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Charge and Mass of Singly Ionized Uranium

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Before entering a mass spectrometer, ions pass through a velocity selector consisting of parallel plates separated by 1.9 mm and having a potential difference of 120 V. The magnetic field between the plates is 0.42 T. The magnetic field in the mass spectrometer is 1.2 T.

    (b) Find the difference in the diameters of the orbits of singly ionized 238U and 235U. (The mass of a 235U ion is 3.903x10-25 kg.)

    2. Relevant equations

    Radius = mv/qB

    3. The attempt at a solution

    In order to compare the radii, I need to evaluate the radius expression twice, once for each ion. I'm given the mass of the 235U ion, but I'm unsure as to the charge on either, as well as the mass of the 238U ion.
     
  2. jcsd
  3. Nov 7, 2011 #2
    You need to know the expression for the velocity, v, of ions emerging from the velocity selector.
    In the velocity selector ions experience a deflecting force due to an electric field and a deflecting force due to a magnetic field.
    The electric field strength is given by E =V/d where V is the voltage between the parallel plates separated by a distance d.
    The ions that emerge from the velocity selector do so because the electric force equals the magnetic force.
    You should be able to show that they emerge with the same velocity (hence 'velocity selector')
    given by v = E/B
    Singly ionised means that each atom has lost 1 electron charge.
    The mass must be given in kg and is given by the mass number of each isotope.
    Hope this helps
     
  4. Nov 7, 2011 #3
    In part a, I have evaluated to find the value of v (which would obviously be the same in both cases).

    I calculated to the mass of the U238 ion to be 3.985 x 10-25, which I believe to be correct.

    Since each atom has lost 1 electron charge, does this mean that the charge on each ion is +e?

    What I have so far is:

    RU235 = [(3.903x10-25)(150,375.94)]/[e*1.2]

    RU238 = [(3.985x10-25)(150,375.94)]/[e*1.2]

    What's wrong here?
     
  5. Nov 8, 2011 #4
    You are almost there. The 'e' in the equation is the charge on an electron. Easy to look up.
    One small point, I got the mass of U238 to be 3.953 x 10-25.
    I got radii of 0.305m for U235 (diameter .610m) and 0.309m for U238 (0.618m)
     
  6. Nov 8, 2011 #5
    It turns out they were very particular regarding accuracy and trailing digits, so it accepted 7.9mm but not 8mm as the answer. All the same, thanks for your help!
     
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