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Mass Spectrometer Problem

  1. Oct 4, 2005 #1
    Hi,

    My Question:

    Suppose the electic field between the two electric plates in the mass spectrometer is 2.48 x 10^4 V/m and the magnetic fields B=B'=0.68 T. The source contains carbon isotopes of mass number 12,13,14 from a long dead piece of tree(To estimate atomic masses multiply by 1.67 x 10^.27 kg). How far apart are the lines formed by the singlely charged ions of each type on the photographic film? What if the ions were doubly charged?

    My work:

    For carbon isotope 12

    m=(qBB'r)/E

    12(1.67 x 10^.27 kg)=((1.6 x 10^.19 coul)(0.68 T)(0.68T)(r))/(2.48 x 10^4 V/m)

    r=.0067m

    Is this correct :confused:

    As for the the ions being doubly charged wouldn't I just double, q.

    Thank You :smile:
     
  2. jcsd
  3. Oct 5, 2005 #2

    Astronuc

    User Avatar

    Staff: Mentor

    Your answer of r = 0.0067 m is correct for C12, but remember that if a particle travels a semicircle, the separation between lines will be the difference in the diameter (or 2*r) of the ion trajectories.

    If the atoms are doubly ionized, then q is doubled. The electric force is twice, but so is the magnetic field.

    Remember, if B is the field in the electrostatic field, and the forces of the electric field and magnetic field balance, then

    q E = q v B, or v = E/B, thus v is independent of charge.

    B' is the deflection field and r = mv / qB' .
     
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