Mass spectrometer with a Lead element

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The discussion centers on the operation of a mass spectrometer using lead isotopes, specifically how to load lead into the device and the implications of ionization. It clarifies that the ions being accelerated are lead atoms missing an electron, requiring reversed plate polarity for acceleration. The conversation also addresses the significance of a 1 μA ion current, equating it to the number of ionized atoms passing through the accelerator per second. Participants discuss the choice of magnetic fields for isotope separation, emphasizing that the field strength should be significantly below 2.5 Tesla for safety and effectiveness. The overall focus is on understanding the calculations necessary for determining velocities and magnetic fields in the spectrometer setup.
  • #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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