The relativistic Doppler effect

AI Thread Summary
The discussion centers on the relativistic Doppler effect and its application to determining the source frequency of light from a receding galaxy. The observed frequency was calculated, and an attempt was made to "un-shift" it to find the source frequency, resulting in a wavelength of 94 nm. However, using the Rydberg equation led to nonsensical values for the energy level. Participants suggest calculating in electron volts (eV) and emphasize the importance of understanding the energy ratio of the photon as generated versus observed. The conversation highlights the need for careful calculations to resolve discrepancies in the results.
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Homework Statement
An electron in a hydrogen atom drops from an excited state n=n’ to the n=1 level and emits a photon in a distant galaxy. When the photon reaches earth, its final energy is observed to be 2.117 eV. If the galaxy is receding at 0.95c, what was the excited state of the electron? Hint: Look at section 1.14 on the Doppler effect in your book.
Relevant Equations
E=hc
B=V/c
Fobs=(((1-(B/C))/(1+(B/C)))^(1/2)
I found the observed frequency from the energy. Then I used the receding Doppler shift formula to find, the source frequency but after that when i tried to use the Rydberg equation I got a value for the energy level less than one. and I'm pretty sure my work is right, any help is greatly appreciated, thank you.
 
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The light you are seeing has already been Doppler shifted to lower energy. You need to "un-shift" it to find the original energy!
 
Yes, I have done this. I found the f(obs) = (5.1189*10^14)hz
and then I "un-shifted" to find the f(source)=(3.19675*10^15)hz
which gave me lambda=94nm for the source light, but when I use the Rydberg equation from there I get a nonsense answer. Sorry, but can someone check if the problem has an answer that works?

1/(94*10^-9)=(1.096*10^7)((1/(n^2))-1) n=non-sensical
 
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The observed frequency will be less than the emitted frequency because the galaxy is receding.
 
help I have 12 hours said:
Yes, I have done this. I found the f(obs) = (5.1189*10^14)hz
and then I "un-shifted" to find the f(source)=(3.19675*10^15)hz
which gave me lambda=94nm for the source light, but when I use the Rydberg equation from there I get a nonsense answer. Sorry, but can someone check if the problem has an answer that works?

1/(94*10^-9)=(1.096*10^7)((1/(n^2))-1) n=non-sensical
Why not do the calculation in ##eV##? The ground state of hydrogen is ##-13.6eV## and the nth state has energy ##-\frac{13.6}{n^2}eV##.

And, yes, I do get a whole number answer (without converting to photon frequency or wavelength).
 
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help I have 12 hours said:
Yes, I have done this. I found the f(obs) = (5.1189*10^14)hz
and then I "un-shifted" to find the f(source)=(3.19675*10^15)hz
So far, so good.
Now, sticking to frequency, what is the ratio of energies of the photon as generated and as seen on Earth?
What is the energy of the photon as generated?
Solve for n.
 
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