# Questions about emission spectra and Rhydberg equation for H-atom

1. Sep 24, 2011

### lolzwhut?

1. The problem statement, all variables and given/known data
I'm just curious to if I have answered this questions properly, and was wondering if someone can see if I should add more to what I have written, or completely change my answer. I'm pretty confident with what I've answered, but would like someone to check :)

So to begin....
1)Why are no two emission spectra for different elements ever the same?
2)What are the wavelengths of the lines in the hydrogen emission spectrum?
3) Use ryhdberg equation to calculate transitions in the H-atom that are taking place to give rise to each line.

1) Excited atoms emit spectral lines at certain frequencies. Because each element has different energies, no two emission spectra for different elements for the emission spectra can ever be the same.......That's my answer, how ever. I'm a bit unsure on maybe I should elaborate more...Was I right about the part on where I say "each element has different energies"?

2) I'm a bit confused for this. For this experiment, I found the different waves for hydrogen with a spectroscope. Should I use my experimental data, or research, and find the actual data? Cause I do have a feeling that my data MIGHT be incorrect.

3) Obvious the Rhydberg equation is 1/pi = R(1/n2f-1/n2i). Since R = 109737 cm, should I convert my values into CM as my final answer or NM? I'm not sure how to even do this equation at all...Here's a picture of the wavelengths of the H-atom I drew:

[PLAIN]http://s4.postimage.org/8dax1o0on/hydrogen_wavelength.png [Broken]

That's the data I got according to my spectroscope. What exactly should I be plugging into the rhydberg equation for this?

I almsot forgot to ask, but this questions asks "Are the lines seen representatives of all the transitions that are taking place in the atom"? This is assuming withe spectroscope, so wouldn't that be YES because those lines show lights projected at all the different levels?

Last edited by a moderator: May 5, 2017
2. Sep 25, 2011

### lolzwhut?

Bump.

I've figured everything out. But can't still solve this question:

"Using the Rhydberg equation, calculate the transitions in the H-atom that are taking place to give rise to each line."

If you look at my data, it has 7 lines....Shouldn't I just be calculating it from the highest line I found (750 nm) and the lowest (490 nm).

If someone can clarify this question for me, it would help me a ton.

So in my case, NF = 750, NI = 490. I will convert those into CM and basically plug n chug?

3. Sep 25, 2011

### Staff: Mentor

Transition is between two levels, so you need to find two integers that will yield the correct value. Other than trial and error I don't see how.

4. Sep 25, 2011

### lolzwhut?

There's no way to predict that..Here's what I'm thinking is right:

1/pi = R(1/nf - 1/ni)

R = 109737 cm

nf = 750 nm = 7.5*10^-5 CM
ni = 490 nm = 4.9*10^10-9 CM

So...

1/pi = 109737 (1/(7.5*10^-5) - 1/(4.9*10^-9)) =

1/pi = -2.23*10^13

UGHhhh...what the heck!!! I JUST DONT GET THIS :(

5. Sep 26, 2011

### Staff: Mentor

Looks to me like you have no idea what are nf and ni.

6. Sep 26, 2011

### lolzwhut?

I actually figured it out....

Here's my work:

a. R((1/2^2)-(1/3^2)) = .00152 = 1/.00152 = 657.89 nm
b. R((1/2^2)-(1/4^2)) = .00205 = 1/.00205 = 487.80 nm
c. R((1/2^2)-(1/5^2)) = .00230 = 1/.00230 = 434.78 nm
d. R((1/2^2)-(1/6^2)) = .00243 = 1/.00243 = 410.21 nm

7. Sep 26, 2011

### Staff: Mentor

Much better now, although still neither fits 750 nm.

8. Sep 26, 2011

### lolzwhut?

My experimental data must have been wrong..What i've calculated is 100% right.