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Dopefish1337
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Homework Statement
An electron in a hydrogen atom moves in a circular orbit of radius 5.10×10-11 m at a speed of 2.80×106 m/s. Suppose the hydrogen atom is transported into a magnetic field of 0.70 T, where the magnetic field is parallel to the orbital angular momentum. What is the change of frequency of the motion of the electron?
Homework Equations
frequency=v/(2*pi*r)
F=qvB
acircle=v2/r
The Attempt at a Solution
Well, the initial frequency would be 2.800*106/(5*10-11*2*pi)= 8.7379*10^15 Hz.
If I then take qvB=mv2/r, and rearrange for r and stick that result into the frequency formula, I get qB/(2*pi*m)= 1.959*1010, 5 orders of magnitude less and thus essentially insignificant compared to the intial amount. Clearly this is wrong. However, I don't know what else to try from here.
Help?
(Oh, and although I doubt it'd matter looking at the initial speed, but I'm fairly confident any relativistic effects can be safely ignored.)