter particle velocities. It is essentially a plate capacitor placed in a magnetic field $ \class{violet}{B} $. An aperture is attached to one side of the capacitor to allow only particles with a specific velocity to pass through.
Detector plate: This is placed at the back of the pinhole to register the particles that land on it.
External Homogeneous Magnetic Field: It is located between the capacitor plates and serves to filter the velocity of the particles. However, it also extends beyond the aperture and causes charged particles to be deflected in a semicircular path before landing on the detector plate.

A particle source provides charged particles, which are then shot between the two electrodes (plates) of the plate capacitor. When the capacitor is turned on, an electric field $ \class{purple}{E} $ forms between the plates. Each time a charged particle enters this electric field, it experiences an electric force $ F_{\text e} $. This force causes the particle to be deflected either towards the positively charged plate or the negatively charged plate.

The strength of the deflection, that is, the magnitude of the electric force experienced by a charge $ q $, is related to the electric field as follows:

$$ \begin{align} F_{\text e} ~=~ q \, \class{purple}{E} \end{align} $$

We can change the electric field and thus the electric force experimentally with the help of the voltage $ U $ between the capacitor plates. The electric force on the plate capacitor is related to the voltage as follows:

$$ \begin{align} F_{\text e} ~=~ q \, \frac{\class{green}{U}}{d} \end{align} $$

Next, we place the plate capacitor in an external homogeneous magnetic field $\class{violet}{B}$. Our goal is not only to understand the functioning of the mass spectrometer but also to derive a specific formula that allows us to calculate the mass of the incoming particles. To ensure that the formula does not become too complicated, we align the magnetic field in such a way that its field lines are perpendicular to the direction of particle motion and perpendicular to the electric field lines of the plate capacitor (see Illustration 1).

Now, between the plates of the plate capacitor, there exists not only an electric field but also a magnetic field. As a result, the particle experiences an additional magnetic force $ F_{\text m} $ (also known as the Lorentz force) between the plates. The direction of this magnetic force and, therefore, the direction in which the particle is deflected can be determined using the Right Hand Rule.

Since we have assumed that the magnetic field $ \class{violet}{B} $ and the velocity $ \class{blue}{v} $ of the particle are perpendicular to each other, the magnetic force $ F_{\text m} $ on the charged particle can be calculated as follows:

$$ \begin{align} F_{\text m} ~=~ q \, \class{blue}{v} \, \class{violet}{B} \end{align} $$

To get a desired velocity $ \class{blue}{v} $ of the particles landing behind the aperture, the magnetic field and the electric field must be adjusted.

In the lesson on the velocity filter, we learned that a particle that passes through the aperture of the velocity filter is not deflected by either the electric force $ F_{\text e} $ or the magnetic force $ F_{\text m} $ within the velocity filter. The particle continues to travel straight ahead as these two forces on the particle are equal: $ F_{\text e} = F_{\text m} $. This allows the particle to pass through the aperture of the plate capacitor unhindered.

We express this balance of forces mathematically by equating the electric force 2 with the magnetic force 3:

$$ \begin{align} q \, \class{blue}{v} \, \class{violet}{B} ~=~ q \, \frac{\class{green}{U}}{d} \end{align} $$

Next, we need to find out the velocity of the particle landing behind the aperture. For this we rearrange equation 4 for $\class{blue}{v}$:

$$ \begin{align} \class{blue}{v} ~=~ \frac{1}{d} \, \frac{\class{green}{U}}{\class{violet}{B}} \end{align} $$

The magnetic field behind the aperture is the same as in the plate capacitor. When the particle leaves the velocity filter, it enters this magnetic field and experiences the same magnetic force $ F_{\text m} $ as in the velocity filter. Here, outside the capacitor, there is no electric force acting on the particle. The magnetic force deflects the particle either downwards or upwards. Let's assume it is deflected downwards, as shown in Illustration 1 (the exact direction of deflection is not relevant in our case).

After a short semicircular flight, the particle lands on the detector plate. By observing the point of impact on the detector plate, the distance between the aperture and the point of impact can be determined. This distance corresponds exactly to the diameter $ 2r $ of the circular path that the particle traverses. Here, $ r $ represents the radius of the circular path, which is half the distance from the point of impact to the aperture.

In a circular motion, a centripetal force acts on the particle (even if the particle only traverses a semicircle):

$$ \begin{align} F_{ \text z} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{ r } \end{align} $$

The magnetic force acts to the center of the circle in the same way as the centripetal force. Thus, the magnetic force IS the centripetal force. Set magnetic force 3 equal to centripetal force 6:

$$ \begin{align} q \, \class{blue}{v} \, \class{violet}{B} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{r} \end{align} $$

Perfect, because with this we have brought into play the easily measurable radius $ r $ and the mass $ \class{brown}{m} $ of the particle.

Transform equation 7 with respect to the mass:

$$ \begin{align} \class{brown}{m} ~=~ \frac{q \, \class{violet}{B} \, r}{\class{blue}{v}} \end{align} $$

Unfortunately, it is not possible to directly measure the velocity $ \class{blue}{v} $ of the particle. That is precisely why we established Equation 5 for the velocity. We can substitute $ \class{blue}{v} $ from Equation 5 into Equation 8 to eliminate the velocity:

$$ \begin{align} \class{brown}{m} ~=~ \frac{ q \, r \, d \, \class{violet}{B}}{ \class{green}{U} } \end{align} $$

We are done! With Equation 9, we can determine the mass of charged particles. From the equation, you can also observe that charges landing farther away from the aperture have a greater mass.

There are two points to keep in mind:

To determine the mass of the particle, it is necessary to know the charge $q$ of the particle. Otherwise, only the ratio of charge to mass, known as the specific charge $ q/\class{brown}{m} $, can be calculated.
Furthermore, it is not possible to determine the mass of uncharged particles, such as neutrons. Neutral particles do not experience deflection in the magnetic field and therefore their mass cannot be determined by this method.

With this knowledge, you can begin with forensic analytics. In the next lesson, we will explore another exciting application of the Lorentz force, namely the Hall effect.

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise: Diameter of the Electron Orbit in the Mass Spectrometer

An electron flies into a plate capacitor with an applied voltage of $ 100 \, \mathrm{V} $ and a plate separation of $ 0.01 \, \mathrm{m} $. It passes through the capacitor without deflection. Both the plate capacitor and the rest of the setup of the mass spectrometer are located in a magnetic field with a field strength $ B ~=~ 1 \mathrm{mT} $. After leaving the plate capacitor, the electron is directed into a circular orbit in this magnetic field.

What is the diameter of this orbit?

Solution

Substitute the given values into the formula for diameter, which was derived in the lesson for mass spectrometers, and look up the rest mass as well as the charge of the electron (elementary charge) in your formula book. Then you get: $$\begin{align}d ~&=~ \frac{2m \, U}{q \, a \, B^2} \\\\ ~&=~ \frac{2 \cdot 9.109 \cdot 10^{-31} \, \mathrm{kg} \cdot 100 \mathrm{V}}{1.6 \cdot 10^{-19}\mathrm{C} \cdot 0.01 \, \mathrm{m} \cdot (10^{-3}\mathrm{T})^2} \\\\ ~&=~ 0.114 \, \mathrm{m}\end{align}$$