With what speed does the ion enter the magnetic field in a mass spectrometer?

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Homework Help Overview

The discussion revolves around a mass spectrometer problem involving a positively charged ion accelerated by a potential difference of 120 V before entering a magnetic field of 0.17 T. Participants are exploring how to determine the speed of the ion as it enters the magnetic field, given its mass and the radius of its circular path.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the potential difference, charge, and the resulting speed of the ion. There is confusion regarding the relevance of various parameters such as the magnetic field strength and mass in the equations. Some participants question the calculation of charge and its implications for solving the problem. Others suggest using conservation of energy principles to relate potential and kinetic energy.

Discussion Status

There is an ongoing exploration of the relationships between energy forms and forces acting on the ion. Some participants have provided calculations and insights into the potential energy and its conversion to kinetic energy, while others are questioning assumptions and seeking clarification on the equations involved. No consensus has been reached yet, but productive lines of reasoning are being developed.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information they can use or the methods they can apply. There is a focus on understanding the physical principles at play, particularly regarding the role of electric potential energy in the context of the problem.

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Homework Statement



A mass spectrometer has a potential difference of 120 V applied to its accelerator plates. It is accelerating a positively charged ion that has a mass of 2.32 x 10^-26 kg. the ion then is directed into a uniform magnetic field of 0.17 T. The magnetic field is perpendicular to the direction that the particle is traveling. The radius of the circular path that the ion takes, while in the magnetic field, is 2.5 cm.
with what speed does the ion enter the magnetic field?

Potential difference = 120 V
Mass = 2.32 x 10^-26 kg
Magnetic field=0.17 T
Theta=90 degrees
Radius= 0.025m
Charge=3.08 x 10^-19 C

Homework Equations


The Attempt at a Solution



since it's traveling in a circular path a=v^2 / r
I also am confused about the free body diagram (do you even need me to figure this question out?)

edit: Is it that the magnetic force = the force of gravity or something like that?
 
Last edited:
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So for your equations, you don't want to use the strength of the B-field, voltage or mass? Do you need any of them? ohhh charge also?
 
Last edited:
pgardn said:
So for your equations, you don't want to use the strength of the B-field, voltage or mass? Do you need any of them?

Some of the information was used to solve for the charge so some of the given is not necessary. I'm pretty sure you need the charge to figure this one out.
 
pinkyaliya said:
Some of the information was used to solve for the charge so some of the given is not necessary.

Yea that is a bit of a strange charge. How did you get it? Its close to what I would expect, but not quite...
 
pgardn said:
Yea that is a bit of a strange charge. How did you get it? Its close to what I would expect, but not quite...

I used the equation

m=qβ^2R^2 / 2ΔV

q=m2ΔV / β^2R^2

q=(2.32 x10^-26kg)(2)(120V) / (0.17T^2)(0.025m^2)
q=3.08...x10^-19 C

Is that right?
 
pinkyaliya said:
I used the equation

m=qβ^2R^2 / 2ΔV

q=m2ΔV / β^2R^2

q=(2.32 x10^-26kg)(2)(120V) / (0.17T^2)(0.025m^2)
q=3.08...x10^-19 C

Is that right?
Yes but...
Probably rounded a bit short since you have a very wide range of numbers some much smaller than the others. Since the amount of charge in 1 electron or proton is 1.6 x 10^-19C.

That would mean 2 extra protons in your ion should have about 3.2 x 10^-19C of charge.

Anyways...

Since you went through all that and got close, how about using cons. of Energy? What do you know about this situation. Any thoughts on kinetic and potential energy?
.
 
pgardn said:
Yes but...
Probably rounded a bit short since you have a very wide range of numbers some much smaller than the others. Since the amount of charge in 1 electron or proton is 1.6 x 10^-19C.

That would mean 2 extra protons in your ion should have about 3.2 x 10^-19C of charge.

Anyways...

Since you went through all that and got close, how about using cons. of Energy? What do you know about this situation. Any thoughts on kinetic and potential energy?
.

Is it maybe that

Ek + Ee = Ek' + Ee'

I don't see how potential energy would play role since it's not changing height.
 
pinkyaliya said:
Is it maybe that

Ek + Ee = Ek' + Ee'

I don't see how potential energy would play role since it's not changing height.

Potential Energy comes in many forms. Sometimes unit analysis can help you. I am going to tell you that before the ion is accelerated it has electric potential energy. Do you have an equation for this, or can you just look a the units on what you have and come out with Joules?

What does that Ee mean for example?
 
pgardn said:
Potential Energy comes in many forms. Sometimes unit analysis can help you. I am going to tell you that before the ion is accelerated it has electric potential energy. Do you have an equation for this, or can you just look a the units on what you have and come out with Joules?

What does that Ee mean for example?

I was working around with it and the only problem I have is finding out the velocity in the end

What I mean is
v=?
ΔV=120 V
β=0.17 T
q=3.08...x10^-19 C
m=2.32 x10^-26 kg

v'=?

ΔE = qΔV
ΔE = (3.08...x10^-19 C)(120 V)
ΔE = 3.696...x10^-17 J

So I know that there was a potential energy of 3.696...x10^-17 J. If it started at rest, I know that this energy would be transformed into kinetic energy and that
Ek'=3.696...x10^-17 J
but I don't know what kinetic energy it ended with.
 
  • #10
The ion is accelerated by U=120 V potential difference. Its final KE is equal to the initial potential energy qU, the accelerating voltage multiplied by the charge of the ion: mv2/2 =qU

The magnetic field exerts a force F, which is perpendicular to both the magnetic field B and the velocity of the ion. The magnitude of the force when v is perpendicular to B is F=qvB. The ion moves along a circle of radius R, and the centripetal force is equal to the magnetic force, mv2/R=qvB. Substitute 2qU for mv2, you get 2qU/R=qvB. q cancels. Isolate v.

ehild
 
  • #11
So as Ehilde stated:

All potential energy starting from rest

to all kinetic as the ion enters perpendicular to the B field at which point it turns a circle.

Sorry but I was attempting to "draw" it out of you.

All electrical pot qV = all kinetic 1/2mv^2
 

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