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Mass of Atom in a Mass Spectrometer

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data
    A mass spectrometer yields the following data for a beam of doubly ionized atoms of a single element.

    B = 8.0*10^-2 T
    q = 2(1.60*10^-19 C)
    r = 0.077 m
    V = 156 V

    I am asked to solve for Mass (m) in kg.

    2. Relevant equations
    q/m=((2V)/((B^2)*(r^2)))

    I am only partially sure that this is the correct equation. It is the equation for the "Charge-to-Mass Ratio of an Ion in a Mass Spectrometer".


    3. The attempt at a solution
    I tried to rearange the equation, but my Algebra skills aren't very good. This is what I ended up with:

    m=q/(2V)/((B^2)*(r^2))

    I tried to work this out with my calculator, but my answer was incorrect.

    The number I got was very small: ~3.89*10^-28

    EDIT:

    Tried to use this instead: m=(q(B^2)(r^2))/(2V)

    Still, my answer was far too small, at ~3.89*10^-26


    My comprehension of both Physics and Algebra is limited at best. I would really appreciate some help. Thank you.
     
    Last edited: May 9, 2010
  2. jcsd
  3. May 10, 2010 #2

    ehild

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    Show how do you calculate. And do not forget the units. What do you think, how much is the mass of an atom in kg-s?

    Your original formula is :

    [tex]\frac{q}{m}=\frac{2V}{r^2 B^2}[/tex]

    Plug in the numbers and simplify.



    ehild
     
    Last edited: May 10, 2010
  4. May 10, 2010 #3
    I was wondering how you guys obtained that formula.
     
  5. May 10, 2010 #4

    ehild

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    The ion is accelerated by an electric field. Flying through a potential difference V, it gains kinetic energy equal to qV

    1/2 mv^2=qV.

    It enters into the chamber of the mass spectrometer, where the magnetic field is perpendicular to the velocity. The magnetic force is

    F=qvB.

    This is a constant force, normal to the velocity, so the ion will move along a circle of radius R.
    The centripetal force for the circular motion is equal to the force of the magnetic field:

    (*) mv^2/R = qvB

    but you know that mv^2= 2qV, so

    2qV=RqvB --->v=2V/(RB).

    (*) can be written as q/m=v/(RB). Plug in the expression for v.

    ehild
     
  6. May 10, 2010 #5
    oh ok that makes sense. I hadn't thought of kinetic energy/work relationship as a way to substitute for speed. thanks!
     
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