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Physics Olympiad 1999 Q29

  1. Sep 28, 2009 #1
    http://www.princeton.edu/~jdpeters/docs ... piad99.pdf

    I can't understand why the answer is E.

    The velocity with which the ion enters the magnetic field is determined by

    1/2 m v^2 = q V

    ans so v is proportional to sqrt(q/m).

    The radius of the arc in a magnetic field of strength B is proportional to (m * v) /q which is proportional to sqrt(m/q).

    Since the oxygen is double the charge and quadruple the mass, we expect oxygen to have a radius which is increased by a factor of sqrt(2)
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Sep 29, 2009 #2

    gabbagabbahey

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    I get a "403-forbidden" error when I click on your link...can you type out the problem instead?
     
  4. Sep 29, 2009 #3
    A mass spectrograph separates ions by weight using simple concepts from physics. Charged
    ions are given a specific kinetic energy by accelerating them through a potential difference. The ions then move through a perpendicular magnetic field where they are deflected into circular paths with differing radii. How would the radius of a singularly ionized common helium atomcompare to the radius of a doubly ionized common oxygen atom if they were
    accelerated through the same potential difference and were deflected by the same magnetic field?

    [A] The radius of the He ion path is 4 times the radius of the O ion path.
    The radius of the O ion path is 2 times the radius of the He ion path.
    [C] The radius of the O ion path is 4 times the radius of the He ion path.
    [D] The radius of the O ion path is 8 times the radius of the O ion path.
    [E] The radius of the He ion path is equal to the radius of the O ion path.

    Answers is [E] supposedly.
     
  5. Sep 29, 2009 #4

    gabbagabbahey

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    Unless the helium atom has 6 neutrons, I'd have to agree with you.
     
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