The relativistic Doppler effect

[help I have 12 hours]

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.

 

[hutchphd]

The light you are seeing has already been Doppler shifted to lower energy. You need to "un-shift" it to find the original energy!

 

[help I have 12 hours]

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

 

[hutchphd]

The observed frequency will be less than the emitted frequency because the galaxy is receding.

 

[PeroK]

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).

 

[rude man]

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.