Atomic Spectra Question  [ENDORSED]

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Nikhita_Jaaswal_2H
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Atomic Spectra Question

Postby Nikhita_Jaaswal_2H » Fri Oct 07, 2016 8:04 am

One of the problems asks: In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron that leads to this spectral line. How would you solve this if you have two unkowns: n1(initial n level) and n2(final n level)?

Katrina_Galian 1J
Posts: 22
Joined: Wed Sep 21, 2016 2:57 pm

Re: Atomic Spectra Question

Postby Katrina_Galian 1J » Fri Oct 07, 2016 10:06 am

For this question, I used a variation of the Rydberg equation found in the textbook on page 7. It accounts for the initial and final energy levels (n). Since the question is dealing with the ultraviolet spectrum, we know that it's going to be the Lyman series, which means that the initial n will be 1. So now you just have to find frequency and the final n.

Taravat_Lakzian_1C
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Joined: Wed Sep 21, 2016 2:56 pm

Re: Atomic Spectra Question  [ENDORSED]

Postby Taravat_Lakzian_1C » Fri Oct 07, 2016 10:23 am

You are given that the ultraviolet spectrum of atomic hydrogen is observed at 102.6 nm. Since the ultraviolet spectrum of atomic hydrogen is also known as the Lyman series, it has a final energy level of n1=1. So, you use the equation v(frequency)=c/lambda to solve for the frequency (2.998 x 10^8 m s^-1 / 102.6 x 10^-9 m) which will give you a frequency of 2.922 x 10^15 s^-1. Then you use the Rydberg equation v(frequency)=R(1/n^2final−1/n^2initial) to solve for n2.
2.922 x 10^15 s^-1 = 3.29 x 10^15 Hz(1/1^2 - 1/n^2 initial)
n^2 = 8.9285 = 9
n=3, so n2=3. The initial energy level is n2=3 and the final energy level is n1=1.
I hope that helps!

Katrina_Galian 1J
Posts: 22
Joined: Wed Sep 21, 2016 2:57 pm

Re: Atomic Spectra Question

Postby Katrina_Galian 1J » Fri Oct 07, 2016 2:39 pm

Wait, wouldn't the final energy level be 3? Since initial n is 1 because of the Lyman series.


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