# Homework Help: Orbitting electron moved to Magnetic Field

1. Mar 25, 2009

### Dopefish1337

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

frequency=v/(2*pi*r)
F=qvB
acircle=v2/r
3. 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.)

2. Mar 26, 2009

### Redbelly98

Staff Emeritus
You wrote:
F=qvB​
That is the force due to the magnetic field. What other force acts on the electron (and thus contributes to the overall net force)?

3. Mar 26, 2009

### Dopefish1337

Whatever force that was keeping it orbitting in the first place would still be there I suppose.

That force would be the electric force I guess, so F=qE=mvi2/r would be the initial force.

I suppose I could work out that initial force numerically, getting some number Fe.

Would it be possible to then have Fe+QvB=mv2/r, and solve that for v, substituting that result in to the frequency formula?

Or would something else change complicating matters? (Or, am I barking down the wrong tree altogether?)

edit: Fe+QvB=mv2/r would need to be solved via the quadratic formula would it not?

Last edited: Mar 26, 2009
4. Mar 27, 2009

### Redbelly98

Staff Emeritus
I haven't actually solved it, so I'll just say that yes, it appears like the quadratic formula is the way to go here.

That being said, there might be some approximation that would simplify things, based on the QvB term is much, much smaller than both Fe and mv2/r. I don't know for sure if that is useful, just thought I'd mention it.

5. Mar 27, 2009

### Dopefish1337

Doing it by hand, this method didn't work.

However, with some help from excel to carry decimals all the way through, it did work (my 'by hand' was close though...). Silly computer being picky about answers...

Anyway, thanks!

Incidently, the change in v was about 3.1 m/s, so that field really didn't make much of a difference relatively speaking...