# Sun question: Determine the velocity of the gas

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1. Dec 13, 2018

### artworkmonkey

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

From near to the center of the solar disc, Fe I line emission from point X shows a spectral line with components 666.823, 666.830 and 666.837 nm, around the rest wavelength of 666.800 nm. The central component is not as bright as those at the longer and shorter wavelengths. Measurements at Y around 12,000 km away from X on the Sun’s surface show three components at 666.826, 666.829 and 666.832. In this case, the central component is brighter than the components at longer and shorter wavelengths.

For X and Y determine the velocity of the gas containing the iron relative to the observer.

Calculate the magnetic field strength at each point

2. Relevant equations
I think the Zeeman effect is involved.

3. The attempt at a solution
I don't want to be told the answer. I would like to work it out myself. However, the course notes assume prior knowledge, and I don't know where to start. If I could just be pointed in the right direction, that would be a great help. Thank you.

2. Dec 13, 2018

### Staff: Mentor

What is the central wavelength for Fe I at rest?
Why does it change with velocity, and how much?

3. Dec 13, 2018

### artworkmonkey

I do appreciate your help, but I am still confused. Thank you for answering though.

4. Dec 13, 2018

### Staff: Mentor

Which part is unclear?

5. Dec 13, 2018

### artworkmonkey

Pretty much all of it. I'm out of my depth.
To calculate the velocity of a wavelength, don't we need to know the frequency as well? I don't know any other way of doing it.

To calculate the magnetic field strength, I think I start with Zeeman effect formula and re-arrange it so B is the subject:

delta λ= (e/4πcme ) λ02 B

I'm not certain though.
Thank you

6. Dec 13, 2018

### Staff: Mentor

There is no "velocity of a wavelength". You can convert wavelengths to frequencies if you like - you know the speed of light.
Looks good. Not sure about the prefactor but that is something the source for the formula will cover.

7. Dec 13, 2018

### artworkmonkey

I think I understand the first part now. It uses the Doppler Effect. First I convert from nm to "meters of Å". 0.1Ån = 1nm
Spectral line now has components 6668.23, 6668.30 and 6668.37 Åm, around the rest wavelength of 6668 Åm.
Speed of light is 299792458 m / s
Dopplar equation is:

V= (c * delta wavelength) / rest wavelength

V= (299792 458 m/s * 0.14) / 6668

V=6294m/s

Does this make sense?
Thank you

8. Dec 13, 2018

### Staff: Mentor

Why is delta wavelength 0.14Å, not (6668.30-6668)Å?

Apart from that: Looks good.

9. Dec 13, 2018

### artworkmonkey

Ah, I can see now that I subtracted to wrong figure. I think I can have a good go at these questions now.
Thank you, mfb, for all your help and guidance on this. It has been very much appreciated. :)