Mass spectrometer with a Lead element

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Summary:: How to load the plates with the Lead element in the spectrometer

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I have a mass spectrometer with lead element which has an electronic configuration 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 4f^14 5d^10 6s^2 6p^2. It has 2 free electron, so the ejected electrons go through the ion accelerator and here, to speed the electrons I thougth the plates' charges must be like that (above must be +, below must be -). It's right or not?
And also I have a table like :

Lead IsotopesRelative AbudanceDecaysIsotope Mass(AMU)
Pb2041.4%Stable203.978
Pb20624.1%Stable205.974
Pb20722.1%Stable206.975
Pb20852.4%Stable207.976
Lead100%207.2

I have a statement like "The ionized isotopes are in Pb^- form. That is only a single electron is missing in lead ions. The number of ionized isotopes entering into the accelerator region corresponds to a 1μA ion current.". What does it mean that Pb^- form and 1μA? I also wondering that which isotopes will I use to find Vacc, Eacc etc.

Note: If this thread resides question threads, please transport there. I don't start this there because of I wasn't sure about. Please don't delete the thread.

[Moderator's note: Moved from a technical forum and thus no template.]
 
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Answers and Replies

  • #2
phyzguy
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I think you don't understand how the mass spectrometer works. It is not electrons which are being accelerated through the accelerator region, it is ionized lead atoms. The ionized lead atoms are missing one electron, so they have a positive charge. So you need the polarity on the plates to be reversed in order to accelerate the ionized lead atoms. The 1 μA beam current is telling you how many atoms/second are passing through the accelerator region. Do you know how to convert 1 μA into atoms/second? What is the definition of an Ampere?
 
  • #3
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Thank you for informations. I'm going to study on all of them.
Do you know how to convert 1 μA into atoms/second? What is the definition of an Ampere?
I also study on it. Do you know any source about spectrometers? If not I just continue with Serway
 
  • #5
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I read a lot of articles and texts about spectrometers. Even though the system I had is different than a classic spectrometer, I understood the logic except 1μA. I think it is a significant thing for solving the whole problem because of there is no enough information (for instance, the voltage which will create a current, electric field in ion accelerator and magnetic field,electric field, voltage of current in velocity selector). I wish there are some more informations :) Could you please explain what is 1μA and where can I use this. Thanks in advance..
 
  • #6
phyzguy
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An Ampere or amp is defined as 1 Coulomb/second. A Coulomb is a measure of charge. The unit of elementary charge (like on 1 electron, or one lead atom from which one electron has been stripped) is 1.602x10^(-19) Coulombs. So 1 amp is
[tex]\rm 1 amp = \frac{1 \frac{Coulomb}{second}}{1.602 \times 10^{-19} \frac{Coulombs}{elementary \, charge}} = 6.24 \times 10^{18}\frac{elementary \, charges}{second}[/tex]
So 1 μA is 1 millionth of this or 6.24x10^(12) charges/second. Does this make sense?
 
  • #7
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An Ampere or amp is defined as 1 Coulomb/second. A Coulomb is a measure of charge. The unit of elementary charge (like on 1 electron, or one lead atom from which one electron has been stripped) is 1.602x10^(-19) Coulombs. So 1 amp is
[tex]\rm 1 amp = \frac{1 \frac{Coulomb}{second}}{1.602 \times 10^{-19} \frac{Coulombs}{elementary \, charge}} = 6.24 \times 10^{18}\frac{elementary \, charges}{second}[/tex]
So 1 μA is 1 millionth of this or 6.24x10^(12) charges/second. Does this make sense?
Yeah it makes sense but where can I use this?In spectrometer, I haven't seen a place where 'time' is used
 
  • #8
phyzguy
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Yeah it makes sense but where can I use this?In spectrometer, I haven't seen a place where 'time' is used
I don't know what you are trying to do, so how could I know? Is this a homework problem? What problem are you trying to solve?
 
  • #9
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aas.PNG

The question is here. I thought someone with knowledge of spectrometers could easily answer the question I asked at #7, but it's OK.

1.The mass difference between two isotopes is sometimes just a neutron mass. The spectrometer should separate them very well. For such an isotope combination, the difference in radius should be around 1 cm. That is r2-r1=1 cm. In order to achieve this, choose a magnetic field with a magnitude in Tesla (maximum magnetic field you can obtain from a conventional magnet is around 2.5 T so be far away from this value) and choose the direction also. Then determine the velocity of isotope you need. Last calculate radius r of a smallest isotope Pb-204.
2.In order to produce this magnetic field, determine the current and number of turns, radius and other parameters of a solenoid you need. Also show how you will position this solenoid.
 
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  • #10
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For such an isotope combination, the difference in radius should be around 1 cm. That is r2-r1=1 cm.
Does this statement give us an information about which isotopes we will choose?
 
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  • #11
phyzguy
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Does this statement give us an information about which isotopes we will choose?
It does when I read it. What do the sentences before that say?
 
  • #12
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What do the sentences before that say?
An ionizing laser is directed onto the lead block under test such that the ionized atoms (or isotopes) leaves the lead block with an initial kinetic energy of 1 eV. The direction is conically upward but sure that the cone is larger than the entrance hole to the accelerator section. The ionized isotopes are in 𝑃𝑏− form. That is only a single electron is missing in lead ions. The number of ionized isotopes entering into the accelerator region corresponds to a 1 𝜇𝐴 ion current.Then the ions enter to velocity selector region. In this region both electric and magnetic fields are applied. Next they enter into the mass spectrometer region. In this region there is magnetic field different than the previous region. Depending of the charge of the ions, they travel in a circular path with a well-defined radius either to the left or to the right.
It does when I read it.
How could it be?
 
  • #13
phyzguy
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I meant the three sentences immediately before the one that reads,"For such an isotope combination, the difference in radius should be around 1 cm. That is r2-r1=1 cm."
 
  • #14
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I meant the three sentences immediately before the one that reads,"For such an isotope combination, the difference in radius should be around 1 cm. That is r2-r1=1 cm."
Okay I get it now, so it must be either Pb206-Pb207 or Pb207-Pb208 right? Supposing that I chose second combination, then ;
r2 - r1 = (m2-m1)*(v2-v1)/e*B.
We know that m𝑛𝑒𝑢𝑡𝑟𝑜𝑛=1.67×10−27 𝑘𝑔 , 𝑄𝑒=1.60×10−19 𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑠 and
r2-r1=1 cm.
.
But we don't know (v2-v1) and B. If we are right thus far, what is the next step then?
 
  • #15
phyzguy
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Okay I get it now, so it must be either Pb206-Pb207 or Pb207-Pb208 right?
Right! Good, you're getting there.
Supposing that I chose second combination, then ;
r2 - r1 = (m2-m1)*(v2-v1)/e*B.
We know that m𝑛𝑒𝑢𝑡𝑟𝑜𝑛=1.67×10−27 𝑘𝑔 , 𝑄𝑒=1.60×10−19 𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑠 and
.
But we don't know (v2-v1) and B. If we are right thus far, what is the next step then?
You need to read the problem more carefully. You are supposed to choose the magnetic field. It says, "choose a magnetic field", and later it says, "Then determine the velocity of isotope you need".

Edit: Also, note that (m1*v1 - m2*v2) is not equal to (m1-m2) * (v1-v2)!
 
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  • #16
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You are supposed to choose the magnetic field.
"In order to achieve this, choose a magnetic field with a magnitude in Tesla (maximum magnetic field you can obtain from a conventional magnet is around 2.5 T so be far away from this value) and choose the direction also."
I consider the bold statement, for instance in spectrometer or velocity selector regions I want to choose a magnetic field 300 Tesla inside the page, is it okay? Then I can set a solenoid to produce these magnetic fields and I can determine which plate is positively charged and negatively charged easily. But I don't know how far I must be away 2.5 Tesla. 300 is exaggerated ? :)
Also, note that (m1*v1 - m2*v2) is not equal to (m1-m2) * (v1-v2)!
Yea, it just escaped me, thanks even so
 
  • #17
phyzguy
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"In order to achieve this, choose a magnetic field with a magnitude in Tesla (maximum magnetic field you can obtain from a conventional magnet is around 2.5 T so be far away from this value) and choose the direction also."
I consider the bold statement, for instance in spectrometer or velocity selector regions I want to choose a magnetic field 300 Tesla inside the page, is it okay?
No! It is most certainly not OK. Read the above statement again, very carefully. Perhaps English is not your native language? Whether it is or not, you need to read more carefully.
 
  • #18
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No! It is most certainly not OK. Read the above statement again, very carefully. Perhaps English is not your native language? Whether it is or not, you need to read more carefully.
I don't know what you are talking about. I read repeatedly but couldn't find the clue. Maybe you can throw some light on this.
 
  • #19
phyzguy
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I don't know what you are talking about. I read repeatedly but couldn't find the clue. Maybe you can throw some light on this.
How about the word "maximum"? What does it mean?
 
  • #20
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maximum
it again escaped me. So now can I take the magnetic field as 1.0 Tesla or maybe less?
be far away from this value
I don't actually get what it says. How far must I choose?
edit: The magnetic field of a typical refrigerator magnet is;
5 × 10−3 T (5 mT) – the strength of a typical refrigerator magnet
Which type of magnet is expected us to use?
 
  • #21
phyzguy
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OK, good. So it means your magnetic field should be much smaller than the 2.5T maximum. 1.0T sounds about right. Or maybe 0.5T. Either of those two values should work. 5 mT sounds low for a permanent magnet, I think a typical permanent magnet has a maximum field of around 5000 Gauss = 0.5T (1 Tesla is 10,000 Gauss).
 
  • #22
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maybe 0.5T
OK I'll do with this. There is a lot of work, that is I'll be back soon. Thanks you for attention up to this time!
 
  • #23
phyzguy
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OK I'll do with this. There is a lot of work, that is I'll be back soon. Thanks you for attention up to this time!
Glad to help. Keep working, you're making progress!
 
  • #24
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After a short break, I'm here again. I would like to tell the process;
I selected Pb207-Pb206 isotopes combination and did the calculation as selecting the isotopes masses(amu). So we know the difference in radius between two isotopes r=0.01m, B=0.5T, Qe = 1.60*10-19 , and r= (m.v)/(q.B). I found 8*10^-22 = 3.44*10^-25*v2-3.42*10^-25*v1. So I can't found v1 or v2 for now.
Does it help us to find the velocity of Pb204 that we are asked to find velocities of these besides? The radius and velocity of Pb204 is unknown afterall.
"Then determine the velocity of isotope you need"
I don't understand this statement. does "determine" mean "choose" here again or it's expected us to calculate? Almost every sentence of homework "determine" had been used. For instance, it says "In order to produce this magnetic field, determine the current and number of turns, radius and other parameters of a solenoid you need. Also show how you will position this solenoid."

edit: which isotope is this "isotope you need"? All of these enough clear to you? It's like a puzzle for me :/
 
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  • #25
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I wonder if it is expected to create a mass spectrometer which have all things from us and if ;
"determine" mean "choose"
whole things in the spectrometer will be defined from us. but relative to what?
 

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