Questions about emission spectra and Rhydberg equation for H-atom

In summary: Please check the nf and ni values, as they seem to be off.In summary, no two emission spectra for different elements ever the same. The wavelengths of the lines in the hydrogen emission spectrum are determined by the energy of the atom that is emitting the light. The Rhydberg equation is used to calculate the transitions in the H-atom that are taking place to give rise to each line. My data was incorrect, and experimental data must have been used instead.
  • #1
lolzwhut?
32
0

Homework Statement


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. My answers
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

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:
Physics news on Phys.org
  • #2
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
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
Borek said:
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.

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

Please check them.
 
  • #6
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
Much better now, although still neither fits 750 nm.
 
  • #8
My experimental data must have been wrong..What I've calculated is 100% right.
 

FAQ: Questions about emission spectra and Rhydberg equation for H-atom

What is the emission spectrum of an atom?

The emission spectrum of an atom is a unique pattern of light emitted by the atom when its electrons transition from higher energy levels to lower energy levels. This pattern is specific to each element and can be used to identify the element.

What is the Rydberg equation?

The Rydberg equation is a mathematical formula that describes the wavelengths of light emitted by hydrogen atoms. It is given by 1/λ = R(1/n12 - 1/n22), where λ is the wavelength, R is the Rydberg constant, and n1 and n2 are the energy levels of the electron.

What is the significance of the Rydberg equation?

The Rydberg equation is significant because it provides a way to calculate the wavelengths of light emitted by hydrogen atoms without having to observe each transition individually. It also helped to explain the patterns seen in the emission spectrum of hydrogen.

How does the Rydberg equation relate to the Bohr model of the hydrogen atom?

The Rydberg equation is derived from the Bohr model of the hydrogen atom, which states that the energy levels of the electron in a hydrogen atom are quantized. The Rydberg equation uses this concept to calculate the energy levels and wavelengths of light emitted by the hydrogen atom.

Can the Rydberg equation be applied to other elements besides hydrogen?

Yes, the Rydberg equation can be applied to other elements, but it is not as accurate as it is for hydrogen. This is because other elements have more complex electron configurations and interactions between electrons, making their emission spectra more complex. Other equations, such as the Balmer series equation, may be used to calculate the wavelengths of light for these elements.

Back
Top