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Mass Spectrometer Questions

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider the mass spectrometer shown schematically in Figure 19.30 (don't have a scanner so I can't put the picture). The electric field between the plates of the velocity selector is 950 V/m, and the magnetic fields in both the velocity selector and the defelection chamber have magnitudes of 0.930 T. Calculate the radius of the path in the system for a singly charged ion with mass m = 2.18x10^-26kg.

    2. Relevant equations

    I'm not sure what to do with the 950 V/m. I also do not know whether a "singly charged ion" means it is a proton or an electron.

    3. The attempt at a solution

    I have attempted to try to find out where I use 950 V/m and what a "singly charged ion" means. I don't think I can go any further without knowing those two pieces of information. I'm not trying to ask for the answer (I've spent 1 hour on this question and I'm willing to spend as much time as it will take to answer it), I just don't know where to start.

    Thanks. :smile:
    Last edited: Apr 8, 2007
  2. jcsd
  3. Apr 8, 2007 #2
    They give you the mass of the particle. Also, this is a mass spectrometer so the ion is positive.

    Now you simply need to calculate the velocity of the particles that get through the velocity selector, right? These particles are undeflected by the B and E fields, so the force due to the B-field is equal to the force due to the E field. Or, qVB = qE. or, v = E/B. Take the ratio of the E and B fields given to find the velocity.

    Now, you know the centripetal force due to the B-field acting on the mass when it passes into the curved tube is F = qVB = mv^2/r, right? You know q, V, B, and m, so you can easily find r.

    "singly charged" simply means that the charge of the ion is the same as the charge of an electron (but positive).
  4. Apr 8, 2007 #3
    Also, notice why the velocity selector is a crucial component of the spectrometer?
  5. Apr 8, 2007 #4
    Thank you very much leright for the quick and informative response.

    Why is the Electric Field given in V/m instead of N/C though?

    Plugging 950 in for the E/B ratio gets me around 1021m/s for the velocity, which seems really low for this type of situation.
  6. Apr 8, 2007 #5
    Both V/m and N/C are suitable units for electric field.

    Regarding the accuracy of the velocity, I am not sure...
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