What Determines the Arc Radius of an Ion in a Magnetic Field?

jdstokes
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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)
 
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I get a "403-forbidden" error when I click on your link...can you type out the problem instead?
 
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.
 
Unless the helium atom has 6 neutrons, I'd have to agree with you.
 
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