What I do not understand about mass spectrometers

In summary, the velocity selector lets through only one isotope at a time with a specific velocity and the spectrometer detects the m/z ratios of the ions by the distance from the aperture at which they land on the detector.
  • #1
TheGmc
17
1
Homework Statement
Mass spectrometer
Relevant Equations
qE=ma
I try to do my assignment which is based on mass spectrometer entirely. The mass spectrometer i am working on has these parts below:

1.Accelerator region
2.Velocity selector region
3.Spectrometer

The elements i am working on are isotopes of the same element and they all enter the accelerator region with same 1eV kinetic energy.

My question is the energy that the ions gain during acceleration is same since they all have same charge. So the heavier isotopes will have smaller velocity and the lighter will have greater velocity. But velocity selector let pass only the ions with certain velocity. So how the spectrometer detects the other isotope ions? I am so confused about that. Thanks in advance.
 
Physics news on Phys.org
  • #2
A velocity selector consists of a magnetic field and an electric field perpendicular to each other, like so:

1591782146531.png


In order for an ion to pass through, the magnetic force bust balance the electric force inside the velocity selector so that it doesn't accelerate sideways! That means you have ##qvB = qE \implies v = \frac{E}{B}##. Hence by varying either the magnetic field or the electric field in that region, you can selectively choose ions with a specific velocity.

The accelerated particles, for any given accelerating voltage, have a range of velocities depending on the masses of the various isotopes. That means that the velocity selector only let's through one isotope (strictly, one m/z value) at a time!
 
Last edited by a moderator:
  • Like
Likes TheGmc
  • #3
Can you explain it like mass spectrometer for dummies :D
I know i can vary the magnitude of magnetic and electric fields to choose the isotopes with a specified velocity. But the thing i am trying to understand i will never be able to pass all of the isotopes together since different isotopes will never have the specified velıcity under the same accelerator region conditions. My professor wants me to determine these accelerator conditions like voltage myself. So for example if i need particles with 10 m/s, for which isotope i will create the accelerator conditions.
 
  • #4
The accelerator has a fixed potential difference across the plates, ##V##. As you correctly deduce, this means that the kinetic energy of all isotopes are equal when they leave the accelerator (and equal to ##qV##). That gives you ##\frac{1}{2}mv^2 = qV \implies v = \sqrt{\frac{2qV}{m}}##.

You will notice that isotopes of different masses ##m## will then have different velocities. This is where the velocity selector comes in! A specific combination of electric and magnetic field strengths permits ions of velocity ##\frac{E}{B}## to enter the detection region. This means that we can effectively consider each isotope one at a time (or more generally, ions with the same m/z value)!

You detect the m/z ratios of the ions by the distance from the aperture at which they land on the detector. You slowly vary the electric field or the magnetic field of the velocity selector, and when a certain isotope is allowed to pass through you will see a spike on the data logger plot, corresponding to that isotope.
 
Last edited by a moderator:
  • Like
Likes TheGmc
  • #5
"One at a time! " from this i understand that i will do the same operations for 4 times if the element has 4 isotopes but by only changing E/B according to the velocity each isotope gain when they exit acceleratır region. If i understood right then why my professor wants me to determine the magnitudes of electric field and magnetic field if they all will be different for different isotopes :(
 
  • #6
TheGmc said:
"One at a time! " from this i understand that i will do the same operations for 4 times if the element has 4 isotopes but by only changing E/B according to the velocity each isotope gain when they exit acceleratır region. If i understood right then why my professor wants me to determine the magnitudes of electric field and magnetic field if they all will be different for different isotopes :(

Do you have the complete problem statement?
 
  • Like
Likes TheGmc
  • #7
IMG20200610132405.jpg
IMG20200610132458_07.jpg
 
  • #8
I was at the 6th wuestion when this question came to my mind
 
  • #9
Right, yes I see. These instructions are very confusing! Here's how I would approach it.

Suppose the accelerator has a voltage ##V = Ed## across it, then the ions leave with velocity ##v = \sqrt{\frac{2qV}{m}}##. This ion will land a distance $$d = 2r = 2\frac{mv}{Bq} = 2\frac{m}{Bq} \sqrt{\frac{2qV}{m}} = \frac{2}{B} \sqrt{\frac{2mV}{q}}$$ from the aperture, where ##B## is the magnitude of the magnetic field in the detector region.

If we suppose all ions are singly charged, you finally have that $$\frac{2}{B}\sqrt{\frac{2V}{q}}(\sqrt{m_1} - \sqrt{m_2}) \approx 1\text{cm}$$ Where ##m_1 = m_2 + m_n##, if ##m_n## is the mass of the neutron (and ##m_1## and ##m_2## are of two isotopes of the type of ion you are using, separated by one mass number unit). You will have to play around with some values for ##V## and ##B## to see what is reasonable.
 
  • Like
Likes TheGmc
  • #10
I start to solve from first prboem by not making a correlation your approach is giving a bigger picture thanks after I resolve every problem can I ask your advice again ?
Thank you very much for your time i was really desparate about this since every problem bound to each other.
I have a question about solenoids too. Everywhere i read including my textbook it is written that "the length of the solenoid should be significanlty bigger than the diameter for a uniform magnetic field" I read that is because where the length is going to infinity magnetic field lines are parallel but in real worl like the problem 2 and 5 what should be the diameter/length ratio so i can calculate something.
 
  • #11
By the way, I realized after reading my messages my punctuation is terrible. Thanks for understanding me under these conditions :D
 
  • #12
You are correct that you can generate a nearly uniform magnetic field with a long solenoid. The formula follows from Ampère's law; take a closed curve consisting of "4 sides", one of which passes through the solenoid. Then, $$\oint_C \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$ $$Bl = \mu_0 I_{enc} = N\mu_0 I$$ Where ##N## is the total number of coils. Now if we define ##n = \frac{N}{l}## as the number of coils per unit length, then we finally see $$B = n \mu_0 I$$So I wouldn't worry about the diamete/length ratio, just know that the above formula is how the magnetic field and current through the solenoid are related!
 
  • Like
Likes TheGmc
  • #13
Okey i will keep everything in mind. Thanks for make me comprehend these things :)
 
  • Like
Likes etotheipi
  • #14
@etotheipi Hello again! I made progress I think. But there is one thing again. At the beginnig of the assignment my proffessor indicated that " The number of ionized isotopes entering into the accelerator region corresponds to a 1 𝜇𝐴 ion current. ". When I try to solve problems 4 and 7 I read that the formula for power is P=VI where I is the current in the circuit. I don't know where to use the information proffessor gave me but the ions have nothing to do with circuit they just pass through the plates. Can you lighten me please ?
 
  • #15
You can think of it like this. If an amount of charge ##dQ## passes through the accelerator, it has gained kinetic energy ##VdQ##. So the rate of change of kinetic energy, or the power, is ##V\frac{dQ}{dt} = IV##.

So if the voltage across the accelerating plates is ##V##, then the power transferred to the ions is ##(\text{10}^{-6}V) \text{W}##.
 
  • Like
Likes TheGmc
  • #16
So there is no constant power it depends on the ions passing through the plates , is it ? I thought that it should be constant so I could not relate the ions with it. Thanks for your answer :)
 
  • #17
You don't have to worry about any of the charge carriers if it already gives you the current. The current ##I = nqAv## is already dependent on the number density ##n## and the charge of the ions ##q##.
 
  • Like
Likes TheGmc
  • #18
So I will multiply the voltage I calculated and the given current of ions 1 𝜇𝐴. Thanks, you helped me a lot about the whole assignment :)
 
  • #19
@etotheipi does this current of ions has something to do with solenoid ? I chose the values for solenoids and found current values but if the ions are affect them too I should change things
 
  • #20
TheGmc said:
@etotheipi does this current of ions has something to do with solenoid ? I chose the values for solenoids and found current values but if the ions are affect them too I should change things

In the solenoid you'll only have electrons as charge carriers; but no, it's completely distinct. The solenoid is only there to produce the ##\vec{B}## field, and you can think of it as existing totally independently of the rest of the setup.
 
  • #21
Thanks :)
 
  • #22
@etotheipi in one question the ions are perpendicular to the electric field and in the other they are parallel does this affect the power?
 
  • #23
In the accelerator, the electric field does work on the ions so it makes sense to speak of the power. In the velocity selector, the electric field is orthogonal to the ions so it does no work, and the power is effectively zero.
 
  • Like
Likes TheGmc
  • #24
@etotheipi when I reread the question while rewriting I get confused again since it says "What should be the power of the voltage supply". in velocity selector I found it zero since its perpendicular in accelerator region I found a value according to the my voltage and the current of ions given in the assignment. but since it says "power of the voltage supply" I imagine something like battery and if its power zero like in the velocity selector part how can it create an electric field ?
 
  • #25
Perhaps it means the power of the battery connected across the solenoid? If the current through the solenoid is ##I##, the number of turns of radius ##r## and cross-sectional area ##A## is ##N##, and the resistivity is ##\rho##, then the total resistance of the solenoid is ##\frac{2\pi r N \rho}{A}## and the power dissipated in the solenoid (which equals the power in via the battery) is that multiplied by ##I## in the solenoid.

The electric field you can think of as being created by a capacitor; in a steady state no current flows, so no power is dissipated here.
 
  • Like
Likes TheGmc
  • #26
but there is no solenoid in accelerator region
 
  • #27
No I was talking about the velocity selector region. In the accelerator region the power is just ##I_{acc}V_{acc}##
 
  • Like
Likes TheGmc
  • #28
Since the questions are identical I related the power of voltage supply with electric field again :/
 
  • #29
You just have to be clear on the different parts of the problem. There is an electric field in the accelerator region, it does work on the ions and as such has a calculable power of ##I_{acc}V_{acc}##.

In the velocity selector region, there is a magnetic field and an electric field. Neither of these does work on the ions that pass through, so they transfer no power to the ions.

You might however want to analyse the circuits that produce the electric and magnetic fields in the velocity selector region. The magnetic field is produced by a solenoid; since we can work out the resistance of the solenoid, we can figure out the power dissipated in it assuming we know the current. The electric field is produced by a capacitor, which in a steady state has no current flowing across it and hence no power is dissipated in the capacitor circuit in steady state.

Finally you have the deflection/detection region, with another magnetic field. Again, if you wanted, you could calculate the power dissipated across the solenoid producing the magnetic field.

The questions your professor is asking are quite arbitrary, and my advice would be to just try and separate out all the different pieces and think about it as logically as you can.
 
  • Like
Likes TheGmc

Related to What I do not understand about mass spectrometers

1. What is a mass spectrometer and how does it work?

A mass spectrometer is a scientific instrument used to measure the mass of atoms and molecules. It works by ionizing a sample, separating the ions based on their mass-to-charge ratio, and then detecting and measuring the abundance of each ion. This information can be used to determine the molecular weight and chemical composition of the sample.

2. What are the different types of mass spectrometers?

There are several types of mass spectrometers, including time-of-flight, quadrupole, magnetic sector, and ion trap. Each type has its own unique design and method of ion separation, but they all operate on the same basic principles.

3. What is the purpose of a mass spectrometer?

Mass spectrometers have a wide range of applications in various scientific fields, including chemistry, biology, and physics. They are used to identify unknown compounds, analyze the composition of a sample, and study the structure and behavior of molecules.

4. How is data interpreted from a mass spectrometer?

The data from a mass spectrometer is typically presented as a mass spectrum, which plots the abundance of each ion as a function of its mass-to-charge ratio. The peaks in the spectrum correspond to different ions, and their relative heights can provide information about the composition of the sample.

5. What are the limitations of mass spectrometry?

While mass spectrometry is a powerful analytical technique, it does have some limitations. It may not be able to detect very small or very large molecules, and it may not be able to distinguish between molecules with similar masses. Additionally, the sample must be in a gaseous state, which can be challenging for some substances.

Similar threads

  • Introductory Physics Homework Help
2
Replies
40
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
9K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
458
  • Introductory Physics Homework Help
Replies
1
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top