• Support PF! Buy your school textbooks, materials and every day products Here!

Doppler shift of a hydrogen line on sun

  • Thread starter Krashy
  • Start date
  • #1
30
5

Homework Statement


Problem.png


Homework Equations


[/B]
delta lambda/lambda = velocity/speed of light

radius of the sun= 696 *10^6m


The Attempt at a Solution



DSC_0232.JPG


Hello,
the solution states the correct answer for this problem is 5.8*10^-12m. This is exactly 2x my answer but i dont really know what i did wrong. There is also a possibility that the solution is wrong because there are quite a few cases in which it is wrong so i cant be 100% sure.
So i hope someone can tell me where i messed up. Thanks for every answer.
 

Attachments

Last edited:

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,815
6,609
radius of earth = 696 *10^6m
You mean the radius of the Sun.

This is exactly 2x my answer but i dont really know what i did wrong.
What is your ##\Delta \lambda## the wavelength difference between? Is this the wavelength difference you are interested in?
 
  • #3
30
5
Oh yeah sorry i meant the radius of the sun. I am searching for the change in wavelength i think and so the shift between blue and red should be 2.9*10^-12m, right?
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,815
6,609
I am searching for the change in wavelength i think and so the shift between blue and red should be 2.9*10^-12m, right?
No, think a bit about what you are computing the wavelength shift relative to. What is the original wavelength the wavelength of and what is the shifted wavelength the wavelength of?
 
  • #5
30
5
I dont think i really get it. As the sides of the sun move away and towards the earth, the spectral line splits into two, right? So the "real" wavelength is 434nm and the shift towards either side should be +/- 2.9*10^-12m, because the doppler shifts are the same only in opposite directions, or is that wrong?
 
  • #6
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,815
6,609
As the sides of the sun move away and towards the earth, the spectral line splits into two, right? So the "real" wavelength is 434nm and the shift towards either side should be +/- 2.9*10^-12m, because the doppler shifts are the same only in opposite directions, or is that wrong?
No, that is correct. So what is the difference between the wavelengths at the edges in opposite directions? (Which is what the problem is asking for.)
 
  • Like
Likes Krashy
  • #7
30
5
All right i think i understand now. The change in wavelength from one side to the middle, so to the "real" wavelength is 2.9*10^-12m. Thus the shift from one side to the other side is 2x that value: 2 * 2.9*10^-12m = 5.8 *10^-12m. This should be correct now, right? Thank you very much for the clarification.
 
  • #8
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,815
6,609
Yes, you are being asked to find the difference between the wavelengths at the edges, not that between the edge and the middle (which is what you computed). Since the speeds are low enough to use the approximation you applied, the difference is a factor of two larger than what you computed.
 
  • #9
30
5
I see, next time i will pay more attention to this. Thanks again.
 

Related Threads on Doppler shift of a hydrogen line on sun

  • Last Post
Replies
1
Views
847
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
4
Views
4K
Top