# Zeeman effect physics problem

1. Oct 29, 2012

### interdinghy

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

"The normal Zeeman effect splits a spectral line at frequency $\nu_{0}$ and two satellite lines at $\nu_{0}$ ± $eB/(4\pi m_{e})$. By what amount (in angstroms) are the satellite lines of the hydrogen Balmer $\alpha$ line ($\lambda_{0} = 6562.81 Å$) split from the central component in a typical sunspot?

Given value for B in a sunspot: .1 T

2. Relevant equations

$\lambda$ = c/$\nu$
d$\lambda$ = c d/d$\nu$

3. The attempt at a solution

I've tried plugging things into $eB/(4\pi m_{e})$ to find the change in frequency for the satellite lines, but I'm not getting a value in hertz, so I'm not exactly sure what I'm doing wrong. I'm pretty sure that once I get an actual frequency out of this I can just use the relevant equations to find the difference in wavelength.

2. Oct 29, 2012

### frogjg2003

Did you calculate the individual wavelengths and take their difference? or did you calculate the difference in frequency and calculate a wavelength with that?

3. Oct 29, 2012

### interdinghy

If I calculate the frequency based off the given wavelength, I get 4.56*10^14 hertz, but that doesn't get me any further I don't think.

The problem is the difference between the initial frequency and the satellite lines. I can't add or subtract the difference because the difference isn't a frequency, it's some nonsense units (seconds^-2 ampere^-1).

4. Oct 29, 2012

### frogjg2003

Try writing out the units of each piece in the SI base units. See what happens.

5. Oct 29, 2012

### interdinghy

I think this is maybe where I'm missing something?

A tesla divided by an electron mass is giving me 1 per second per ampere, and wolfram seems to agree with this.

$\nu_{0} + s^{-1}A^{-1} = Hz + s^{-1}A^{-1}$ is adding incompatible units, so I'm pretty sure I can't do it.

6. Oct 29, 2012

### frogjg2003

What about the electron charge?

7. Oct 29, 2012

### frogjg2003

Also, I'm getting per second squared in Wolfram.

8. Oct 29, 2012

### interdinghy

Oh okay I see now. I was so sure e was the base of the natural log. I actually tried looking around for other things it could stand for, but putting it in as the electron charge on wolfram made it work.

Thanks!