Mass spectrometer with a Lead element

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SUMMARY

The forum discussion focuses on the operation of a mass spectrometer utilizing lead isotopes, specifically how to load lead into the spectrometer and the implications of ionization. The participants clarify that the ions, specifically Pb^- ions, are accelerated through the ion accelerator, requiring the plates to be charged oppositely to what was initially suggested. A current of 1 μA indicates the flow of ionized lead atoms, and the discussion emphasizes the importance of understanding the magnetic field and velocity selection in the spectrometer's operation.

PREREQUISITES
  • Understanding of mass spectrometry principles
  • Knowledge of ionization processes and Pb^- ions
  • Familiarity with electric and magnetic fields in particle acceleration
  • Basic concepts of current measurement in amperes
NEXT STEPS
  • Research the calculation of ion current in mass spectrometry
  • Study the relationship between magnetic fields and particle velocity in accelerators
  • Learn about the properties and applications of lead isotopes in mass spectrometry
  • Explore the design and function of solenoids in creating magnetic fields
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This discussion is beneficial for physicists, engineers, and researchers involved in mass spectrometry, particularly those working with lead isotopes and ion acceleration techniques.

  • #31
Was the difference between m1 and m2 supposed to be 1.67x10-27? I wonder if I made the wrong calculation.
m2 - m1 = 0.02x10-25 = 2x10-27. Does a 0.33 slip cause a problem? I hope does not :)
 
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  • #32
requied said:
Was the difference between m1 and m2 supposed to be 1.67x10-27? I wonder if I made the wrong calculation.
m2 - m1 = 0.02x10-25 = 2x10-27. Does a 0.33 slip cause a problem? I hope does not :)

In Post #27 you wrote:
m1 = 3.42*10-27x205.974 = 3.42*10-25
Where did 3.42*10^-27 come from? Also, the result is not what my calculator says. And you should be including units.
 
  • #33
phyzguy said:
Where did 3.42*10^-27 come from?
Oh, I thought I fixed it when I see it first time. So, actually 1AMU = 1.66X10-27 and m1 = 205.974*1.66x10-27 = 3.4191684×10-25 = 3.42x10-25. But I can't edit the post #27 now.
phyzguy said:
Also, the result is not what my calculator says.
It may says the same result now I guess.
phyzguy said:
And you should be including units.
I'm going to take care of it. Thanks for notice.

edit: I think only latest posts can be edited. So #27 will not be able to be changed.
 
  • #34
So then m2 - m1 = 1 AMU = 1 neutron mass, right?
 
  • #35
I was talking about the same thing at #31. And because of we can't use these as m2-m1 in ##
(r_2 - r_1) = r_2 (1 - \sqrt{\frac{m_1}{m_2}})## I calculated these m2 and m1 one by one. So in this way, we don't have to use actual mneutron I think. What do you think?
 
  • #36
You're right. The original problem didn't ask you to use the neutron mass, it just said, "The mass difference between two isotopes is sometimes just a neutron mass. The spectrometer should separate them very well."
 
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  • #37
phyzguy said:
You're right. The original problem didn't ask you to use the neutron mass, it just said, "The mass difference between two isotopes is sometimes just a neutron mass. The spectrometer should separate them very well."
Would you like to add something else? Because I am starting over to do the homework.
 
  • #38
requied said:
Would you like to add something else? Because I am starting over to do the homework.
I think you've got the idea. A couple of things to remember:
(1) Clearly state the assumptions you are making.
(2) Make sure to keep enough significant figures so the error in your final answer is less than 10%, especially when subtracting two similar sized numbers like r2-r1.
(3) Always include the units!
 
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  • #39
phyzguy said:
I think you've got the idea. A couple of things to remember:
(1) Clearly state the assumptions you are making.
(2) Make sure to keep enough significant figures so the error in your final answer is less than 10%, especially when subtracting two similar sized numbers like r2-r1.
(3) Always include the units!
Thanks for everything ! All the best
 
  • #40
In velocity selector region, the question says "determine" but I did it like this below. I chose the voltage necessarily. Is there another way to determine these values?
1591802065885.png
 
  • #41
And also, how can I calculate power of this voltage supply? This can be thought of a capacitor?
 

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